WBCHSE Class 11 Chemistry Chemical Thermodynamics Questions And Answers

Chemical Thermodynamics Long Answer Type Questions

Chemical Thermodynamics Question 1. Classify the following systems into open, closed, or isolated:

1. Living cell,
2. A gas is enclosed in a cylinder fitted with a movable piston.
3. The walls ofthe container and the piston are impermeable and thermally insulated.
4. The substances present in a soda water bottle. The chemicals participated in a chemical reaction occurring in a closed glass container.
5. Hot tea is kept in a thermos flask.

Answer:

1. An open system: A living cell exchanges matter and energy with its surroundings to maintain itself.
2. A closed system: The walls of the container and the piston are impermeable and thermally insulated. So, the system cannot exchange matter and heat with its surroundings.  However, as the piston is movable, the system can do work or work can be done on it if the pressure on the die piston is decreased or increased. So, the system can exchange energy in the form of work with its surroundings,
3. A closed system: Here, the components present in the bottle constitute the system. As the bottle is closed, the system cannot exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings.
4. A closed system: Here, the chemicals constitute the system. As the reaction container is closed, the system is unable to exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings,
5. An isolated system: The walls of a thermos flask are made up of insulating materials. Again, the mouth of the flask is closed. So, the system can exchange neither matter nor energy with its surroundings.

Question 2. What is an adiabatic system? Is this an isolated system?
Answer:

An adiabatic system Is a closed system which can exchange several forms of energy (e.g., work) but not heat with its surroundings.

It is not an isolated system as an isolated system cannot exchange either matter or energy with its surroundings.

Question 3. Identify the following an extensive or intensive property: Enthalpy, internal energy, pressure, viscosity, heat capacity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, number of moles, refractive index, entropy.
Answer:

An extensive property ofa system depends upon the mass ofthe substance present in the system. Its value increases as the amount of substance in the system increases. Enthalpy, internal energy, heat capacity, number of moles, and entropy are extensive properties.

An intensive property of a system is independent of the amount ofthe substance present in the system. Pressure, viscosity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, and refractive index are intensive properties

Question 4. A closed container with impermeable diathermal walls contains some amount of gas. If the gas is considered to be a system, what type of system will it be X is a state function of a thermodynamic system. How are its infinite and infinitesimal changes denoted?
Answer:

As the walls of the container are impermeable, flow of matter into or out ofthe system is not possible. Again, the walls are diathermal. Thus, the exchange of heat between the gas and the surroundings is possible. Hence, the gas is a closed system.

Question 5. Thermodynamic state functions are path-independent quantities. Explain with an example.
Answer:

A state function is a path-independent quantity. This means that when a system undergoes a process, the change in any state function depends only on the initial and final states of the system, and not on the path of the process.

To make it clear, let us consider the following process in which the mol of an ideal gas changes its state:

2 atm, 4L, 273K→1 atm, 8L, 273K

We can make this change by either of the following two processes. However, in each of these processes, the change in a state function, viz., P or V is the same.

Question 6. Give two examples of path-dependent quantities. Are they properties ofa system?
Answer: Two path-dependent quantities are heat (q) and work ( w). These are not the properties ofa system.

Question 7. Under what conditions will a system be in thermodynamic equilibrium?
Answer: A system will be in thermodynamic equilibrium if it simultaneously maintains mechanical equilibrium, thermal equilibrium and chemical equilibrium.

Question 8. Why is the change in any state function in a cyclic process zero? Is the change in any function both reversible and irreversible cyclic processes zero?
Answer: The value of any state function of a system depends only on the present state of the system. In a cyclic process, the initial and the final states of a system are the same, and so are the values ofa state function at these two states.

Hence, the change in a state function will be zero in a cyclic process. The change in any state function depends only on the initial and the final states ofthe system.

It does not depend on the path followed to carry out the change. This means that a state function undergoes the same change in a process with the specified initial and final states irrespective of whether the process is carried out reversibly or irreversibly.

Now, in a cyclic process, the change in any state function is always zero. Hence, for both reversible and irreversible cyclic processes, the change in any state function will be zero.

Question 9. A closed system participates in the following process: A→B→C. In step A→B heat absorbed by the system = q cal and in step B→C, heat released by the system = qcal. Therefore, in this process, the sum of the heat absorbed and heat released by the system is zero. Is this an adiabatic process? Give reason.
Answer: In an adiabatic process, the system does not exchange heat with its surroundings at any step in the process. In the given process, the system absorbs heat in one step and releases heat in the other step. So, this process cannot be regarded as an adiabatic process.

Question 10. 1mol of an ideal gas participates in the process as described in the figure.

1. What type is the overall process?
2. Is this an isothermal process?
3. Mention the isobaric and isochoric steps in this process.

Answer:

1. It is a cyclic process because the system returns to its initial state after performing a set of consecutive processes.
2. This is not an isothermal process, because the temperature of the system does not remain constant throughout the process although the initial and the final temperatures are the same.
3. Step BC represents an isobaric process because the pressure ofthe system remains constant in this step. Step CA represents an isochoric process because the volume ofthe system remains constant in this step.

Question 11. Are the following changes reversible or irreversible? Give proper explanations:

1. Melting of ice at 0°C and 1 atm pressure
2. The pressure ofa gas enclosed in a cylinder fitted with a piston is 5 atm.
3. The gas is expanded against an external pressure of 1atm.

Answer:

1. At 0°C and 1 atm pressure ice remains in equilibrium with water. Ifthe temperature ofthe system is increased by an infinitesimal amount, ice melts into the water slowly. Again, if the temperature of the system is decreased by an infinitesimal amount, water freezes into ice slowly.
2. Thus, an infinitesimal increase or decrease in temperature causes a change in the direction of the process. Hence, the melting of ice at 0°C and 1 atm pressure can be considered as a reversible process,
3. An irreversible process: The external pressure is considerably less than the pressure of the gas. So, the gas will expand rapidly without maintaining thermodynamic equilibrium during the process. Hence, this expansion will occur irreversibly.

Question 12. One mole of an ideal gas participates in a cyclic reversible process as described. Indicate the type of processes the system undergoes in the steps AB, BC and CA. Assume T2>T1.
Answer: AB: It is an isochoric process as the volume of the system remains unaltered in this step. BC: It is given that T1<T2- Again, the given indicates V2 > V1. This means that the volume of the system increases with a decrease in temperature. This happens in case of an adiabatic expansion of a gas. Therefore, BCstep indicates an adiabatic process.

Question 13. Why is a process occurring in an open container considered to be an isobaric? What is the origin of the internal energy of a system? Why cannot the absolute value of internal energy be determined?
Answer: A process in an open container takes place under constant atmospheric pressure. Thus, it is an isobaric process.

Question 14. Is the internal energy of a system at 25°C greater or less than its internal energy at 50°C?
Answer: The internal energy of a system increases with the temperature rise. So, the internal energy ofa system will be greater at 50 °C than that at 25 °C.

Question 15. In the process A→ B→ C, the change in internal energy of the system in the steps A→ B and B→ C are -x kJ.mol¯1 and y kj-mol¯1, respectively. What will be the change in the internal energy ofthe system in step C→ A?
Answer: A→ B→ C

Given, ΔU_{A→ B} = -x KJ. Mol^-1 and ΔU_{B→ C} = -x KJ. Mol^-1

∴ ΔU_C→A  = -ΔU_A→C = (x-y) kJ. Mol-1 [Since U is a state function, its change in die forward direction of a process is the same as that in the backward direction but opposite in sign].

Question 16. Which is not a state function: (q+w), w, H, G?
Answer: ‘ w’ is not a state function it is not a property of the system.

Question 17. In a process, a system absorbs 500J of heat and performs 800J of work. In the process, q =__________ w =__________
Answer: As the system absorbs heat. and performs work, q is + ve and w is -ve. Hence, q = +500 J and w – -800 J.

Question 18. In a process, a system releases 500J of heat and work done on the system is 300 J. In the process, q = w =
Answer: The system releases heat and work is done on the system i.e., q is -ve and w is +ve so, q = -500J, w = +300J.

Question 19. A certain amount of a gas participates in the cyclic process ABCD (follow figure). Calculate the total work done g in the process
Answer: Pressure volume work, w = -P_ex(V₂-V₁)=-P_exΔV

In step: AB: P_ex = x atm, ΔV= (2y-y)L=yL

∴ w₁ = -P_ex ΔV = -xy L . atm

In steps BC and DA. work done is zero because the system remains constant in these steps. Instep CD: P_ex = 0.5x atm and AV = (y- 2y) L = -y L.

∴ ω₂ = -P_exΔV = —0.5x (-y L) = 0.5.xy} L . atm

So, the total work done in the process, ω = ω₁ + ω₂

= (—xy + 0.5xy) L .atm =-0.5xy L.atm

= -0.5xy x 101.3 J [since 1Latm = 101.3 J] =-50.65xy J

Question 20. Under which condition will the pressure-volume work be, \(w=-\int_{V_1}^{V_2} P d V?\) pressure of the gas.
Answer: \(w=-\int_{V_1}^{V_2} P_{e x} d V\)

So, in case ofa reversible process, \(w=-\int_{V_1}^{V_2} P d V\)

Question 21. Calculate the work done in the following process which an
ideal gas undergoes
Answer: \(\text { 1st step: } w_1=-n R T_1 \ln \frac{V_2}{V_1} ; 2 \text { nd step: } w_2=-n R T_1 \ln \frac{V_1}{V_2}\)

So, total work \(w_1+w_2=-n R T_1 \ln \frac{V_2}{V_1}-n R T_1 \ln \frac{V_1}{V_2}\)

⇒ \(=-n R T \ln \frac{V_2}{V_1}+n R T \ln \frac{V_2}{V_1}=0\)

Question 22. A particular amount of gas participates separately in the two processes given below: Process-1 For which process, the work done is maximum?
Answer:

⇒ Process-1: Step-1: ω₁ = -P_ex-ΔV = -P₁(V₂– V₁)

⇒ Step-2: ω₂ = 0 [Since the volume of  the system is constant]

⇒ Total work, ω = ω₁ + ω₂ = -P₁(V₂– V₁)

⇒ So, |ω| = |P₁(V₂-V₁)|

⇒ Process-2: Step-1: ω₁ = 0 [∴ the volume of the system is constant]

⇒ Step-2: ω₂ = —P_ex ΔV = -P₂(V₂-V₁)

⇒ Total work, ω’ = ω₁+ ω₂ = -P₂(V₂– V₁)

⇒ So, |ω’| = |P₂(V₂-V₁)| [Since P₂ < P₁ , |ω’| < |ω’| ]

Question 23. In which of the following reactions is the work done zero? Assign the sign of w (+ ve or- ve)for the cases in which work is involved.
Answer: In a chemical reaction, pressure-volume work, ω = -P_exΔV = -ΔnRT; where Δn = total number of moles of gaseous products – total number of moles of gaseous reactants.

In reaction 3 Δn = 0Δ, So, w =0

In reaction 1 Δn = 2-(1+2)=-1

So, w = -ΔnRT = RT, i.e., w> 0

In reaction Δn=l. So, ω= -ΔnRT =-RT, i.e., w < 0.

Question 24. Write down the mathematical form of the first law of thermodynamics for an infinitesimal change that involves only pressure-volume work. Write down the form of this equation ifthe above change occurs reversibly.
Answer:

In case of an infinitesimal change, the mathematical form of the first law of thermodynamics is: dU= δq + δw; where δq = heat absorbed by the system, δw = work done on the system and dU is the change In Internal energy of tire system.

For an infinitesimal change involving only P-V work, δw=-P_exdV. So, for an infinitesimal change Involving only P-V work, the form of the first law of thermodynamics will be, δw=-P_exdV

Question 25. According to the first law of thermodynamics, AU = q + w. Write down the form of this equation for the following processes: Cyclic process Adiabatic process Isothermal expansion of an ideal gas Process occurring in an isolated system.
Answer: An isolated system does not exchange energy or matter with its surroundings. So, for a process occurring in an isolated system, q = 0 and w = 0. Therefore, ΔU = q + w or, ΔU= 0 + 0 or, ΔU = 0

Question 26. Among the following processes identify those In which the change in internal energy (A U) Is zero: Isothermal compression of ideal gas Adiabatic expansion of ideal gas Free adiabatic expansion of an ideal gas Reversible cyclic process Irreversible cyclic process.
Answer: The change in internal energy of an ideal gas in its isothermal compression is zero, When an ideal gas undergoes an adiabatic expansion, its internal energy decreases. In the adiabatic free expansion of an ideal gas, the internal energy of the gas remains the same, Since U is a state function, its change in any cyclic process (reversible or irreversible) will be zero.

Question 27. A closed system undergoes a process A→B. If it occurs reversibly, then the system absorbs qy amount of heat and performs a amount of work. However, if it occurs irreversibly, then the system absorbs the q₂ amount of heal and does the w2 amount of work. Is (q₂ + w₁) greater than, less than or equal to (q₂+ w₂)?
Answer: For a reversible process: ΔU₁ = q₁ + w₁ and for the irreversible process: ΔU₂ = q₂+w₂.

In both processes, the initial state (A) and final state (B) of the system are identical. Since U is a state function, its change in a process depends only on the initial and final states ofthe system, and not on the nature of the process.

As the initial and final states in both processes are identical, the change in internal energy in both cases will be the same. Therefore, ΔU₁ = ΔU₂ and q₁+ ω₁ = q₂ + w₂

Question 28. For an ideal gas, the isothermal free expansion and adiabatic free expansion are the same processes— Explain For chemical changes, why is the change in enthalpy more useful than the change in internal energy?
Answer: According to the first law of thermodynamics, ΔU = q+ω. In the free expansion of a gas, w = 0. Again, in an adiabatic process, q = 0 Hence, in an adiabatic free expansion of an ideal gas, the change in internal energy ΔU =q + 0=0.

Also, in an isothermal process, the change in the internal energy of an ideal gas is zero. Again for free expansion of an ideal gas, w = 0. So, in an isothermal free expansion of an ideal gas, ΔU = q + w or, 0 = q+ 0 or q = 0. Therefore, it can be concluded that both processes are the same.

Question 29. The definition of enthalpy shows that for n mol of an ideal gas H = U + nRT.
Answer: If the enthalpy, internal energy, pressure and volume of ‘n’ mol of anideal gas at a temperature of TK are, U, P and V respectively, then H = U+PV. For the ‘n’ mol of an ideal gas, PV = nRT. So, H = U+nRT.

Question 30. Prove that for an Ideal gas undergoing an isothermal change, AH = 0.
Answer: The change in enthalpy of an ideal gas undergoing a process, ΔH = ΔU+nRAT. In an isothermal process, ΔT = O. So, ΔH = A{Again, in an isothermal process of anideal gas, A U = 0 and hence AH = 0.

Question 31. Under what conditions are(I) A U = qv &(li) AH = qP?
Answer: AU = qv; Conditions: Closed system, constant volume, only P-V work is considered (ii) AH = qp; Conditions: Closed system, constant pressure, only P-V work is considered.

Question 32. The heat required to raise the temperature of1 mol ofa gas by 1°C is q at constant volume and q’ at constant pressure. Will q be better than, less than or equal to q’? Explain
Answer: The heat required to raise the temperature of1 mol of gas by 1 ° at constant pressure is greater than that required at constant volume. At constant pressure, the heat absorbed by a gas is used up in two ways.

One part of it is used by the gas for doing external work, and the remaining part is utilised for increasing the temperature of the gas.

At constant volume, the heat absorbed by a gas is completely utilised for increasing the temperature of the gas as no external work (P-Vwork) is possible at constant volume. Therefore, q’ must be greater than q.

Question 33. Cp-Cy = x J-g-1.K-1 and Cp-Cy = x J-g-1.K-1 ] mol-1. K-1 for an ideal gas. Ifthe molecular mass ofthe gas is M then establishes a relation among x, X and M.
Answer: For a substance, molar heat capacity = specific heat capacity x molar mass.

Therefore, \(C_{P, m}=C_P \times M \text { and } C_{V, m}=C_V \times M\)

Given: Cp,m- Cv,m = XJ mol-1. K-1

∴ CpxM-CvxM = X

or, (Cp-CV)M = X; hence, X = Mx

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Question 34. Why is the sign of ΔH negative for an exothermic reaction and why is it positive for an endothermic reaction?
Answer: In a chemical reaction, the change in enthalpy, AH = sum of the total enthalpies of products – Sum of the total enthalpies of-reactants = \(\Sigma H_P-\Sigma H_R\) In case of exothermic reaction, \(\Sigma H_P-\Sigma H_R\), and henace ΔH<0; while for an endothermic reaction \(\Sigma H_P-\Sigma H_R\) resulting ΔH>0.

Question 35. A 0.5 mol sample of H₂(g) reacts with a 0.5 mol sample of Cl₂(g) to form 1 mol of HCl(g). The decrease in enthalpy for the reaction is 93 kj. Draw an enthalpy diagram for this reaction.

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) ; \Delta H=-93 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The reaction is associated with a decrease in enthalpy. So, it is an exothermic reaction. In such a reaction, the total enthalpy of the product(s) (2Hp) is less than that of the reactant(s)(£flfi). Therefore, in the enthalpy diagram for the reaction, ZHp lies below ZHR.

Question 36. A 1 mol sample of N₂ (g) reacts with 1 mol of O₂(g) to form 2 mol of NO₂(Og), where the increase in enthalpy is 180.6kj. Draw An enthalpy diagram for this reaction.
Answer: \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) ; \Delta H=+180.6 \mathrm{~kJ}\)

In the reaction, the enthalpy increases. So, it is an endothermic reaction. In such a reaction the total enthalpy ofthe product(s)(ΣH_R) is greater than that ofthe reactant(s) (ΣH_p). Therefore, in the enthalpy diagram for the reaction, ΣH_p lies above ΣH_R

Question 37. Identify the exothermic and endothermic changes:
Answer: 

⇒ \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g)+57.0 \mathrm{~kJ}\)

Answer: \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+57.0 \mathrm{~kJ}\) Heat is realed in this reacton,. so it’s an exothermic reaction.

H₂O(s) + 6.02 kJ→H₂O(l) . Heat is absorbed in this reaction. So, it is an endothermic process.

⇒ \(\begin{aligned}
& \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)-130 \mathrm{~kJ} \\
& \text { i.e. } \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g)+130 \mathrm{~kJ} \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)
\end{aligned}\)

Question 38. Write down the thermochemical equations for the following reactions:

1. A 1mol sample of methane gas reacts with 2 mol of oxygen gas to form lmol of carbon dioxide and 2mol of water. In this reaction, 890.5 kl of heat is produced.
2. A 1mol sample of carbon (graphite) reacts with 1 mol ofoxygen to form 1 mol of carbon dioxide gas. The heat evolved in this reaction is 393.5 kj.
3. 6 mol of carbon dioxide gas reacts with 6 mol of water to form 6 mol of oxygen gas and lmol of glucose. The heat absorbed in this reaction is 2200 kj.

Answer: \(\begin{aligned}
& \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.5 \mathrm{~kJ} \\
& \text { i.e., } \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta H=-890.5 \mathrm{~kJ}
\end{aligned}\)

⇒ \(\begin{aligned}
& 6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) \\
& \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g)-2800 \mathrm{~kJ}
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { i.e., } 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \text {; } \\
& \Delta H=-2800 \mathrm{~kJ} \\
&
\end{aligned}\)

Question 39. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l); \dot{2} \Delta \mathrm{H}=-285.8 \mathrm{~kJ}\)  What will be the value of AH for the reaction: 2H₂O(l)→ 2H₂(g) + O₂(g)?
Answer: \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \dot{2} \Delta H=-285.8 \mathrm{~kJ}\)

If this equation is written in the reverse manner, we have

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) ; \Delta H=+285.8 \mathrm{~kJ}\)

Multiplying this equation by 2, we have

⇒ \(2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=+571.6 \mathrm{~kJ}\)

Question 40. Mention the standard stales of the following elements at 25°C and later: carbon, bromine, Iodine, sulphur, oxygen, calcium, chlorine, fluorine and nitrogen.
Answer: At 25C; and 1 atm, the standard states of the given elements are—

1. Carbon: C(s, graphite);
2. Bromine: Br₂(Z);
3. Iodine: l₂(s):
4. Sulphur: S(s, rhombic);
5. Oxygen: O₂(g);
6. Calcium: Ca(s)
7. Chlorine: Cl₂(g);
8. Pluorine: P₂(g) ;
9. Nitrogen: N₂(g)

Question 41. Why is ΔH = ΔU for the following two reactions? Explain.

1. NaOH(aq) + HCl(aq)→ NaCl(aq) + H₂O(l)
2. CHΔ(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g)

Answer: This reaction occurs In a solution. For a reaction occurring in a solution, All = AH

For this reaction An (total number of moles of gaseous products – total number of moles of gaseous reactants) = (1 +2)-(l + 2) = 0. So, Δ7 = ΔU according to the relation ΔH = ΔU + ΔT.

Question 42. Why does the value of All for a chemical reaction depend on the physical states of the reactant (s) and produces)?
Answer: In the case of solids and liquids, ΔH = ΔH for a chemical reaction (as AV is negligible here). If the participating substances are gases, then ΔH = ΔH + ΔnRT. Hence, AH for a chemical reaction depends on the physical states of the reactant(s) and product(s)

Question 43. Give an example of a reaction for each of the following relations between AH and ΔH: (I) ΔH<ΔH (11) ΔH > ΔH (111) ΔH = AH.
Answer: In a reaction, if a gaseous substance (either as a reactant or as a product or both) participates, the change in enthalpy in the reaction at constant pressure and temperature is given by ΔH= AU + ΔnRT. K An (total number of moles of gaseous products – total number of moles of gaseous reactants) >0, < 0 or =0, then AH > AH, AH < AH or All = AH respectively.

⇒ \(\begin{aligned}
& 2 \mathrm{H}_2 \mathrm{O}(g) \rightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(\mathrm{~g}) ; \\
& \Delta n=(2+1)-2=+1 \text {. So, } \Delta H>\Delta U
\end{aligned}\)

⇒ \(\begin{aligned}
& 2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta n=0-(2+1)=-3 \text {. So, } \Delta H<\Delta U
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) ; \\
& \quad \Delta n=2-(1+1)=0 . \text { So, } \Delta H=\Delta U
\end{aligned}\)

Question 44. Give an example of a physical change for each of the following relations between ΔH Anil ΔH: Δ11 > ΔU; ΔH < ΔU; ΔH K ΔH
Answer: The equation ΔH= ΔH + ΔnUT can be used in case of a process involving phase change ofa substance.

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(g) \text {. Here, } \Delta n=+1 \text {. So, } \Delta H>\Delta U \text {. }\)

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \text {. Here, } \Delta n=-1 \text {. So, } \Delta H<\Delta U \text {. }\)

Question 45. Which element in each of the following pairs has the standard heat of formation to zero? [O₂(g), O₃(g)] (It) [Cl₂(g), Cl(g) ] [S (s, rhombic), S (s, monoclinic)]?
Answer: The enthalpy of formation of an element in its standard state is zero, at 25 the standard state of oxygen, chlorine and sulphur are O₂(g)> Cl₂(g) and S (s, rhombic). So, at 25 °C, the standard heats of formation of O₂(8)1 CI₂(g) and S(s, rhombic) will be zero.

Question 46. In which of the following reactions At 25°C, does the standard enthalpy change correspond to the standard enthalpy of formation of H₂O(Z)? Give reasons.
\(\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)\(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)\(2 \mathrm{H}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
Answer: The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z) because, the standard state of oxygen at 25 °C, is O₂(g) and not O₃(g).

The standard enthalpy change in this reaction refers to the standard heat of formation of H₂O(Z) because 1 mol of H₂O(Z) is formed from its stable constituent elements.

The standard enthalpy change in this reaction does not indicate the standard heat of formation of H₂O(Z). This is because the stable state of hydrogen at 25 °C, is not H(g).

Question 47. Which one of the given reactions indicates the formation reaction ofthe compound produced in the reaction?

1. S (monoclinic) + O₃(g)→ SO₃(g)
2. C (graphite, s) + 2H₂(g)→CH₄(g)
3. N₂(g) + O₂(g)→2NO(g)

Answer:

1. In the formation reaction of a compound, 1 mol of the compound is formed from its constituent elements. S(s, monoclinic) +O₂(g)→SO₂(g), at 25 °C this reaction does not represent the formation of S02(g) because, the stable form of sulphur is S(s, rhombic) at 25°C.
2. C(s, graphite) + 2H₂(g)→CH₄(g), at 25 °C reaction represents the formation reaction of CH₄(g) because lmol of CH4(g) is formed from its stable constituent elements.
3. N₂(g) + O₂(g)→2NO(g), at 25°C this reaction does not represent the formation reaction of NO(g) because 2 mol of NO(g) are formed in the reaction.

Question 48. “At 25°C the standard heat offormation ofliquid benzene is + 49.0 kj.mol-1—What does it mean
Answer: At 25°C, the standard enthalpy of formation of liquid benzene is +49.0 kj.mol-1. This means that at 25 °C and 1 atm when 1 mol liquid benzene forms from its constituent elements, the enthalpy change that occurs is +49.0 kj.

In other words, at 25 °C and atm pressure, the change in enthalpy in the following reaction is +49.0 kj

6C(s, graphite) + 3H₂(g)→C₆H₆(l)

Question 49. Give an example ofa combustion reaction whose standard enthalpy change is equal to the standard enthalpy of formation ofthe compound formed in the reaction.
Answer: At 25 C the standard heat of combustion of solid naphthalene [C₁₀H₈(s)] is 5147 kj. mol-1. this means that at 25 c and 1 atm pressure when 1 mol of solid naphthalene is completely burnt in the presence of oxygen the enthalpy change that occurs is 51747kj.

Question 50. The standard enthalpy of combustion of Cxliy(l) at 25°C is Q kj. mol-1 . Write down the thermochemical equation for the combustion reaction of this compound.
Answer: The combustion reaction for C(s, graphite) is: This reaction also C(s, graphite) + 0₂(g)→CO₂(g). represents the formation reaction of CO₂(g). Therefore, at 25°C, the standard heat of combustion of C(s, graphite) = the standard heat formation of CO₂(g)

Question 51. The standard heats of combustion of CH₄(g) and C₂H₆(g) are -890 kj.mol-1 and -1560 kj. mol-1 respectively. Why is the calorific value of C₂H₆(g) lower than that of CH₄ (g)?
Answer: The enthalpy of combustion of a compound is always negative. So, the standard enthalpy of combustion of the given compound, ΔHº_C = -Q kj .mol-1. The thermochemical equation for this combustion reaction

⇒ \(\begin{aligned}
\mathrm{C}_x \mathrm{H}_y(l)+\left(x+\frac{y}{4}\right) \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H_c^0=-Q \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 52. Identify whether the enthalpy of the initial state is greater than, less than or equal to that of the final state in the following changes: solid;→ liquid; vapour→liquid →vapour
Answer: Solid→Liquid: It is an endothermic process. In this process

⇒ \(\Delta H>0 \text {, i.e., } H_{\text {liquid }}-H_{\text {solid }}>0 \text { or, } H_{\text {liquid }}>H_{\text {solid }} \text {. }\)

Therefore, the enthalpy of the final state will be greater than that ofthe initial state.

Vapour→Solid: It is an exothermic process. So, in this process

Question 53. What docs Mi signify in each of the following equations?

1. HCl(g) + 5H₂O(l)→HCl(5H₂O); ΔH = -64 kJ
2. HCl(g) + aq→HCl(aq); ΔH = -75 kJ
3. HCl(5H₂O)+20H₂O(l)—>HC1(25H₂O); Δ7=-8.1 kJ

Answer: In this process, 1 mol of HC1 dissolves in 5 mol of water, forming a solution of definite concentration. The enthalpy change in such a process is known as the integral heat of the solution. So, AH, in the process Indicates the integral heat of the solution.

In this process, 1 mol HC1 dissolves in a large amount of water, forming an infinitely dilute solution. The enthalpy change in such a process is called the heat of solution. So, AH, in process 2, indicates the heat of the solution.

This is a dilution process because a solution with a definite concentration is diluted by adding solvent to it. So, AH, in this process, indicates the heat of dilution.

Change in enthalpy remains the same whether a reaction is carried out in one step or several steps under similar reaction conditions Explain the rearms.

Question 54. Given (at 25°C and1 atm pressure):

1. C (s, diamond) + O₂(g)→CO₂(g); ΔH0=-393.5 kj-mol-1
2. C (s, graphite) +O₂(g)→CO₂(g); ΔH0=-391.6 kj-mol-1

Find the standard heat of transition from graphite to diamond.
Answer: \(\begin{aligned}
\mathrm{C}(s, \text { diamond })+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \\
\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \\
& \Delta H^0=-391.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

[Subtracting equation 1 From equation We get C(s, graphite)→ C(s, diamond); AH° = 1.9 kj. mol-1 So, the heat of transition for this process is 1.9 kj.mol-1

Question 55. Consider the given enthalpy diagram, and > calculate the unknown AH by applying Hess’s law.
Answer: Following the given diagram, we have—

1. A+B→ C + 2D ; ΔH = ?
2. A + B→ E + 2D; ΔH = +27kJ
3. E + 2D →C+ 2D;ΔH = -13kJ

Adding equation 2 and equation 3, we get A + B→C+2D; ΔH = (27- 13)kJ = 14 kJ

Question 56. For the reaction, A + B →D, AH is -30 kj. Suppose, D is prepared from A and B and then it is again converted into A and B by following the stages D —> E —* A + B. Calculate the total enthalpy change in these two stages.
Answer: A + B → D; ΔH = -30 kj ⋅⋅⋅⋅⋅(1) The process D→E →A + B comprises the following two steps: D→E ⋅⋅⋅⋅⋅(2) and E→A + B Overall reaction: D→A + B The reaction [4] is the opposite ofthe reaction [1]. So, the enthalpy change in reaction [4] is +30 kj. Hence, the total enthalpy change insteps [2] and [3] is +30 kJ

Question 57. At 25°C, if standard enthalpies of formation o/MX(s), M+(ag) and X-(aq) are -x, y and -zkj-mol-1 respectively, then what will the heat of reaction before the reaction M+(aq) + X~(aq)-+ MX(s).
Answer: M+(aq) + X-(aq)→MX(s)
The standard heat of reaction, \(\begin{aligned}
\Delta H^0 & =\Delta H_f^0[\mathrm{MX}(s)]-\Delta H_f^0\left[\mathrm{M}^{+}(a q)\right]-\Delta H_f^0\left[\mathrm{X}^{-}(a q)\right] \\
& =(-x-y+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 58. The standard heat of sublimation of sodium metal is 108.4 kj.mol-1. What is its standard heat of atomisation?
Answer: At 25 °C, the standard state of sodium is Na(s). The sublimation process of Na(s) is Na(s)→Na(g). At 25 °C the enthalpy change process is equal to the sublimation enthalpy of Na-ihetal. Again, in the above process, I mol of Na(g) is formed from Na(s). So, at 25°C, the enthalpy change in this process is equal to the standard enthalpy of atomisation of sodium. Thus, the standard enthalpy of atomisation of sodium is 108.4 kj. mol-

Question 59. At 25°C, the bond dissociation energy of N₂(g) is 946 kj.mol-1. What does it mean? What would be the standard atomisation enthalpy of N₂(g) at 25°C?
Answer: At 25°C, the standard bond dissociation energy of N₂(g) is 946 kj.mol-1. This means that the energy required to break 1 mol of N=N bonds completely in the gaseous state to form gaseous nitrogen atoms is 948 kj. At 25 °C, the standard state of nitrogen is N₂(g). Now, the formation of 1 mol of N(g) takes place by the following process \(\frac{1}{2} \mathrm{~N}_2(g) \rightarrow \mathrm{N}(g)\)

Since 1 mol of N(g) is produced from N₂(g) in the process [1], the enthalpy change in this process at 25 will be equal to the standard atomisation enthalpy of nitrogen.

In process [1], change in enthalpy \(=\frac{1}{2}\) x bond dissociation energy of N=N = \(=\frac{1}{2} \times\) 946 = 473 k.mol-1 Therefore, at 25 °C the standard atomisation enthalpy of nitrogen is 473 kJ.mol-1.

Question 60. A—B bonds present in AB3(g) molecule undergo stepwise dissociation by the following sequence of steps. (1) AB₃(g)→AB₂(g) + B(g) (11) AB₂(g)→AB(g) + B(g) (HI) AB(g)→A(g) + B(g)
Answer: Bond energy of A — B bond in AB3(g) molecule = the average bond dissociation energy of three A— B bonds in AB3(g) molecule. If the standard enthalpy change in step is AH° kj.mol-1, then the bond dissociation energy of the A-B bond in Ab₃(g) molecule
\(=\frac{x+\Delta H^0+z}{3} \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Now, the bond dissociation energy of the A-B bond = kj.mol-1

⇒ \(y=\frac{x+\Delta H^0+z}{3} \quad \text { or, } \Delta H^0=3 y-(x+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

At 25 °C, if the enthalpy changes in steps and are x and z kj.mol-1, respectively, and the bond dissociation energy of the A— B bond is y kj.mol-1, then what would be the enthalpy change in step (2)?

Question 61. Will the transformation ofice into water be spontaneous -2°C and latm pressure? Will the reverse process be spontaneous at this pressure and temperature?
Answer: No. The transformation of ice into water at -2°C and 1 atm pressure is not spontaneous. This is because the sum ofthe increase in entropy of the system and the decrease in entropy of the surrounding is less than zero.

The reverse process, i.e., the transformation of water into ice is spontaneous. This is because at -2°C and 1 atm pressure the sum of die decreases in the entropy ofthe system and the increase in entropy ofthe surroundings is greater than zero.

Question 62. Water remains in equilibrium with its vapour at 100°C and atm. Will the transformation of water into its vapour be spontaneous at this pressure and temperature?
Answer: No. Since water and its vapour are in equilibrium, neither the forward process (water→vapour) nor the reverse process (vapour →water) is favourable

Question 63. Is rusting of iron a spontaneous process? What is the entropy of a system? Give its mathematical definition. What is its unit?
Answer: Rusting of irony is a spontaneous process.

Question 64. If the process A→B occurs reversibly, then the change in entropy of the system is AS1. When the same process occurs irreversibly, the change in entropy of the system is AS2. Will the value of ASt be greater than, less than or equal to the value of AS2?
Answer: Since entropy is a state function, the change in entropy ofa system in a process does not depend upon whether the process is carried out reversibly or irreversibly.

Question 65. Heat is not exchanged between the system and its surroundings during the free expansion of an ideal gas. Therefore, in this process, q = 0. Will the change in entropy in this process be zero?
Answer: In the free expansion of an ideal gas, no exchange of heat takes place between the system and its surroundings. Because of the expansion, the volume of the gas increases, and the larger space is now available to the gas molecules for their movement. This results in an increase in randomness in the system, and hence the entropy ofthe system increases.

Chemical Thermodynamics Questions  

Question 66. What will be the sign of ASsys (+ve or -ve ) in the process of—

1. The vaporisation of a liquid
2. Condensation of a vapour
3. Sublimation of a solid.

Answer: In the vaporisation of a liquid (liquid -> vapour) \(\Delta S_{\text {system }} \text { is +ve. }\) This is because a substance in its vapour state possesses greater entropy than its liquid state.

Question 67. The following changes are performed on 1 mol of N₂ gas

1. Pressure Is decreased at a constant temperature
2. Volume is decreased at a constant temperature
3. What will be the sign of Ssys In These changes?

Answer: If the pressure of a gas is decreased at a constant temperature, the volume of the gas increases. At a larger volume of a gas, the gas molecules get a greater chance to move about. As a result, the randomness of the molecules increases, which increases the entropy ofthe gas. Hence, A = +ve.

At constant temperature, the decrease in the volume of a gas reduces the availability of space for the movement of gas molecules. This results in a decrease in the randomness of the molecules; consequently, the entropy ofthe gas decreases. Hence, A = -ve.

Question 68. Write the SI unit of energy.
Answer: The SI unit of entropy is I K-1

Question 69. For a process in an isolated system, what are the conditions of spontaneity and equilibrium?
Answer: For a process occurring in an isolated system, the condition of spontaneity is ASsysytem>0 and the condition of equilibrium is āSSystem +surr = 0

Question 70. Write the relation between ASsys & ASsurr when a process reaches equilibrium. What will be the value of ASuniv?
Answer: For a process at equilibrium \(\Delta S_{\text {system }}+\Delta S_{\text {surr }}=0\text { But } \Delta S_{s y g}+\Delta S_{\text {surr }}=\Delta S_{\text {univ }} \text {. So, } \Delta S_{\text {univ }}=0\)

Question 71. For a reversible process A = -20 J. K-1. What will be the value of ASsun in this process?
Answer: In a reversible process, ΔS_sys = (-) ΔS_surr As, Δ = -20 J⋅K-1.So, ΔS_surr = +20 J.K-1

Question 72. The following two reactions occur spontaneously. What will be the signs of AS and AS_surr in these two reactions?
Answer: At a given temperature and pressure, the change in entropy of the surroundings, \(\Delta S_{\text {surr }}=-\frac{\Delta H}{T}\)

Chemical Thermodynamics Questions

In this reaction, AS = + ve is the gaseous substance produced in the reaction. Since, in this reaction, AH > 0; according to equation AS_surr = – ve.

In this reaction, AS_sys = -ve as the number of gas particles decreases. Again, in this process, AH < 0. So, according to the equation, AS, Ufr > 0

Question 73. For a reaction ΔH > 0, and another ΔH < 0. For both the reactions ΔSsys < 0. Which one is likely to occur spontaneously? Which one always occurs nonspontaneous? define

1. Gibbs free energy
2. The standardfree energy of formation of a substance
3. The standard free energy change in a chemical reaction.

Answer: For the reaction with ΔH>0, ΔS_surr. Again, for this reaction ΔS_sys < 0, and hence AS_sys, + ΔS_univ < 0 i.&, < 0. So, this reaction will be non-spontaneous

For the reaction with AH »<8ir ΔS_surr is -ve. Again, for this reaction ΔS_surr < 0, and hence ΔS_sys, + AS_surr < 0, i.&, < 0. So, this reaction will be non-spontaneous

Question 74. For a process at a particular temperature and pressure G = ΔH- TΔS. It can be written as: – What does the quantity within the bracket indicate?
Answer: In a process at a given pressure and a temperature of T K, the change in Gibbs free energy (AG) and the change in entropy ofthe universe is expressed by the relation

⇒ \(\begin{aligned}
& \Delta G=-T \Delta S_{\text {univ }} \text { or, }-\Delta G=T \Delta S_{\text {univ }} \\
& \text { Given equation: }-\Delta G=T\left[-\frac{\Delta H}{T}+\Delta S\right]
\end{aligned}\)

Comparing equations [1] and [2] we get that the quantity within the bracket in equation [2] is ΔS_univ.

Question 75. Consider the reaction, A → 2B, if the free energy per mole of A is GA and that of B is GB then what will the relation be between GA and GB when reaction 1 occurs spontaneously and 2 is at equilibrium?
Answer: For a spontaneous reaction at a given temperature and pressure, ΔG < 0. Therefore, 2G_B< G_A

At a given temperature and pressure, for a reaction at equilibrium, AG = 0

Therefore, ΔG = 2G_B-G_A = 0 or, G_A = 2G_B

Question 76. At a particular temperature and pressure, when will the changes in the entropy of the system (ΔS_sys) and the surroundings soundings (ΔS_surr) be equal but the opposite in sign in a chemical reaction? In this condition, what will the value of AG be? Will the reaction be spontaneous in this condition?
Answer: At a particular temperature and pressure, a reaction will be at equilibrium in the reaction AS_univ = 0. Now, ΔS_univ = ΔS_sys + ΔS_surr or, ΔS_sys + ΔS_surr = 0 or, ΔS_surr =

So, at equilibrium ΔS_sys  = -ΔS_surr

We know, ΔG = -TΔS_univ

At equilibrium, ΔS_univ = 0, So, ΔG = 0.

Since the reaction is at equilibrium, neither the forward process nor the reverse process is spontaneous.

Question 77. At a given temperature and pressure, for a chemical reaction AH < 0 & AS <0. Is the spontaneity of this reaction dependent upon temperature? If it depends and if the reaction reaches equilibrium at a temperature of Tg, then will the reaction be spontaneous below or above Te?
Answer: At a given temperature and pressure, a process will be spontaneous in this process ΔG < 0. Again, ΔG = ΔH- TΔS. So, at a given temperature and pressure, ΔH -TΔS < 0 for a spontaneous process. For the given physical change, ΔH < 0 and ΔS < 0. Therefore, for this process. ΔH- TΔS will be less than zero only when |ΔH| is greater than |7ΔS|. Now the value of |TΔS| depends on temperature.

Therefore, the spontaneity of this process will depend on temperature. Ifthe equilibrium temperature ofthe process is Tg, then at equilibrium ΔG = 0 = ΔH- TgΔS or, ΔH = Tg x AS A process occurs spontaneously ΔG < 0 . or, ΔH- TΔS < 0 [ v ΔG = ΔH- TΔS ] Therefore, ΔH- TΔS < 0 or, TgΔS- TΔS <0 or, (Te- T)ΔS < 0 For the given process, AS < 0. So, Tg-T> 0 or, Tg> T Therefore, the given process will be spontaneous at a temperature below the equilibrium temperature.

Question 78. A physical change at a fixed pressure attains equilibrium at 353 K. In this process, ΔH > 0 and AS > 0. What will be the direction (from left to right or right to left) of this process at 350 K and 355 K?
Answer: For a spontaneous process at constant temperature and pressure, AG = AH- TAS and AG < 0. At equilibrium, ΔG = 0 or, ΔH- TΔS = 0 or, ΔH = TΔS [T’ = equilibrium temperature = 353 K]

Chemical Thermodynamics Questions

ΔH = 353 x ΔS and ΔG = ΔH- TΔS = (353- DΔS For this physical change, ΔS > 0. Hence, AG will be negative if 353- T < 0 or, T > 353K. This indicates that the process will be spontaneous above 353K and it will be non-spontaneous when T < 353K (as at this condition ΔG > 0 ). Therefore, at 355K the direction ofthe process will be left to right and at 350 K the direction ofthe process will be from right to left.

Question 79. At a given temperature and pressure, for a chemical reaction AH > 0 and AS < 0. The reaction is non-spontaneous at all temperatures. But the reverse reaction is spontaneous at all temperatures—Give reason.
Answer: At constant temperature and pressure a process will be
spontaneous if ΔG < 0 for the process Again, ΔG = ΔH- TΔS

For the given reaction, ΔH > 0 and ΔS < 0.

Hence, from equation [1] we get, AG is positive for this process. So it is non-spontaneous. But for the same temperature and pressure, the AG value of the reverse reaction must be negative. Therefore, the reverse reaction is spontaneous.

Question 80. The results obtained by applying the mathematical form of the first law of thermodynamics for different processes are given below. Identify the processes (m = P-V work only):ΔU = 0, q + w = 0, q = 0, ΔU < 0. w < 0 q= O,ΔV>O , ΔU =w=0
Answer: According to the first law of thermodynamics, ΔU=q +w…..[1]

In the process, ΔU = 0. U is a state function. So, in a cyclic process, Its change is zero. So, from equation (1], q + w = 0. Thus, the given conditions indicate a cyclic process.

Again in the isothermal process, the change in internal energy for an ideal gas is zero. Thus according to equation [1], for isothermal change of an ideal gas, q + w = 0. Thus, the given conditions indicate an isothermal change.

In the process, q = 0 . So, Δ = w or -ΔU = -w (from equation [1]).

As q = 0 , the process is adiabatic. In such a process, the work done by the system (Δv < 0) is equal to a decrease in internal energy (Δt< 0). So, the given conditions indicate that the process is adiabatic and work is done by the system.

As q = 0 and if (P-V work) = 0, from equation [1], At)1 also becomes zero, q = 0 indicates that the process is adiabatic. Since w =’ 0 and ΔV> 0, the volume of the system increases in process but no work is done by the system. This happens in case of an expansion against zero pressure. Thus, the given conditions indicate the adiabatic expansion against zero pressure

Question 81. The entropy of the system decreases on condensation of a vapour though it occurs spontaneously. Explain.
Answer: On condensation of vapour, the entropy change of the system is, ΔS = S_liquid-S_vapour. ΔS= S_liquid– S_vapour, the entropy of the system decreases in the process. Condensation of a vapor is an exothermic process. Hence, the entropy of the surroundings increases. However, the increase in entropy of the surroundings is greater than the decrease in entropy of the system. So, Δ S_toatal is always positive in this process.

ΔS_total = ΔS_sytem +ΔS_surroundings

So, condensation of a vapor is a spontaneous process.

Question 82. The formation of ice from water is exothermic, but water does not convert into ice spontaneously at ordinary temperature and pressure. Why?
Answer: At a given temperature and pressure, a process will be spontaneous if it occurs with a decrease in free energy.

Again, at a fixed temperature and pressure, the free energy change of a system in a process is, ΔG = ΔH- TΔS; where ΔH and ΔS are enthalpy and entropy change of the system respectively.

During the conversion of water Into Ice, the entropy of the system decreases because the entropy of a liquid Is greater than that of a solid. So, for the conversion of water Into Ice, AS Is -ve.

Again, All Is also- ve as It Is an exothermic process. Thus, both ΔH and TΔS are -ve. But at ordinary temperature and pressure, |TΔS| Is greater than|TΔS]. Hence, the value of ΔG becomes |ΔH|. So, the conversion of water Into Ice at ordinary temperature and pressure Is not spontaneous.

Question 83. “The amount of heat present in hot water is greater than that in cold water”—explain whether the statement is correct or not.
Answer: The statement is wrong because heat can never be stored in any system as it is a form of energy In transit. During a process beat appears at the boundary of a system. Heat does not exist before and after the process.

Question 84. Give examples of two processes involving only P-V work, where the system does not perform any work.
Answer: During the expansion of a gas against zero external pressure, work done is zero.

If a process, involving only pressure-volume work, is carried out at constant volume, then work done in the process will be zero. For example, in the case of the vaporization of water in a closed container of fixed volume, the work done is zero.

Chemical Thermodynamics Questions

Question 85. For the reaction \(\mathrm{CH}_4(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) ;\) AH0 = negative at 25°C. If the standard enthalpies of combustion of CH₄(g) and CH₃OH(g) at 25°C are -x kj mol-1 & -y kl.mol-1 respectively, then will the value of x be less than, greater than, or equal to the value of y?
Answer: \(\begin{array}{r}
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-x \mathrm{~kJ} \\
\mathrm{CH}_3 \mathrm{OH}(\mathrm{g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=-y \mathrm{~kJ}
\end{array}\)

Subtracting equation [2] from equation [l], we obtain

\(\mathrm{CH}_4(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CH}_3 \mathrm{OH}(g) ; \Delta H^0=(y-x) \mathrm{kJ}\)

According to the question, (y-x) = negative. Thus, x > y.

Question 86. A plant is growing. What do you think of the entropy changes of the plant and its surroundings?
Answer: The entropy decreases during the growth of the plant (i.e., system) because the ordered structure of the plant is formed during 1(8 growth. However, the entropy of the surroundings increases during the process. The increase in entropy of the surroundings is much greater than the decrease in entropy of the system. As a result, the net entropy change of the system and its surroundings is always positive during the growth of a plant.

Question 87. When does an adiabatic process become isentropic?
Answer: In a process, if the entropy of a system remains unchanged, then the process is called isoentropic.

In a reversible adiabatic process \(\delta q_{r e v}\) =0 So the entropy change \(d S=\frac{\delta q_{r e \nu}}{T}=0\) Therefore, a reversible adiabatic process is isentropic.

Question 88. At a certain temperature and pressure, ΔH = 0 for the reaction: X + X = X₂. Show that the reaction is non-spontaneous at this temperature and pressure.
Answer: At a certain temperature and pressure, for a chemical reaction, ΔG = AH- TAS.

Now, AH = 0 for given the reaction. So, AG = -TAS In the given reaction, X₂  is produced from two atoms of X. So, the entropy of the system decreases. So, AS < 0.

AG =-T x (negative quantity) = positive quantity The positive value of AG indicates that the reaction is non-spontaneous.

Question 89. The magnitude of work done by a gas in a reversible expansion is greater than that in an irreversible expansion provided the initial and final states are identical in both processes —explain
Answer: Suppose, an ideal gas expands isothermally from its initial state P1 V1 to its final state P₂ V₂.

A reversible process consists of an infinite number of small steps, and at each step, the driving force, is infinitesimally greater than die opposing force. Suppose, in a reversible expansion of an ideal gas, die external pressure of the gas at a step of the process is decreased by a very small amount of dP, causing the increase in volume of the gas by dV. The work done by the gas will be, dw = ~{P_ex~dP)dV = -PdV [v Pv- dP = P (pressure of the gas) and dP x dV is very small as in the P-V diagram the points, m, and n are very close and pressure in these two points are not much different]. In the above diagram, PdV is given by the area of the strip, mnqp. The total work done in the reversible process will be equal to the sum of the die areas of all such strips. Thus, in such a process, work is done.

Chemical Thermodynamics Questions

Absolute value of work done,|w| = area ABNM Now, in isothermal irreversible expansion, suppose, the gas is expanded from volume to V2 against an external pressure which is equal to the final pressure of the gas. In this process work done,

⇒ \(w=\int_{V_1}^{V_2} P_2 d V=-P_2 \int_{V_1}^{V_2} d V\left[P_2=\text { constant during expansion }\right]\)

The absolute value of work done,|w|=P₂(V₂– V₁) =area ABM: From the graphs, it is clear that work done in reversible isothermal expansion is greater than that in irreversible isothermal expansion since the area of ABNM is greater than that of ABNE. This also indicates that the work involved in a process depends upon the path (or nature) of the process.

Question 90. mol of an ideal gas is freely expanded at a constant temperature. In this process, which of the quantities or quantities among w, q, AU, and AH are 0 or >0 or <0?
Answer: During the isothermal free expansion of a gas the work done is zero. So w = 0. Again internal energy and enthalpy remain the same during the isothermal expansion of an ideal gas. Therefore, AU = 0 and AH = 0.

According to the first law of thermodynamics, AU = q + w. For the given process AH = 0 and w = 0.

Therefore, q = 0. Thus, for isothermal free expansion of lmol of an illegal gas q = 0, w = 0, AU = 0, AH = 0.

Question 91. A gas cools down during adiabatic expansion and it becomes hot during adiabatic compression —explain
Answer: In an adiabatic process, heat is not exchanged between the system and its surroundings. In this process, if the gas is expanded, then the gas does work at the expense of its internal energy. As a result, the internal energy as well as the temperature of the gas decreases due to the adiabatic expansion of an ideal gas.

As no heat is exchanged between the system and its surroundings in an adiabatic process, any work done on the gas during adiabatic compression increases the internal energy of the gas. As a result, the temperature of the gas increases.

Question 92. Given: A(s) + B₂(g)→AB₂(g);ΔHº = -x kJ \(\mathrm{A}(\mathrm{s})+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) \rightarrow \mathrm{AB}_3(\mathrm{~g}) ; \Delta H^0=-y \mathrm{~kJ}\)
Answer: Multiplying the equation by 2, we get,

2A(s) + 2B₂(g) 2AB₂(g); ΔH0 = -2xkJ………[3]

Writing the equation in a reverse manner and multiplying both sides by 2, we get,2AB₃(g) 2A(s) + 3B₂(g); AH0 = +2y kJ Adding equations [3] and [4], we get,

A(5) + 2B2(g) + 2AB3(g-2A(s) + 2AB2(g) + 3B2(g); AH0 = (2y-2jc)kJ

2AB₃(g)→2AB₂(g) + B₂(g); AH0 = +2ykJ

Therefore, the standard reaction enthalpy for the given reaction = 2(y- x) kJ.

Question 93. At 25°C the standard enthalpy of formation of freon gas (CHC1F₂) is -480.0 kj.mol-1. Write down the thermochemical equation representing the formation reaction of the compound.
Answer: The constituent elements of the compound, are CHC1F₂ carbon, hydrogen, chlorine, and fluorine. The standard states of these elements are C(graphites), H₂(g), Cl₂(g), and F₂(g). Thus, the thermochemical equation for the formation reaction-

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{Cl}_2(g)+\mathrm{F}_2(g) \rightarrow \mathrm{CHClF}_2(g) ;\)

⇒ \(\Delta H_f^0\left[\mathrm{CHClF}_2(g)\right]=-480.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 94. For each of the given changes, state whether the final enthalpy Is greater or less than the initial enthalpy: H₂Q(r)→ H₂O(l), H₂O(g) H₂O(l)
Answer: Change is a melting process. So, it is an endothermic process. Hence, in this process, ΔH = +ve.

ΔH=H[H₂0(l)]- H[H₂0(s)]. As ΔH > 0, H[H₂O(l)] > H[H₂O(S)], indicating greater enthalpy for the final state than the initial state.

Change is a condensation process and so, it is an exothermic process. Hence in this process AH = -ve. AH = H[H20(/)]-H[H20(g)] Since AH < 0, H[H20(/)] < H[H20(g)]. Thus enthalpy of the final state is less than that of the initial state.

Question 95. Give two examples of spontaneous processes in which the disorderliness of the system decreases.
Answer: Transformation of water into ice at 1 atm pressure and below (T’C temperature. Condensation of water vapor at1 atm pressure and below 100 C temperature.

Question 96. There occurs no exchange of heat between a system and its surroundings in an adiabatic process. So, the change in entropy of the system is zero in this process. Justify the statement.
Answer: The statement is incorrect. It is applicable only for a reversible adiabatic change.

We know, the change in entropy of a system in a reversible process \(d S=\frac{\delta q_{r e V}}{T}\) (where 6qrev is heat lost or gained by the system in the process at 7’K). Now, for the reversible adiabatic process, our = 0.

Therefore, dS – 0. So, the entropy change Is zero in a reversible adiabatic process. In the case of an irreversible process, there is no relation between the entropy change and the heat lost or gained by the system in the process.

However, it can be shown that the entropy change in an irreversible process is \(d S>\frac{\delta q_{l r}}{T}\) (where oq[rr is the heat lost or gained by the system in the process at T K). Now, in an irreversible adiabatic process 6q(rr = 0. So, dS (or AS) > 0. Tilus the entropy of the system increases in an irreversible adiabatic process.

Question 97. Mg) + A(g)→A₂(g); these types of reactions are generally exothermic. Explain.
Answer: In the reaction, one molecule A₂(g) is formed from two atoms of A(g). Thus, the entropy of the system decreases. For any chemical reaction at fixed temperature and pressure, ΔG =ΔH- TΔS. For the given reaction, ΔS<0. The reaction will be spontaneous if ΔG < 0.

So, negative quantity = ΔH- (T X negative quantity) = ΔH + positive quantity

∴ ΔH = (-ve) quantity- (+ye) quantity ; So, ΔH < 0

∴ The given reaction is exothermic.

Chemical Thermodynamics Questions

Question 98. What do you mean by a perpetual motion machine of the second kind?
Answer: A machine working in a cyclic process absorbs heat from a single thermal reservoir and completely converts the heat into the equivalent amount of work, is called a perpetual motion machine of a second kind. This type of machine contradicts the second law ofthermodynamics, & it is impossible to construct.

Question 99. At ordinary temperature and pressure, solid NH₄C1 dissolves in water to form NH4(aq) and Cl(aq) ions. The process is endothermic. Indicate the signs
(+ or-) of ΔS_sys, ΔS_surr, ΔH And ΔG
Answer: As the process is endothermic, AH > 0.

At ordinary temperature and pressure, the dissolution of solid NH₄C1 occurs spontaneously, so AG < 0.

NH+₄ and Cl- ions in the crystal lattice of NH₄C1 are held at fixed positions and consequently, they are not able to move. But in an aqueous solution of NH₄C1, NH, and Cl- ions are distributed throughout the solution. Naturally, the randomness of the ions in aqueous solution of NH₄C1 is greater than that in solid NH₄C1. Hence, in the dissolution process of NH4CI, ΔS_sys > 0.

As the process is endothermic, the system absorbs heat from the surroundings. Consequently, the surroundings lose heat. As a result, the entropy of the surroundings decreases (AS_surr < 0).

However, the increase in the entropy system is greater than the decrease in the entropy of the surroundings. So, the net entropy change of the system and its surroundings is always positive, and the process occurs spontaneously.

Chemical Thermodynamics Short Answer Type Questions

Question 1. Classify as an open, closed, or isolated system:

1. A cup of coffee placed on a table
2. Water in a beaker is boiled by heating
3. Lead nitrate is heated in a test tube
4. Solid NH₄Cl is heated in a closed vessel
5. Substances present in a soda-water bottle
6. Mercury enclosed in the thermometer

Answer:

1. Open system
2. Open system
3. Open system
4. Closed system
5. Closed system
6. Closed system

Question 2. A certain amount of gas is enclosed in a container with permeable and diathermal walls. Which type of system does the gas belong to?
Answer: The walls of the container are diathermal. So the system can exchange heat with its surroundings. Again, the walls ofthe container are permeable. So, the system can also exchange matter with its surroundings. Hence, the gas belongs to an open system.

Question 3. Does the volume of a closed system remain fixed?
Answer: If the walls of a closed system are non-rigid or movable, then the volume of the system does not remain fixed. For example, a gas enclosed in a cylinder fitted with a movable piston is considered a closed system. Here, the volume of the gas (system) can be increased or decreased by altering the pressure of the gas (system)

Question 4. Give an example of a thermodynamic quantity which is not a state function. Is it a property of a system?
Answer: Heat is not a state function because heat absorbed or by a system in a process depends upon the path of the realised It is not a property of the system.

Question 5. A occurring in an isolated system does not have any effect on its surroundings and the process is ft influenced by its surroundings. Why?
Answer: An isolated system does not interact with its surroundings

Question 6. A system participates in the following process What will be the change in enthalpy of the system in this process—positive, negative or zero?
Answer: Since it is a cyclic process and enthalpy is a state function, the change in enthalpy ofthe system in this process will be zero.

Question 7. If the value of internal energy for 10 g water at a particular temperature is x J, then what will be the value of internal energy for 20 g water at the same temperature?
Answer: Internal energy is an extensive property i.e., the value of the internal energy of a system is proportional to the amount of the substance present in the system. So, the value ofthe internal energy for 20 g water will be 2x J.

Question 8. In process A →B →C → D, the heat absorbed by the system in steps A → B and B C are q₁ and q₂, respectively, and the heat released by the system in step C→ D is q₃. If q₁ + q₂ + q₃ = o, then will the process be adiabatic?
Answer: In an adiabatic process, heat is not exchanged between a system and its surroundings at any stage of the process. In the given process, heat is being exchanged between the system and the surroundings. Thus, the process is not adiabatic though ihe sum of the amounts of heat absorbed and released is zero for the process.

Question 9. Why is infinite time required for the completion of an ideal reversible process?
Answer: In an ideal reversible process, the system maintains equilibrium at every intermediate step and the process is extremely slow. Thus, from a theoretical point of view, an ideal reversible process should require an infinite time for Its completion.

Question 10. In a process, if the heat released by the system and 80 work done on the system are 90 ) and 120), respectively, then what will he of q and w?
Answer: q=-90J ad w=+120J.

Question 11. If the specific heat capacity and molecular mass of a gas are cp and M respectively then what will be the molar heat capacity of the gas?
Answer: Molar heat capacity of tyre gas, C_p,m = c_p x M.

Question 12. Give an example of a process which Is simultaneously isothermal and adiabatic.
Answer: Adiabatic free expansion of an ideal gas (or isothermal free expansion of an ideal gas). The reason is that no exchange of heat occurs between the system and surroundings in this process and the temperature of the tire system remains constant throughout the process.

Question 13. At 25°C, the standard reaction enthalpy for the reaction AB₃(g)→1/2A2(g)+3/2(g) is. find the standard reaction enthalpy for the reaction.
Answer: Writing this equation. in reverse manner and multiplying both sides by 2, we get,
A2(g) + 3B2(g)→2AB3(g); -2AH°. So, at 25°C, the standard

Question 14. Mention the standard state of sulphur and iodine at 25
Answer: At 25°C, the standard state of sulphur is solid rhombic sulphur [S(rhombic, s)] and that of iodine is solid iodine [1₂(s)].

Question 15. At 25°C, is the standard reaction enthalpy for the reaction 2H(g) + O(g) → H₂O(I) the same as the standard enthalpy of formation of H₂O(f)
Answer: The given reaction does not indicate the formation reaction of H₂O(f) because the standard states of hydrogen and oxygen at 25°C are H₂(g) and O₂(g), respectively. Hence, the standard reaction enthalpy of the given reaction is not the same as the standard enthalpy of the formation of H₂O(l).

Question 16. Assuming experimental conditions are the same, compare (ΔH-ΔU) values for the given reactions

1. \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Answer: In case of reaction (1) \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Therefore , \(\Delta H-\Delta U=\Delta n R T=-\frac{3}{2} R T\)

In case of reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Chemical Thermodynamics Questions

Therefore, \(\Delta H-\Delta U=\Delta n R T=-\frac{1}{2} R T\)

Hence, the magnitude of H-U is greater in the case of 2

In case of reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Hence, the magnitude of (AH- AU) is greater in the case of 2

Question 17. What do you mean by ‘the enthalpy of solidification of water at 0°C and 1 atm pressure = -6.02 kj-mol-1′?
Answer: This means that 6.02 kj of heat is released when one mole of water completely freezes to ice at 0°C and 1 atm pressure.

Question 18. Why are spontaneous/natural processes irreversible?
Answer: The spontaneous or natural processes are irreversible because the thermodynamic equilibrium of the system is not maintained in such types of processes.

Question 19. Give examples of two processes where AS is zero
Answer: Adiabatic reversible process Cyclic process

Question 20. In which of the following two processes, the change in entropy of the system will be negative?

• Fusion of Ice.
• Condensation of water vapor

Answer: The change in entropy of the system is negative during the condensation of water vapor. In this process \(\Delta S_{\text {sys }}=S_{\text {water }}-S_{\text {water vapour }}<0\)

As the molecules in water vapor have more freedom of motion than they have in the water, the molecular randomness is higher in water vapor than in water. Thus, \(S_{\text {water }}<S_{\text {water vapour }}\) Consequently, the value of AS becomes negative.

Question 21. What will be the change in entropy in an irreversible cyclic process?
Answer: As entropy is a state function, the change in entropy in either a reversible or an irreversible cyclic process is zero.

Question 22. Are the given statements correct? If bent Is absorbed In a chemical reaction, the reaction cannot be spontaneous. The entropy of the system may decrease In a reaction, but the reaction can occur spontaneously
Answer: The statement is incorrect is correct.

Question 23. Which of the following will have a greater entropy? 1 mol of H₂ gas (T = 300 K, V = 5ml, )0 1 mol of H₂ gas (T = 300 K, V = 10mL).
Answer: As the entropy of a gas increases with the increase in its volume, the entropy of mol of H₂ (T = 300 K, V = 10 mL) will be greater than that of l mol of H₂ (T = 300 K, V = 5 mL).

Question 24. For a process, ΔS_sys = -15 J.K-1. For what value of ASsurr will the process be non-spontaneous?
Answer: The condition of non-spontaneity of a process is \(\Delta S_{s y s}+\Delta S_{s u r r}<0.\) \(\Delta S_{s y s}=-15 \mathrm{~J} \cdot \mathrm{K}^{-1}\), then will be non-spontaneous

Question 25. Which condition does not satisfy the spontaneity criteria of a reaction at constant temperature and pressure: ΔH<0, ΔS<0, ΔH>0, ΔS<0, ΔH>0, ΔH<0, ΔS>0?
Answer: At constant temperature and pressure, a reaction will be spontaneous if AG = -vc for the reaction at that temperature and pressure. If AG = +vc, the reaction will be nonspontaneous. At a constant temperature and pressure, AG = AH-7’AS. If AH > 0 and AS<0, then AG = +i/e. Thus, a reaction will be noil-spontaneous if AH > 0 and AS < 0.

Question 26. Which of the following conditions favours the spontaneity of a reaction at a constant temperature and pressure? ΔH > 0 , ΔH < 0 , ΔS > 0 , ΔS < 0
Answer: The conditions of ΔH < 0 and ΔS > 0 favour a reaction to be spontaneous at a constant temperature and pressure.

Question 27. According to the definition of a thermodynamic system, which system do living beings belong to, and why?
Answer: According to the definition of a thermodynamic system, every living being in nature belongs to an open system.

Explanation: All living beings (systems) take food (matter) from the surroundings and excrete waste materials (matter) to the surroundings. They also exchange heat (energy) with the surroundings.

Question 28. The change in internal energy in different steps of the process A→ B → C→ D are given: A→Bx.kJ-mol-1; B→ C, -y kj-mol-1; C-D,z kj-mol-1. What will the value of AH be for the change A → D?
Answer: \(A →{\Delta U_1} B→{\Delta U_2} C →{\Delta U_3} D\)

Δ Ux = UR- UA = x kj.mol-1 ;

ΔU2 = Hc.— UB = -ykjmol-1;

ΔU3 = UD-uc = zkj.mol-1

∴ For the change A D

A U = UD-UA =(UD-UC) + (UC-UB) + (UB-UA)

=(z-y + x) kj-mol-1.

Question 29. A gas is allowed to expand against zero external pressure. Explain with reason whether the process is reversible or irreversible.
Answer: The expansion ofa gas against zero external pressure is an irreversible process. As the opposing pressure is zero, the gas expands rapidly, and it cannot maintain thermodynamic equilibrium during its expansion.

Question 30. Change in enthalpy in different steps of the process A→B→C→A are given: A→B, x kj-mol-1; C→A, y kj-mol-1. Find the value of AH for step B→C.
Answer: The initial and final states (A) are the same in the given process. So it is a cyclic process. Since H is a state function, AH = 0 for the process. Thus in this process \(\begin{aligned}
& \Delta H=\left(H_B-H_A\right)+\left(H_C-H_B\right) \Psi\left(H_A-H_C\right)=0 \\
& \left.x+\left(H_C-H_B\right)+y=0 \text { or }\left(H_C\right)_B H_B\right)=-(x+y) \mathrm{kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

So change in enthalpy for BC =- (x + y) kj-mol-1.

Question 31. Due mole of ideal gas is expanded isothermally. In this process, which of the quantity (or quantities) among w. q, AH , AH is(are) zero or >0 or <0?
Answer: During isothermal expansion, heat is absorbed and work is done by the gas. So q > 0 and w < 0. Again internal energy and enthalpy remain the same during isothermal expansion of an ideal gas. Thus, AU = 0 and AH = 0 . For the isothermal expansion of lmol of ideal gas q>0, w< 0, AU = 0, AH = 0.

Chemical Thermodynamics Questions

Question 32. Give examples of two processes by which the internal energy of a gas can be increased.
Answer: The internal energy of a gas can be increased by increasing the die temperature ofthe gas.

If a gas is compressed adiabatically (considering only P-V work) its internal energy increases.

Question 33. Give examples of three processes in which the change in internal energy of the system is zero.
Answer: The change in internal energy of any cyclic process is zero.

The change in internal energy of an ideal gas during isothermal expansion or compression is zero. In adiabatic free expansion of an ideal gas, the change in internal energy is equal to zero.

Question 34. Why are the standard reaction enthalpies of the following two reactions different?
Answer: \(\begin{aligned}
& \mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-393.5 \mathrm{~kJ} \\
& \mathrm{C}(\text { diamond }, s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-395.4 \mathrm{~kJ}
\end{aligned}\)
Anszwer: Graphite and diamond are two allotropes ofsolid carbon. Different allotropic forms have different enthalpies. As the given U reactions involve different allotropes, the standard reaction enthalpies ofthese reactions are different.

Chemical Thermodynamics Questions

Question 35. At 25°C, the standard reaction enthalpy for the reaction is -221.0 kj. Does this enthalpy change indicate the standard enthalpy of the formation of CO(g)? If not, then what would be the value of the enthalpy of j formation of CO(g) at 25°C?
Answer: In the given reaction 2 mol CO(g) is formed from its stable constituent elements. Sd,’ definition, AH0 of this reaction does not represent standard enthalpy of formation of CO(g). 2C(graphite,s) + O₂(g) 2CO(g) ; AH0 = -221.0 kj

Dividing both sides by 2, we get

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g) ; \Delta H^0=-110.5 \mathrm{~kJ} \cdots[1]\)

In reaction [1], 1 mol of CO(g) is formed from its stable constituent elements. Thus, in this reaction AH0 = standard enthalpy of formation of CO(g).

So, at 25°C the standard enthalpy of formation of CO(g) =-110.5 kj-mol-1.

Question 36. What do you mean by the standard enthalpy of atomisation of chlorine at 25°C = + 121 kj.mol-1?
Answer: This means that at 25°C and 1 atm pressure, 121 kj of heat is required to produce 1 mol of gaseous Cl-atom from Cl₂(g). Thus, the change in enthalpy for the process, \(\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Cl}(\mathrm{g}); \Delta H_{\text {atom }}^0=+121 \mathrm{~kJ}\)

Question 37. Determine the standard reaction enthalpy for the reaction: \(\mathrm{A}_2 \mathrm{~B}_3(s)+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(s)+3 \mathrm{CB}_2(g)\) Given: \(2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow \mathrm{A}_2 \mathrm{~B}_3(s); \Delta H^0=-x \mathrm{~kJ}\)
Answer:  Reversing equation 1 we get \(\mathrm{A}_2 \mathrm{~B}_3(s) \rightarrow 2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) ; \quad \Delta H^0=+x \mathrm{~kJ}\)

Chemical Thermodynamics Questions

Multiplying the equation by 3, we get

⇒  \(3 \mathrm{CB}(g)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow 3 \mathrm{CB}_2(g) ; \Delta H^0=-3 y \mathrm{~kJ}\)

Adding equations [3] and [4], we get

⇒ \(\mathrm{A}_2 \mathrm{~B}_3(\mathrm{~s})+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(\mathrm{~s})+3 \mathrm{CB}_2(g) ; \Delta H^0=(x-3 y) \mathrm{kJ}\)

Therefore, the standard reaction enthalpy for the given reaction =(x- 3y)kJ.

Question 38. In which ofthe following reactions does AH0 at 25 °C indicate the standard enthalpy of formation \(\left(\Delta \boldsymbol{H}_f^0\right)\) of the compound formed in each of the reactions?
Answer: Here, 2 mol of NH₃(g) is produced from the stable constituent elements, H2(g) and N2(g). Hence according to the definition, at 25°C the standard reaction enthalpy of this reaction is not equal to the standard enthalpy of formation of NH3(g).

The standard state of oxygen at 25°C is O₂(g). So, the given equation does not represent the formation reaction of NO(g). Consequently, at 25°C the standard reaction enthalpy of this reaction is not equal to die standard enthalpy offormation of NO(g).

Here, 1 mol of solid NaCl is formed from its stable constituent elements. Hence at 25°C, the die standard reaction enthalpy of this reaction is equal to the die standard enthalpy of formation of NaCl(s).

Question Discuss the change in die degree of randomness for the following cases— Combustion of kerosene, Sublimation of dry ice, Extraction of salt from seawater, Condensation of water vapour, Crystallisation of a solid from its aqueous solution,
Answer: Combustion of kerosene: Kerosene is a mixture of liquid hydrocarbons. The combustion of kerosene produces C02 and water vapour. As in the reaction, a liquid converts into a gaseous mixture, the molecular randomness ofthe system increases.

Sublimation of dry ice: In dry ice, (solid carbon dioxide), the molecules of C02 exist in an orderly state. When this solid sublimes, the gaseous molecules formed move randomly, i.e., the degree of randomness ofthe molecules increases.

Extraction (Crystallisation) of salt from seawater: In seawater, the attractive forces between Na+ and Cl- ions are weak as the distance between these ions is large. Hence, these ions are virtually free to move randomly. NaCl extracted from the seawater is in a solid state with a crystal structure in which Na+ and Cl- are arranged in a definite order. Hence, the extraction of salt from seawater is associated with a decrease in the degree of randomness.

Condensation of water vapour: In water vapour, the intermolecular forces of attraction between H2O molecules are weak, so the molecules remain in a state of randomness. However, water obtained by condensation of water vapour has less freedom of motion and hence less degree of randomness because of the stronger intermolecular forces of attraction compared to water vapour. Hence, in the case of condensation of water vapour, the degree of randomness decreases.

Crystallisation of a solid from its Aqueous solutions:

In an aqueous solution, the solute molecules exist in a state of random motion. When the solute is crystallised from its solution, the solute molecules in the die crystal remain at fixed positions and become almost motionless. So, the crystallisation of a solid from its solution causes a decrease in randomness.

⇒ \(\mathrm{Cr}^{3+}+6 \mathrm{H}_2 \mathrm{O}(a q) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)\): In this process, one Cr3+ ion combines with six water molecules to form a single complex ion, [Cr(H20)6]3+. Consequently, the number of particles in the system reduces (from 7 to 1 for each combination), thereby decreasing the degree of randomness.

⇒ \(\mathrm{NH}_4 \mathrm{NO}_2(s) \xrightarrow{\Delta} \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}):\) On decomposition, 1 formula unit of solid ammonium nitrite produces 1 molecule of N2(g) and 2 molecules of water vapour i.e., H20(g). In this process, a solid (in which the molecules are orderly arranged) converts into gaseous substances. Furthermore, the number of molecules also increases. Naturally, this process increases the randomness ofthe system.

Chemical Thermodynamics Questions

Question 40. For a spontaneous or irreversible process occurring in an isolated system, when does the entropy of the system attain maximum value? Under this condition, what will be the change in entropy of the system?
Answer: when a spontaneous or irreversible process occurs in an isolated system, the entropy of the system increases with the progress of the process towards equilibrium. The value of entropy becomes maximum when the process attains equilibrium, and there occurs no further change in the entropy ofthe system. Thus, the value of entropy is maximum at the equilibrium state ofthe process and under this condition, the change in entropy ofthe system is zero.

Question 41. Is the entropy change of a system influenced by the change in temperature? Explain.
Answer: The entropy of a system is highly dependent on temperature. With the increase or decrease in temperature, the randomness of the constituent particles (atoms, molecules or ions) of a system increases or decreases. Now, the entropy of a system is a measure of the randomness of its constituent particles. Thus, the entropy change ofa system is influenced by the change in temperature.

Question 42. Mention if the entropy of the system increases or decreases In each of the following cases: Bolling of water, Sublimation of solid iodine,

⇒ \(\begin{aligned}
& 2 \mathrm{O}_3(g) \rightarrow 3 \mathrm{O}_2(g) \\
& \mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_3(g)+\mathrm{HCl}(g) \\
& \mathrm{Hg}(l) \rightarrow \mathrm{Hg}(g) \quad \text { (6) } \mathrm{I}_2(g) \rightarrow \mathrm{I}_2(s) \\
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \text { (8) } 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \\
& \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2(g) \\
& \mathrm{PCl}_5(g) \rightarrow \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
& \text { Haemoglobin }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \text { Oxyhaemoglobin } \\
& \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightarrow \mathrm{Ni}(\mathrm{CO})_4(g) \\
& 4 \mathrm{Fe}(s)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \\
& \mathrm{C}(\text { diamond }) \rightarrow \mathrm{C} \text { (graphite) } \\
&
\end{aligned}\)

Answer: The change in entropy in a phase transition or a reaction depends on the following factors.

For a given amount of substance, the entropy of the substance in different physical states varies in the order: of S_gas > S_liquid > S_soild where S is the molar entropy. Because of this, the entropy of the system increases in the phase changes such as liquid → vapour, solid → vapour, while it decreases in the phase changes such as liquid → solid, vapour → liquid and vapour → solid.

In a reaction, indie reactants are solids or liquids or in a dissolved state in solution and they convert into gaseous products, then the entropy of the die system increases, (b) If the number of gaseous particles (atoms or molecules) in a reaction increases or decrease, then the entropy of the system increaser or decreases.

Increases (liquid — vapour transition), Increases (solid-gas transition), Increases (the number of gaseous particles increases) Increases (gaseous substances are formed from a solid reactant) Increases (liquid — gas transition) Decreases (gas-solid transition) Increases (the number of gaseous particles increases).

Decreases (the number of gaseous molecules decreases) Increases (gaseous substance is formed through the reaction of the reactants, one of which is solid and the other is in a dissolved state in solution) Increases (the number of gaseous molecules increases) Decreases (the number of gaseous molecules decreases)

Decreases (the number of gaseous molecules decreases)

Decreases (the number of gaseous molecules decreases) (g) Increases (the crystal structure of diamond is more compact than that of graphite. As a result, the molecular randomness in graphite is more than that in diamond.

Chemical Thermodynamics Questions

Question 43. Why does the entropy of the gaseous system Increase with the temperature rise?
Answer: Due to large intramolecular distance and weak intermolecular forces, the molecules in a gas can move about freely. The motion ofthe molecules becomes more random and disordered with the rise in temperature as the average speed of the molecules increases. Now, the entropy ofa system is a measure of the disorderliness of the constituent molecules. Therefore, the entropy ofa gas increases with the temperature rise.

Question 44. Give an example of a process for each of the given cases: ΔG = 0, ΔS <0, ΔG= 0, ΔS > 0 ΔG < 0, ΔS > 0 ΔG<0, ΔS<0 in a system.
Answer: Fusion of ice at 0°C and 1 atm pressure.

Condensation of water vapour at 100°C and 1 atm.

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm}\)

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm} .\)

Question 45. At 25°C and 1 atm pressure, for the reaction 3O₂(g) → 20₂(g); H = 286kJ and AS = -137.2 J.K-1. Is this reaction spontaneous? Does the spontaneity of this reaction depend on temperature? Is the reverse reaction spontaneous? If so, then why? Does the spontaneity of the reverse reaction depend on temperature?
Answer: No. (2) For the given reaction ΔH > 0 and ΔS < 0. So, according to the equation ΔG = ΔH-TΔS, AG will be positive at any temperature. Hence, the spontaneity of this reaction is independent of temperature.

The reverse reaction is spontaneous.

In the reverse reaction [20₃(g)→ 3O₂(g)] ; AH = —286 kj and AS = +137.2 J K- . So, according to the equation, AG = AH- TAS, AG is negative.

The spontaneity of the reverse reaction is also independent of temperature as AG is negative.

H<0 and S> 0 at any temperature.

Question 46. The standard enthalpy of formation of C7H5N3O6 is -x kj.mol-1 at 25c. write the thermochemical equation for the formation reaction of the compound.
Answer: The constituent elements of the compound C₇H₅N₃O₆(s) whose standard states at 25C are carbon, hydrogen, oxygen and nitrogen C (graphite, s), H₂(g), O₂(g) and N₂(g) respectively. Therefore, the equation representing the formation reaction of the compound should contain C (graphite, s), H₂(g), O₂(g) and N₂(g) as the sources of carbon, hydrogen, oxygen and nitrogen respectively. are Hence, the thermochemical equation for the formation reaction of the given compound is—

⇒ \(\begin{aligned}
& 7 \mathrm{C} \text { (graphite,s) }+\frac{5}{2} \mathrm{H}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g})+\frac{3}{2} \mathrm{~N}_2(\mathrm{~g}) \rightarrow \mathrm{C}_7 \mathrm{H}_5 \mathrm{~N}_3 \mathrm{O}_6(\mathrm{~s}) ; \\
& \Delta H_f^0\left[\mathrm{C}_7 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{O}_6(s)\right]=-x \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 47. A particular amount of an ideal gas participates in a reversible process as given in the figure. What type of process is this? Explain the changes in each step.
Answer: This process is cyclic because the system returns to its initial state after undergoing consecutive processes AB, BC and CA. In step AB, the gas expands reversibly at constant pressure. p A Hence, step AB indicates an isobaric change. In step BC, the temperature of the gas decreases at constant volume. Thus BC indicates an isochoric change. In step CA, the gas is compressed reversibly at a constant temperature. Thus CA indicates a reversible isothermal change.

Chemical Thermodynamics Questions

Question 48. The transformation of A to B can be carried out in the following two ways in which the initial and final states are identical. \(A \xrightarrow{\Delta H=-x \mathrm{~kJ}} B ;\) \(A \xrightarrow{\Delta H=-y \mathrm{~kJ}} C \xrightarrow{\Delta H=?} B ;\) What will be the value of ΔH during transformation of C to B?
Answer: According to Hess’s law, the change in enthalpy in process = total change in enthalpy in process

∴ -xkJ =-yKJ + ΔH = (y-c)kJ

Question 49. Under what conditions the heat of reaction at fixed pressure is equal to that at fixed volume?
Answer: we know \(\Delta H=\Delta U+P \Delta V\)

In case of a reaction involving gaseous substances, the equation [1] can be written as, AH = AU + AnRT [Assuming ideal behaviour of the gases]

If a reaction involves only solid or liquid substances (i.e., no gaseous substance), then the change in volume of the reaction system is negligible i.e., ΔV≈O. According to equation (1), for such type of reaction ΔH ≈ AU.

For example: \(\mathrm{NaOH}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

Again, if in a gaseous reaction, the difference between the total no. of moles of gaseous products and the total no. of moles ofthe gaseous reactants is zero (i.e., Δn = 0 ), then AH = A17.

Again, if in a gaseous reaction, the difference between the total no. of moles of gaseous products and the total no. of moles ofthe gaseous reactants is zero (i.e., Δn = 0 ), then AH = A17.

Question 50. The initial pressure, temperature & volume of 1 mol of gas are P1, T1 and V1 respectively. The state of the gas is changed in the following two ways. Will the internal energy change be the same in both cases?
Answer: The internal energy of a system is a state function. The change in internal energy in a process depends only on the initial and final states of the system. It does not depend on the path used to arrive at the state. Since the initial and final states are the same in 1 and 2, the internal energy change will also be the same.

Chemical Thermodynamics Questions

Question 51. If one mole of an ideal gas is expanded in the following two ways, then will the value of P2 and P2 be greater than, less than or equal to P1?
Answer: The temperature of the gas remains the same during isothermal expansion. Therefore, P2 < P1. On the other hand, the temperature of the gas decreases during an adiabatic expansion. Therefore, in the process, T2 is less than T1. In this process. P2 < P2 since T2<T1

Question 52. Write three differences between reversible and irreversible processes. Melting office at 0°C and 1 atm pressure is a reversible process— explain.
Answer: Second part: It is a reversible process. Ice melts at 0°C under normal atmospheric pressure. Latent heat for the fusion of Ice Is lit) cal .g-1, i.e., 80 cal of heat is required to melt logfile. If 80 cal heat IB is extracted from the surroundings, 1 g of ice gets converted Into water. Therefore, at normal pressure and temperature, ice and water remain in an equilibrium state. By Increasing or decreasing the value of the driving force (by the supply or extraction of heat) the process can be made to move In the forward or backward direction. So, the melting of ice at normal atmospheric pressure and temperature is an example of a reversible process.

Question 53. The boiling point of benzene is 80.1 °C. At ordinary pressure and 70°C, the benzene vapour spontaneously transforms into liquid benzene. In this process, what will the signs of AH, AS and AG be?
Answer: The entropy of the system decreases when a vapour transforms into a liquid. So AS < 0. Again, the condensation is an exothermic process. So, in this process, AH < 0. Under the given conditions, the benzene vapour spontaneously condenses into liquid. So, in this process AG < 0.

Question 54. Calculate AU and AH in calories if one mole of a monoatomic ideal gas is heated at a constant pressure of 1 atm from 25°C to 50°C.
Answer: for an ideal gas, the changes in internal energy (A U) and enthalpy (AH) due to the change in its temperature are given by the relations

⇒ \(\Delta U=n \mathrm{C}_{V, m} \Delta T \text { and } \Delta H=n \mathrm{C}_{P, m} \Delta T\)

For a monoatomic ideal gas \(C_{V, m}=\frac{3}{2} R \text { and } C_{P, m}=\frac{5}{2} R \text {. }\)

The number of mole of the gas, n = 1 ; and the change in temperature (AT) = 25 K

Therefore \(\Delta U=1 \times \frac{3}{2} R \times 25=1 \times 1.5 \times 1.987 \times 25 \mathrm{cal}\) =74.5cal and \(\Delta H=1 \times \frac{5}{2} R \times 25=1 \times 2.5 \times 1.987 \times 25 \mathrm{cal}=124.18 \mathrm{cal}\)

Question 55. How much hard coal is required to produce the same amount of heat as is produced by the combustion of 2.0 I, of gasoline (mainly isooctane, C8H,b)? Given: AjJJ0 of C8H18 = -5460 kj-mok1 , density of isooctane = 0.692 g. mL-1 and the calorific value of hard coal is 32.75 kj g-1.
Answer: The mass of 2.0 L of gasoline = 2000 x 0.692 g = 1384 g For 1384 g of the gasoline, the number of moles =\(\frac{1384}{114}=12.14\) [molar mass of the gasoline = 114 g. mol-1 [ For the combustion of 1 mol of gasoline, the amount of the liberated heat is 5460 kj. Therefore, the combustion of 12.14 mol of gasoline will produce 5460 x 12.14 kJ of heat. The calorific value of the hard coal is 32.75 kJ. g 1. So, the amount ofthe hard coal that will produce \(5760 \times 12.14 \mathrm{~kJ} \text { of heat is } \frac{5760 \times 12.14}{32.75} \mathrm{~g}=2135.16 \mathrm{~g}\)

Question 56. One kg of graphite is burnt in a closed vessel. The same amount of graphite is burnt in an open vessel. Will the heat evolved in the two cases be the same? If not, in which case it would be greater?
Answer: Burning of graphite involves the following reaction \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

In an open vessel, if a reaction is carried out, it occurs at constant atmospheric pressure. So, in the reaction, the heat change is the same as the enthalpy change (ΔH). The heat change in a reaction carried out in a closed vessel at constant volume is the same as the internal energy change ( Δ17).

For the above reaction, An = 1-1 = 0. Thus, according to the relation, ΔH = ΔU+ΔnRT, we have AH = AU. Therefore, the burning of 1kg of graphite will produce the same amount of heat irrespective of whether the reaction is carried out at constant volume or constant pressures.

Question 57. Can ΔH be taken as the sole criterion of the spontaneity of a reaction? Justify with an example.
Answer: In an exothermic reaction, ΔH is negative. This means that the energy of the system decreases in an exothermic reaction. On the other hand, ΔH is positive for an endothermic reaction, indicating the energy of the system increases in such a reaction.

As enthalpy decreases in an exothermic reaction, it was once thought that only exothermic reactions would be spontaneous. However, there are some chemical reactions for which AH is positive although they were found to occur spontaneously. Therefore, AH cannot be regarded as the sole criterion for determining the spontaneity of reaction.

Question 58. An Intimate mixture of Fe₂O₃ and ΔA₁₂O₃ is used in solid fuel for rockets. Calculate the fuel value per gram and fuel value per cm3 of the mixture.
ΔH<Fe₂O₃) = 1669,4 kj- mol-1 ,
ΔH(A₁₂O₃) = 832.6 kJ-mol-1
Answer: The reaction is 2A1 + Fe₂O₃→2Fe + A₁₂O₃

The enthalpy change in this reaction is \(\begin{aligned}
\Delta H^0 & =\Delta H_f^0\left(\mathrm{Al}_2 \mathrm{O}_3\right)-\Delta H_f^0\left(\mathrm{Fe}_2 \mathrm{O}_3\right) \\
& =(1669-832.6) \mathrm{kJ}=836.4 \mathrm{~kJ}
\end{aligned}\)

The total mass ofthe reactants (2A1 + Fe₂O₃)

= [2 X 27 + (2 X 55.85 + 3 X 16)] g =213.7 g

Therefore, the fuel per gram of the mixture

Chemical Thermodynamics Questions

⇒ \(=\frac{836.4}{213.7} \mathrm{~kg} \cdot \mathrm{g}^{-1}=3.91 \mathrm{~kJ} \cdot \mathrm{g}^{-1}\)

⇒ \(=\frac{836.4}{213.7} \mathrm{~kg} \cdot \mathrm{g}^{-1}=3.91 \mathrm{~kJ} \cdot \mathrm{g}^{-1}\)

The volume of the mixture of the reactants

⇒ \(=\left(\frac{2 \times 27}{2.7}+\frac{159.7}{5.2}\right) \mathrm{cm}^3=50.7 \mathrm{~cm}^3\)

Therefore, the fuel value per cm3 ofthe mixture

⇒ \(=\frac{836.4}{50.7} \mathrm{~kJ} \cdot \mathrm{cm}^{-3}=16.49 \mathrm{~kJ} \cdot \mathrm{cm}^{-3}\)

Question 59. In a constant volume calorimeter, 3.5g of gas with molecular mass 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K, due to the combustion process. Given, that the heat capacity of the calorimeter is 2.5 kj – K-1, what will be the value of enthalpy of combustion of the gas?
Answer: Por a combustion reaction carried out in a constant volume calorimeter, the amount of liberated heat is given by the relation, q = Calx AT.

Ccal = 2-5 W’ K_1 = (298.45- 298) K =0.45 K

Thus, q = 2.5 x 0.45 kl = 1.125 kj

For 3.5 g ofthe gas, the number of moles \(=\frac{3.5}{28}=0.125\)

Thus, the burning of 0.125 mol ofthe given gas liberates 1.125 kj of heat Hence, the enthalpy of combustion of the gas is \(\frac{1.125}{0.125} \mathrm{~kJ} \cdot \mathrm{mol}^{-1}=9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 61. If the bond dissociation energies of XY(g), X₂(g) and Y₂(g) are in the ratio of 1: 1: 0.5 and JH for J the formation of XY(g) is -200 kj.mol-1, then what is will be the bond dissociation energy 7 of X₂(g)?
Answer: \(\text { s. } \frac{1}{2} X_2(g)+\frac{1}{2} Y_2(g) \rightarrow X Y(g)\) Suppose, the bond dissociation energy of XY is x. So, the bond dissociation energies of X2 and Y2 will be x and, respectively. For the reaction

⇒ \(\Delta H^0=\Delta H_f^0(\mathrm{XY})-\frac{1}{2} \Delta H_f^0\left(\mathrm{X}_2\right)-\frac{1}{2} \Delta H_f^0\left(\mathrm{Y}_2\right)=-200 \mathrm{~kJ}\)

In terms of bond energies,

⇒ \(\Delta H^0=\left(\frac{1}{2} x+\frac{1}{2} \times \frac{x}{2}\right)-(x)=\frac{3 x}{4}-x=-\frac{x}{4}\)

Therefore, \(-\frac{x}{4}=-200 \mathrm{~kJ}\)

or, x = 800 kJ

Thus, the bond dissociation energy of X₂(g) is 800 kJ.mol-1.

Chemical Thermodynamics Questions

Question 62. For the process H₂O(l)⇌H₂O(g), ΔH = 40.8 kJ⋅mol‾1 at the boiling point of water. Calculate molar entropy change for vaporization from the liquid phase
Answer: Molar entropy change for vaporisation \(\Delta S=\frac{\Delta H_{v a p}}{T_b}\)

Given ΔH_vap = 40.8KJ⋅mol¯1, fr water T_b=373 K

∴ \(\Delta S=\frac{40.8 \times 10^3}{373}=109.38 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

∴ Molar entropy change for vaporization of water = 109.38 I.K-1.mol-1

Question 63. A gas confined in a cylinder with a frictionless piston is made to expand from 1L to 5L under a constant pressure of 1.5 atm. During the process, 800 J of heat is supplied from an external source. Calculate the change in internal energy of the gas. (1L- atm = 101.3 J)
Answer: We know, w = -P_ex ( V₂– V₁)

∴  ω= -1.5(5- 1 ) = -6 L⋅atm = -6 x 101.3 I = -607.8J

It is also given that q = + 800 1

Using the 1st law of thermodynamics, we have ΔU = q + w = (800- 607.8) J = 192.2J

∴ The change in internal energy of the gas Is 192.2 J

Question 64. Calculate AH0 of the following reaction at 298 K:

⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)

Given: \(\begin{array}{r}
\mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)+3 \mathrm{O}_2(g) \rightarrow 3 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=-1368 \mathrm{~kJ}
\end{array}\)

⇒ \(\begin{aligned}
& 2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-2600 \mathrm{~kJ} \\
& 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ} \text { at } 298 \mathrm{~K}
\end{aligned}\)

Answer: ⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)

\(\begin{aligned}
& 2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-2600 \mathrm{~kJ} \\
& 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ} \text { at } 298 \mathrm{~K}
\end{aligned}\) \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ}\)

Multiplying each of the equations [2] and [3] by \(\frac{1}{2}\) and then adding them, we have

\(\begin{array}{r}
\mathrm{C}_2 \mathrm{H}_2(g)+3 \mathrm{O}_2(g)+\mathrm{CO}(g) \rightarrow 3 \mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=\left[\frac{1}{2}(-2600)+\frac{1}{2}(-566)\right] \mathrm{kJ}
\end{array}\)

Subtracting equation [1] from equation [4], we have \(\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l) ; \\
& \Delta H^0=\left[\frac{1}{2}(-2600)+\frac{1}{2}(-566)-(-1368)\right] \mathrm{kJ}=-215 \mathrm{~kJ}
\end{aligned}\)

Question 65. What is meant by the terms change of entropy (ΔS) and change in free energy (ΔG) of a system? Write down the mathematical relation between them. At 0°C, liquid water and ice remain in equilibrium. If lg of liquid water under equilibrium conditions is converted to ice, explain with reason whether the process is endothermic or exothermic.
Answer: Third part: In the conversion of water into ice, the entropy of the system decreases, and Hence AS_sys < 0

we know, ΔG = ΔH- TΔS

For the given process, ΔG< 0. As the process occurs spontaneously, ΔG < 0 for the process. According to the relation [1|, if ΔS<0, then AG will be negative only when ΔH < 0. So, the process is exothermic.

Question 66. Given: C(s) + O₂(g)→CO₂(g); AH = -393.5 kj 2H₂(g) + O₂(g)→2H₂O(g) ; AH = -571.6 kj

Calculate ΔH of the reaction:

\(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g})\)
Answer: C(s) +O₂(g)→CO₂(g); AH = -393.5 kj⋅⋅⋅[1]

2H₂(g) + O₂(g)→2H₂O(g) ; ΔH = -571.6 kJ⋅⋅⋅[2]

Subtracting equation [2] from equation [1], we have C(s) + 2H₂O(g)→CO₂(g) + 2H₂(g);
ΔH = [-393.5- (-571.6)] k] = 178.1J.

Chemical Thermodynamics Questions

Question 67. For the reaction, N₂(g) + 3H₂(g)→+2NH₂(g), ΔH and ΔS are -95.4 kj and -198.3 J. K-1 respectively. Assuming ΔH and ΔS are independent of temperature, will the reaction be spontaneous at 500 K? Explain
Answer: we know ΔG-ΔH-TΔS

Given That ΔH = -95.4 Kj and ΔS =-198.3J.K-1

∴ ΔG= AG =-95.4 kJ- 500 K(-198.3 X 10-3 kj.K-1)

= 3.75kj

As ΔG > 0 at 500 K, the reaction will not be spontaneous at 500 K

Question 68. The bond energy of any diatomic molecule is defined to be the change in the internal energy for its dissociation. At 298 K, O2(g)→2O(g) AH = 498.3 kj.mol¯1 . Calculate the bond energy of O₂ molecule R = 8.314 J-K-1.mol-1
Answer: Given: O₂(g)→2O(g); ΔH = 498.3 kj⋅mol-1
For the above reaction, An = 2-l = l

We know, ΔH = ΔU+ ΔnRT

∴ 498.3 kj = ΔU+1 x 8.314 x 10~3 x 298 kj.

∴ ΔU = 495.82 kJ

Therefore, the bond energy of the O₂ molecule = 495.82 kJ

Question 69. State the first law of thermodynamics. An ideal gas of volume 6.0 L was made to expand at constant temperature and pressure of atm by supplying heat. If the final volume of the gas was 12.0 L, calculate the work done and the heat supplied in joule in the process [1L.atm= 101.3J]
Answer: We know, w = -Pex V₂– V₁

∴ w = -2(12-6) L- atm = -12 L’- atm = -1215.6 J

As the process is isothermal and the system is an ideal gas, AU = 0 for this process. According to the 1st law of thermodynamics, ΔU = q + ω

∴ 0 = q- 1215.61 or, q = + 1215.6J

Question 70. A process is always spontaneous at all temperatures if the enthalpy change is and entropy changes is _______
Answer: —ve and +ve.

Question 71. Two moles of an ideal gas were expanded isothermally against a constant opposing pressure of 1 atm from 20 l to 60 complete w,q, E, and H for the process in joule (given 1L. atm = 101.3J)
Answer: We know, w = -Pex(V2-V1)

∴ \(\begin{aligned}
w=-1(60-20)=-40 \mathrm{~L} \cdot \mathrm{atm} & =-40 \times 101.3 \mathrm{~J} \\
& =-4.052 \mathrm{~kJ}
\end{aligned}\)

For this process ΔE – 0 and ΔH = 0 because the process is isothermal and the system is an ideal gas. As per the first law of thermodynamics, ΔE = q+ w or, 0 = q- 4.052 kJ

∴ q = 4.052 kJ.

Question 72. The latent heat of fusion of ice at 0°C is 80cal/g; Calculate the molar entropy change for the fusion process.
Answer: We know, \(\Delta S=\frac{\Delta H_f}{T_f}\)

where ΔH_f– = latent heat of fusion of a substance and Tf = its melting point.

Now, ΔH_f = 80 cal⋅g-1 x 18g = 1440 cal

∴ \(\Delta S=\frac{1440}{273} \mathrm{cal} \cdot \mathrm{K}^{-1}=5.27 \mathrm{cal} \cdot \mathrm{K}^{-1}\)

Question 73. Calculate AG° for the reaction; \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\) at 298K. Given, at 298 K ΔH_fº for H₂O(1) is -286 kj⋅mol-1 and the molar entropies (S°) for H₂ (g), O₂(g) and H₂O(Z) are 130.7, 69.9 J K-1 . mol-1 respectively.
Answer: For the given reaction, \(\Delta S^0=S_{\mathrm{H}_2 \mathrm{O}(l)}^0-\left[S_{\mathrm{H}_2(\mathrm{~g})}^0+\frac{1}{2} S_{\mathrm{O}_2(g)}^0\right]\)

⇒ \(=\left[69.9-\left(130.7+\frac{1}{2} \times 205.1\right)\right] \mathrm{J} \cdot \mathrm{K}^{-1}=-163.35 \mathrm{~J} \cdot \mathrm{K}^{-1}\)

Given that ΔH_fO[H₂O(l)] = -286 kj-mol-1

So, AH° = -286 kJ for the given reaction.

We know, ΔG° = ΔHO -TΔS0

∴ AG° = [-286 x 103- 298(-163.35)] J = -237.32 kj.

Chemical Thermodynamics Questions

Question 74. Above what temperature the following reaction will be spontaneous?
Answer: We know, ΔG = ΔH-TΔS. A reaction at a given temperature and pressure is spontaneous if AG < 0 for the reaction. Therefore, at a given pressure and a temperature of TK, the reaction will be spontaneous if—

ΔG < 0 or, ΔH- TΔS < 0

∴ \(T \Delta S>\Delta H \text { or, } T>\frac{\Delta H}{\Delta S}\)

Given that, ΔH = 144.6 kj and ΔS = 0.116 kl. K¯1

∴ \(T>\frac{144.6}{0.116} \text { or, } T>1246.55 \mathrm{~K}\).

So, the reaction will be spontaneous above 1246.55 K.

Question 75. Calculate AH ofthe following reaction at 25°C. 4Fe(s) + 3O₂(g)→2Fe₂O₃(s) Given: Fe₂O₃(s) + 3C(graphite)→2Fe(s) + 3CO(g), ΔH = 117.30 kcal C(graphite) + O₂(g)→CO₂(g) ; AH = – 94.05 kcal \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=67.63 \mathrm{kcal} \text { at } 25^{\circ} \mathrm{C}\)
Answer: \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=67.63 \mathrm{kcal} \quad \ldots[1]\)

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=-94.05 \mathrm{kcal} \cdots[2]\)

C(graphite) + O₂(g)→CO₂(g); AH = -94.05 kcal ⋅⋅⋅⋅[2]

Fe₂O₃(s) + 3C(graphite)→2Fe(s) + 3CO(g)

AH = 117.30 kcal

Multiplying equation [3] by 2 and equation [1] by 6 and adding them together, we have

2Fe2O₃(s)+6C(graphite)+3O₂(g)→4Fe(s)+6CO₂(g)

AH = [2(117.30) + 6(67.63)] kcal

Subtracting equation [4] from the equation obtained by multiplying equations [2] by 6, we have

4Fe(s) + 3O₂(g)→2Fe₂O₃(s) ;

ΔH = 6(-94.05)- [2(117.30) + 6(67.63)] =-1204.68kcal

Question 76. Discuss whether the difference between the heat of the reaction at constant pressure and constant volume will depend on the temperature of the following reaction.
Answer: 

2CO(g) + O₂(g)→+2CO₂(g)

We know, AH = AU+AnRT

For the given reaction, An = 2-(2 + 1) = -1

ΔH-ΔU = ΔnRT = -RT

So, the value of (ΔH-ΔU) depends on the temperature.

Question 77. Which ofthe following is an example of a closed system

1. A hot water-filled thermos flask
2. An ice water-filled airtight metallic bottle
3. A water-filled stainless steel bowl
4. A hot water-filled glass beaker

Answer: 2. An ice water-filled airtight metallic bottle

An ice-water-filled airtight metallic bottle is an example of a closed system.

Chemical Thermodynamics Questions

Question 78. Which is an intensive property of a system

1. Internal energy
2. Entropy
3. Mass
4. Density

Answer: 4. Density Density is an intensive property.

Question 79. In a process, 600J of heat is absorbed by a system, and 300J of work is done by the system. Calculate the change in internal energy of the system
Answer: Given: q = +600J, w = -300J. So, change in internal energy, ΔH = q + w = (600- 300)J = 300J

Question 80. The latent heat of the vaporization of water at a normal boiling point is 40.75 kJ. mol-1 . Calculate the change in entropy of vaporization.
Answer: Given: \(\Delta H_{\text {vap }}\) = 4075 kJ. mol-1, Tb =100C = 375k \(\Delta S=\frac{\Delta H_{v a p}}{T_b}=\frac{40.75 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{373}=109.25 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

Question 81. Calculate the enthalpy of the formation of liquid ethyl alcohol from the following data.
Answer: \(\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-1368 \mathrm{~kJ} \\
& \mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H=-393 \mathrm{~kJ} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-287 \mathrm{~kJ}
\end{aligned}\)

Question 82. State the condition of spontaneity and equilibrium in terms of Gibbs free energy change of a system.
Answer: The condition of spontaneity regarding Gibbs free energy is AG < 0. The condition of equilibrium in terms of Gibbs free energy is AG = 0.

Question 83. Which variable is kept constant in an isochoric process?
Answer: Volume is kept constant in an isochoric process.

Question 84. Calculate the change in internal energy of the gas when it expands from 2L to 8L at a constant pressure of 2 atm absorbing 400 J of heat in the process. (1 L-atm = 101.35 J)
Answer: We know, w = -Pex (V2-V1)

∴ ω=_2(8- 2)=-12 L.atm=-12 x 101.3 J =-1215.6 J.

According to the 1st law of thermodynamics, A U=q + ω

Given that q = 400 J

∴ ΔU= (400- 1215.6)J = -815.61

∴ Change in internal energy of the gas =-815.61.

Question 85. At 25 °C the standard heat of formation of liquid H20 is -286.0 kJ mol-1. Calculate the change in standard internal energy for this formation reaction.
Answer: Formation reaction of \(\mathrm{H}_2 \mathrm{O}(l): \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

For this rections \(\Delta n=0-\left(1+\frac{1}{2}\right)=-\frac{3}{2} \text {. }\)

We know, ΔH° = ΔU° + ΔnRT

∴ \(-286.0 \mathrm{~kJ}=\Delta U^0+\left(-\frac{3}{2}\right) \times 8.314 \times 10^{-3} \times(273+25) \mathrm{kJ}\)

or, ΔU° = -282.28 kj

Therefore, the change in internal energy for the formation reaction of H20(/) is -282.28 kJ.

Chemical Thermodynamics Questions

Question 86. The temperature of 4 mol of a gas decreases from 40°C to -60 °C on adiabatic reversible expansion. The molar-specific heat of the gas at a constant volume being 12 J.K-1 mol-1, determines the change in internal energy and work done in this process
Answer: For an adiabatic expansion

\(w=\Delta U=n C_{V, m}\left(T_2-T_1\right)\left[T_2<T_1\right]\)

∴ ΔU = 4 X 12 X (213- 313) J = -4.8 kj and w = Δt = -4.8 kj

∴ In the process, the change in internal energy = -4.8 kJ, and the amount of work done = 4.8 kJ.

Question 87. Write the thermodynamic relation generally used to predict whether a reaction is spontaneous or not For exothermic and endothermic reactions with their change in entropies being positive and independent of temperature variations, comment on the spontaneity of the reactions in both cases for temperature variations. Compute ΔH° at 298 K for: OH(g)→+H(g) + O(g)
Answer: The thermodynamic relation that is generally used to predict the spontaneity of a reaction is AG = AH- TAS, where AG, AH, and AS are the changes in free energy, enthalpy, and entropy in the reaction at a given pressure and temperature of T K. At a given pressure and temperature, for a spontaneous reaction AG < 0. In exothermic reaction, AH < 0.

In such a reaction, if AS> 0, then AG = AH-TAS = -ve-T(+ve). As T is always positive, AG will always be negative at any value of T. Hence, an exothermic reaction will always be spontaneous at any temperature if AS > 0.

In an endothermic reaction, AH > 0. In such a reaction, if AS >0, then AG = AH-TAS =+ve-T(+ve) So, AG will be -ve only when |TASl > |AH|. It happens at high temperatures. So, an endothermic reaction with AS > 0 will be spontaneous at high temperatures.

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{OH}(\mathrm{g}) ; \Delta H^0=10.06 \mathrm{kcal} \quad \cdots[1]\)

⇒ \(\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta H^0=104.18 \mathrm{kcal} \\
& \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) ; \Delta H^0=118.32 \mathrm{kcal}
\end{aligned}\)

By dividing each of the equations [2] and [3] by 2, and then adding them together, we have

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \text {; }\)

ΔHº= (52.09 + 59.161) kcal = 111.25 kcal

Subtracting equation [1] from equation [4], we have

OH(g)→H(g) + O(g) ; ΔHº = (111.25 -10.06) kcal = 101.19 kcal

So, ΔHº for the given reaction is 101.19 kcal.

Question 88. Which one is the correct unit of entropy—

1. k-1. mol-1
2. J.k-1. mol-1
3. J.mol-1
4. J-1.k-1. mol-1

Answer: 2. J.k-1. mol-1

Chemical Thermodynamics Questions

Question 89. For the reversible reaction A + 2B→ C + Heat, the forward reaction will proceed at

1. Low temperature and low pressure
2. Low pressure
3. High pressure and low temperature
4. High pressure and high temperature

Answer: 3. High pressure and low temperature

For the reversible reaction A + 2 BC + A, the forward reaction will proceed at high pressure and low temperature.

Question 90. What is meant by an isolated system?
Answer: We know, AG = AH- TAS

or, AG = (29.3- 298 X 104.1 x 10-3) kj.mol-1

= -1.7218 kj-mol-1

At a particular pressure and 298 K temperature, the free energy change of the given reaction is negative which indicates the spontaneity of the reaction.

Question 91. At 25°C which of the following has an enthalpy of formation zero–

1. HCL(g)
2. O₂(g)
3. O₃(g)
4. NO(g)

Answer: 2. O₂(g)

Question 92. For which of the following reactions, AS > 0 —

1. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
2. \(\mathrm{HCl}(g)+\mathrm{NH}_3(g) \rightleftharpoons \mathrm{NH}_4 \mathrm{Cl}(s)\)
3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
4. \(\mathrm{MgO}(s)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{Mg}(s)+\mathrm{H}_2 \mathrm{O}(g)\)

Answer: 3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)

⇒ \(\mathrm{NH}_4 \mathrm{NO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Question 93. The heat of combustion of benzene is x J-mol-1. The heat of the formation of carbon dioxide and water are y J.mol-1 and k J .mol-1 respectively. Calculate the heat of the formation of benzene.
Answer: \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ;\)

With increasing the number of species in the system, the entropy system increases for which AS > 0.

ΔH = -x J-mol-1

C(s) +O₂(g) CO₂(g) ; AH = -y J. mol-1

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta H=-z \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

6 x eqn.(2) + 3 x eqn.(3)- eqn.(1) we get,

6C(s) + 3H2(g)→CgH6(l) ;

= (x- 6y- 3z) J .mol-1

Chemical Thermodynamics Questions

Question 94. State the second law of thermodynamics based on entropy. The boiling point of ethanol is 78.4°C. The change in enthalpy during the vaporization of ethanol is 96 J- mol-1. Calculate the change in entropy of vaporization of ethanol.
Answer: We know,

⇒ \(\begin{aligned}
\Delta S_{\text {vap }}=\frac{\Delta H_{\text {vap }}}{T_b} & =\frac{96}{(273,+78.4)} \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \\
& =0.2732 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 95. An amount of work w is done by the system and a q amount of heat is supplied to the system. By which the following relations the change in internal energy of the system can be expressed-

1. Δ U = q-w
2. Δ U =q+w
3. ΔU=q
4. ΔU=w-q

Answer: 4. ΔG<0

Question 96. Write the SI unit of entropy
Answer: SI unit of entropy: J. K-1. mol-1

Question 97. For the following reaction at 298 K 2X + Y -> Z AH = 300 kj.mol-1 and AS = 0.2 kj K-1.mol-1 At what temperature will the reaction become spontaneous considering AH and AS to be constant over the temperature range?
Answer: For spontaneous process, AG < 0

∴ \(\Delta H-T \Delta S<0 \text { or, } \Delta H<T \Delta S_1 \text { or, } \frac{\Delta H}{\Delta S}<T\)

⇒ \(\text { or, } \frac{300}{0.2}<T \text { or, } 1500<T\)

Therefore, the given reaction becomes spontaneous above 1500 K temperature.

Chemical Thermodynamics VSAQ’s

Question 1. A thermodynamic state function is a quantity—

1. Used to determine heat changes
2. Whose value is independent of the path
3. Used to determine pressure-volume work
4. Whose value depends on temperature only

Answer: 2. Whose value is independent of the path

A thermodynamic state function of a system is a quantity whose value depends only on the present state of the system. Its value does not depend on the path of a process in which the system participates.

Chemical Thermodynamics Questions

Question 2. For the process to occur under adiabatic conditions, the correct condition is —

1. ΔT=0
2. ΔP=0
3. q=0
4. w=0

Answer: 3. q=0

In an adiabatic process, no exchange of heat takes place between the system and its surroundings.

Question 3. Enthalpies of all elements in standard states are

1. Unity
2. Zero
3. <0
4. Different for each element

Answer: 2. Zero

By convention, the enthalpies of all the elements in their standard states are considered to be zero.

Question 4. ΔU0 for combustion of methane is -XkJ .mol 1. The value of AH° is

1. ΔU0
2. >ΔU0
3. <ΔU0
4. 0

Answer: 3. <ΔU0

The combustion reaction for methane is—

⇒ \(\mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

For this reaction, An = 1- (1 + 2) = -2

we know ΔH0 = ΔU° + ΔnRT

For the above combustion reaction, ΔH° = – X- 2RT

As T is positive, ΔH0 < ΔU0

Chemical Thermodynamics Questions

Question 5. The enthalpy of combustion of methane, graphite, and dihydrogen at 298K are -890.3 kj.mol-1, -393.5 kj.mol-1, and -285.8 kj.mol-1 respectively. Enthalpy of formation of CH4(g) will be—

1. -74.8KJ. mol-1
2. —52.27kl.mol-1
3. +74.8KJ. mol-1
4. +52.26KJ. mol-1

Answer: According to the given data

⇒ \(\begin{aligned}
& \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta H^0=-890.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \\
\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285,8 \mathrm{~kJ} \cdot \text { mol }^{-1}\)

By 1 xeq.(2) + 2xeq.(3)-eq.(l), we get the thermochemical equation involving the formation reaction of CH4(g)

⇒ \(\begin{aligned}
& \mathrm{C}(s \text {, graphite })+\mathrm{O}_2(g)+2 \mathrm{H}_2(g)+\mathrm{O}_2(g)-\mathrm{CH}_4(g)-2 \mathrm{O}_2(g) \\
& \quad \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(g)-2 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

ΔH° =[- 393.5 + 2(-285.8)- (-890.3)]kJ.mol-1 C(s, graphite) + 2H2(g) CH4(g) ;AH° = -74.8 kj.mol-1 This equation represents the formation reaction of CH4(g). Hence, the enthalpy of formation of CH4(g) is -74.8kJ.mol-1

Chemical Thermodynamics Questions

Question 6. A reaction, A + B→+C + D + q is found to have a positive entropy change. The reaction will be— 

1. Possible at high temperature
2. Possible only at low temperature
3. Not possible at any temperature
4. Possible at any temperature

Answer: 4. The Thermochemical equation for the reaction indicates that the reaction is exothermic. So, for this reaction, AH < 0. It is given that AS > 0 for the reaction. So, according to the relation AG = AJT-TAS, AG will be <0 at any temperature. Hence, the reaction is possible at any temperature.

Question 7. In a process, 701 J of heat is absorbed by a system and 394J of work is done by the system. What is the change in internal energy for the process?
Answer: Given: q = +701J and w = -394 J {~ve sign as the work is done by the system)

Now, ΔU = q + w or, A U = (701- 394)J = +307 J.

So, the change in internal energy of the system = +307 J.

Chemical Thermodynamics Questions

Question 8. The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and A U was found to be -742.7 kj.mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K:

⇒ \(\mathrm{NH}_2 \mathrm{CN}(s)+\frac{3}{2} \mathrm{O}_2(g) \rightarrow \mathrm{N}_2(g)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)\)

Answer: For the given reaction. Δn \(=(1+1)-\frac{3}{2}=+\frac{1}{2}\)

We know, ΔH = ΔU+ΔnRT

∴ \(\Delta H=\left(-742.7+\frac{1}{2} \times 8.314 \times 10^{-3} \times 298\right)\)

= -741.64kJ . mol-1

∴ The change in enthalpy for the reaction is -741.46kJ.mol-1

Question 9. Calculate the number of kj necessary to raise the temperature of 60g of aluminum from 35°C to 55°C. The molar heat capacity of A1 is 24J.mol-1. K-1.
Answer: Heat (q) required to raise temp, of m g of a substance (specific heat capacity=c) from T1 to T2 is: q=mc(T2-T1)

⇒ \(c=\frac{\text { Molar heat capacity }}{\text { Molar mass }}=\frac{24 \mathrm{~J} \cdot \mathrm{K}^{-1}}{27 \mathrm{~g}}=\frac{24}{27}\left(\mathrm{~J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\right)\)

⇒ \(\text { So, } q=60 \times \frac{24}{27}[(273+55)-(273+35)] \mathrm{J}=1.066 \mathrm{~kJ}\)

Hence, 1.066kJ heat is required.

Question 10. Calculate enthalpy change on freezing 1 mol water at 10°C to ice at -10°C. A = 6.03kJ. mol1 at 0°C. Cp[H20(J)]=75.3J.mol-1.K-1 , Cp[H20(s)]=36.8J. mol-1.K-1
Answer: The given process can be considered as the sum of the following two processes—

Water(10°C) -4 water(0°C) 4 ice(0°C) 4 ice (-10°C)

To calculate enthalpy changes in steps (1) and (3), we use the relation, qp=AH=Cp(T2-T1) [At constant P, q = AH].

Step 1: AHj =Cp[H2O(l)](T2-T1)

= 75.31. mol-1.K-1 [273- (273 + 10)]K

=-0.753 kl-mol-1

Step 2: Freezing of water takes place. In this step,

AH2 = -AfusH = -6.03kj.mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kj-mol-1

Step 2: Freezing of water takes place. In this step

ΔH2 = -ΔfusH = -6.03kj. mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kl-mol-1

So, the total change in enthalpy in the entire process,
AH =AHJ+AH2 + AH3

=[- 0.753- 6.03- 0.368]kJ. mol-1

=-7.151kj.mol-1

Note: The answer to this problem is given as -5.65kJ.mol-1. The process may be: water at -10°C -> ice at -10°C,

Chemical Thermodynamics Questions

Question 11. Enthalpy of combustion of C to CO₂: -393.5 kj.mol-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer: Given:C(s) + O₂(g)→CO₂(g); AH0 = -393.5 kj.mol-1

According to this equation, 44 g of CO₂ = 12 g of C

∴ \(35.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{12}{44} \times 35.2 \equiv 9.6 \mathrm{~g} \text { of } \mathrm{C} \equiv 0.8 \mathrm{~mol} \text { of } \mathrm{C}\)

From above equation, 1 mol C = 393.5 kJ of heat released

∴ 0.8 mol of C = 393.5 x 0.8kJ = 314.8kJ of heat released Therefore, the heat released on the formation of 35.2 g of CO₂ from C and O₂ is 314.8 kj

Question 12. Enthalpies of formation of CO(g), CO₂(g) , N₂O(g) and N₂O₄(g) are -110, -393, 81 and 9.7kj-mol-1 respectively. Find the value of ArH for the reaction: N₂O₄(g) + 3CO(g)→N₂O(g) + 3CO₂(g)
Answer: \(\begin{aligned}
& \Delta_r H=\Sigma \Delta_f H \text { (Products) }-\Sigma \Delta_f H \text { (Reactants) } \\
&=3 \Delta_f H\left[\mathrm{CO}_2(g)\right]+\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}(g)\right] \\
&-\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}_4(g)\right]+3 \Delta_f H[\mathrm{CO}(g)] \\
&= {[3(-393)+1(81)-1(9.7)-3(-110)]=-777.77 \mathrm{~kJ} }
\end{aligned}\)

Question 13. N₂(g) + 3H₂(g)Δ₂NH₃(g) ; ΔrH°=-92.4 kj.mol-1. What is the standard enthalpy of the formation of NH₃?
Answer: \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) ; \Delta_r H^0=-92.4 \mathrm{~kJ}\)

⇒ \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightarrow \mathrm{NH}_3(g) ; \Delta_r H^0=-46.2 \mathrm{~kJ}\)

Equation [1] represents the formation of NH3(g) from the constituent elements. So, the standard enthalpy change for the reaction represented by equation [1] = the standard enthalpy of formation for NH3(g) = -46.2 kj.mol-1.

Question 14. Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:

⇒ \(\begin{aligned}
\mathrm{CH}_3 \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta_r \mathrm{H}^0=-726 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta_c H^0=-393 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta_f H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
Answer: To get the formation reaction for CH3OH(l), we combine three thermochemical equations provided in the given way:

\(\begin{array}{r}
\left.\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g})+2\left[\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\right]-\mathrm{CH}_3 \mathrm{OH}(l)-\frac{3}{2} \mathrm{O}_2(\mathrm{~g})\right\rceil \\
\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{H}_2 \mathrm{O}(l) ;
\end{array}\)

AH0 =[- 393 + 2(—286)- (-726)]kJ.mol-1 or, C(s) + 2H₂(g) + O₂(g)→CH₃OH(Z); AH0 = -239 kj-mol-1 Eq. [1] represents formation reaction of CH3OH(Z). So, standard enthalpy offormation is -239 kl.mol-1.

Chemical Thermodynamics Questions

Question 15. Calculate the enthalpy change for the process CCl₄(g)→C(g) + 4Cl(g) & calculate the bond enthalpy of C—Cl in CCl₄(g). Given: AvapH°(CCl4) = 30.5 kj-mol1, Af H°(CCl4) = -135.5 kl.mol-1, AaH°(C)=715.0 kj. mol-1, AaH°(Cl₂) =242 kj-mol-1 where AaH° is enthalpy of atomisation. m For the reaction 2Cl(g)→Cl₂(g) , what are the signs of AH and AS?
Answer: Based on the given information, we can write the following thermochemical equations-

⇒ \(\begin{aligned}
& \mathrm{CCl}_4(l) \rightarrow \mathrm{CCl}_4(g) ; \Delta H^0=30.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{C}(s)+2 \mathrm{Cl}_2(g) \rightarrow \mathrm{CCl}_4(l) ; \Delta H^0=-135.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

C(s)-C(g) ; ΔH0 = 715.0 kl. mol-1

Cl2(g)→2Cl(g) ; ΔH0 = 242 kl .mol-1

By, eq. [3]- eq. [2]- eq. [1] + 2 x eq. [4] we have,

⇒ \(\begin{gathered}
\mathrm{C}(s)-\mathrm{C}(s)-2 \mathrm{Cl}_2(g)-\mathrm{CCl}_4(l)+2 \mathrm{Cl}_2(g) \\
\mathrm{C}(g)-\mathrm{CCl}_4(l)-\mathrm{CCl}_4(g)+4 \mathrm{Cl}(g)
\end{gathered}\)

ΔH° = [715.0- (-135.5)- 30.5 + 2 x 242] kj

Or, CCl4(g)C(g) + 4Cl(g); ΔH° = 1304 k

In the CC14 molecule, there are four C—Cl bonds. To break all these bonds, 1304 kj of energy is required. Hence, C—Cl bond enthalpy in CCl4(g) = 1/4x 1304 = 326 kj.mol-1.

Question 16. For 311 isolated systems, ΔU = 0, what will be AS?
Answer: For a spontaneous process occurring in an isolated system AS is positive (i.e., ΔS > 0 ).

Question 17. For the reaction, 2A + B→C; AH = 400 kj. mol 1 & AS = 0.2 kj.K-1 .mol-1 at 298 K. At what temperature will the reaction become spontaneous considering AH, AS to be constant over the temperature range?
Answer: We know, AG = AH- TAS. For a spontaneous reaction at a given temperature and pressure A G < 0.

Given: AH = 400 kj.mol-1 and AS = 0.2 kj. K-1. mol-1 So, AG = (400- Tx 0.2) kj. mol-1 According to this relation, AG will be <0 when Tx 0.2 > 400 i.e., T> 2000K.

Question 18. For the reaction 2Cl(g)→Cl2(g), what are the signs of AH and AS?
Answer: The process involves the formation of a bond, which is always exothermic. Hence, AH < 0 for this process. The no. of gaseous particles decreases in the process. Consequently, the randomness of the system decreases. Hence, AS < 0 for this process.

Question 19. For the reaction, 2A(g) + B(g)→2D(g) ; ΔUº=-l0.5kJ and ASº = -44.11.K-1. Calculate AG° for the reaction and predict whether it may occur spontaneously.
Answer: Temperature has not been mentioned in the problem. Here, the calculation has been done by considering temperature as the normal temperature (298 K). For the reaction, An = 2- (2 + 1) = -1 . So, for this reaction, ΔH0 =ΔU° + ΔnRT =- 10.5-1 x 8.314 X 10-3 x 298 kj.

=-12.98kj

We know, ΔG° = ΔHº -TASº

∴ AGº =(- 12.98 + 298 x 44.1 x 10-3) kl =0.16kl

The positive value. of AG° indicates; that the cannot occur spontaneously.

Chemical Thermodynamics Questions

Question 20. The equilibrium constant for a miction is 10. Find the value of AG0 t R a 0.814 MC 1- mol-1, T = 300K.
Answer: Wekliow, AG° w -2.303/logK

Given K=10 and T = 300k

∴ A (1° a 11.303 X 11.3 1 4 x 300 log10).mol-1

=-5.74kJ.mol-1

Question 21. Comment on (lie thermodynamic stability of NO(g).
Given: \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}(g); \Delta_r H^0=90 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g); \Delta_r H^0=-74 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
Answer: For the first reaction, refer to the standard enthalpy of formation (ΔrH°) for NO because 1 mol of NO forms from its constituent elements. The positive value of Δr-Hº of a compound implies that the compound has enthalpy (or energy) than its constituent elements. Hence, the compound will be unstable. Therefore, the positive value of ΔfH° for the first reaction indicates that NO is unstable.

Question 22. Calculate the entropy change in the surroundings when 1.00 mol of ΔfHº(J) is formed under standard conditions. AH° = -286 kj.mol-1
Answer: For the given process

⇒ \(\Delta S_{\text {surr }}=-\frac{\Delta_f H^0}{T}=-\frac{-286 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{298 \mathrm{~K}}\)

= 959.73 J-K-1- mol-1

Question 22. Which type of system does not interact with its surroundings? Classify the following systems into open, closed, and isolated systems: a plant certain amount of a liquid enclosed in a container with rigid, impermeable, and adiabatic wall
Answer: 1. Isolated System 2. Open 2. Isolated

Chemical Thermodynamics Questions

Question 23. If the volume and density of 5g of pure iron sample are ‘ V’ and ‘ d.’ respectively then what will be the volume and density of 10 g of the same sample?
Answer: volume = 2 V; density = d

Question 24. The specific heat capacity of 10g of a sample of aluminum is x cal. g^-1 K^-1 . Is the value of the specific heat capacity of 5g of that sample\(\frac{x}{2}\) cal .g^-1 K^-1?
Answer: No, because it is a state property of the system.

Question 25. On which factors does the change in a state function depend?
Answer: Upon initial and final states of the system.

Question 26. The initial state of a system is ‘This system participates in the following process: A→B→C→A. What will be the change in the internal energy of the system in this process?
Answer: zero

Question 27. Which of the following is (are) not a state function?

1. Enthalpy
2. Heat capacity
3. Heat
4. Work done

Answer: Heat and work

Chemical Thermodynamics Questions

Question 28. At T K, what will be the value of (H- U) for 1 mol of ideal gas?
Answer: RT

Question 29. In an isothermal expansion of an ideal gas, both ΔH and ΔU are zero. What will be the values of ΔU and ΔH in the isothermal compression of an ideal gas?
Answer: ΔH = 0; ΔU = 0

Question 30. What will be the value of the change in internal energy of a process occurring in an isolated system?
Answer: Zero

Question 31. In a process, q cal heat is absorbed by a closed system. If work done by the system is w cal, then what will be the value of the change in internal energy (ΔU) in this process?
Answer: ΔU = q-w,

Question 32. For a process occurring in a closed system, ΔU = q. If only pressure-volume work is performed by the system then which type of process is this?
Answer: Isochoric

Question 33. The amount of heat released by a system at constant pressure is 20 kj. What will be the value of ΔH in this process?
Answer: ΔH = -20 kj

Question 34. For an ideal gas C V,m = J mol^-1 K^-1. What will be the value of Cp m?
Answer: 20.784 J.mol^-1-K^-1?

Question 35. A liquid in a closed adiabatic container is stirred. Among ΔU, w, and q, which one will be zero?
Answer: q = 0

Chemical Thermodynamics Questions

Question 36. If the temperature of I mol of an ideal gas is doubled then what will be the change in value of internal energy? Will it increase, decrease, or remain the same?
Answer: Will increase.

Question 36. 2H₂ + O₂-2H₂O , AH = -571.6 kj. Does this equation express the thermochemical equation for the formation of H₂O?
Answer: No, because the physical states of the reactants and products are not mentioned in the reaction.

Question 37. What temperature and pressure are usually considered as the standard state of a substance?
Answer: Pressure = 1 atm, any temperature may be considered.

Question 38. Which allotropic form of carbon is considered a source of carbon in the formation reactions of carbon compounds at 25°C and 1 atm?
Answer: C(graphite,s).

Question 39. Which of the following two reactions indicates the formation reaction of HI(g) at 25°C and 1 atm?

⇒ \(\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \rightarrow \mathrm{HI}(g)\)

⇒ \(\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g)\)

Answer: Reaction because the standard state of means than iodine is I2(s).

Question 40. Between O₂(g) and O₃(g), whose standard enthalpy of formation (ΔH0f) is taken as zero at 1 atm and 25°C?
Answer: O₂ (g) because at 25°C the standard state of oxygen is O₂(g)

Question 41. In which of the following reactions the standard heat of reaction is equal to the standard heat of formation of CaBr₂(s) at 25°C?

• Ca(s) + Br₂(l)-CaBr₂(s)
• Ca(s) + Br₂(g)-CaBr₂(s)

Answer: Reaction (1), because at 25X the standard state of bromine is Br2(l).

Chemical Thermodynamics Questions

Question 42. Heat is required to vaporize 1 g of water at 100°C and 1 atm is 2.26kj. What is the enthalpy of vaporization of water at this temperature and pressure?
Answer: 40.68 kj

Question 43. 1 mol of H₂O(Z) is formed when 1 mol of H+ ions and 1 mol of OH- ions react together in aqueous solution. Does this reaction represent the formation reaction of water?
Answer: No, because H₂O(f) is not formed from its constituent elements.

Question 44. Consider the reactions

• A₂(s) + B₂(g)-+A₂(g) + B₂(g); AH = -x kj
• A₂(g) + B₂(g)-+2AB(g) ; AH = -y kj .

What is the value of the change in enthalpy for the following reaction A₂(s) + B₂(g)-+2AB(g)?
Answer: — (X + y) kJ ;

Question 45. S(monoclinic, s)-S(rhombic, s). What is the enthalpy change of this process called?
Answer: Heat of transformation

Question 46. \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) ;\) \(\Delta H^0=-228.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text { if } \Delta H_f^0\left[\mathrm{H}^{+}\right]=0\) then what will be the value of AH0Of[OH-] ?
Answer: -228.5 kj mol^-1

Question 47. The bond dissociation energies of three A-B bonds in ΔB3(g) molecule are x,y, and zkj-mol^-1 respectively. What is the bond energy of the A- B bond?
Answer: l/3(x + y + z)kJ-mol^-1

Question 48. Why is the heat of reaction of a reaction occurring in a bomb calorimeter equal to the change in internal energy of the reaction system?
Answer: The volume of the reaction system remains the same.

Question 49. In the case of diatomic gaseous I molecule 33. -7.43kJ -mol^-1 ;
Answer: In the case of diatomic gaseous molecules.

Chemical Thermodynamics Questions

Question 50. What will be the difference between ΔH and ΔU in the combustion reaction of C₁₀H₈(g) at 25°C?
Answer: -7.43kJ -mol^-1

Question 51. Write down the relation between ΔH and ΔU for the following reaction: CH₃COOH(aq) + NaOH(a<y)-CH₃COONa(ag) + H2O(l)
Answer: zero

Question 52. If the bond energy of the C- H bond is +416.18 kj .mol^-1, then what will be the enthalpy of the formation of the C-H bond?
Answer: 416.18kJ. mol^-1

Question 53. What is the most important feature of Hess’s law?
Answer: It is used to determine the heat of the reaction;

Question 54. The standard bond dissociation energy of A2(g), B2(g) and AB(g) molecules are x,y, and zkj-mol-1 respectively. What will be the standard enthalpy of the formation of AB(g)?
Answer: [z- 1 /2(x + y)]kj . mol^-1

Question 55. At temperature T K the difference between ΔH and ΔU for the following reaction is \(+\frac{1}{2} R T: \mathrm{A}+\frac{1}{2} \mathrm{~B}_2(\mathrm{~g}) \rightarrow \mathrm{AB}(\mathrm{g}).\) What is the physical state of (solid or gas)?
Answer: Soild [A(s)]

Question 56. Give an example of a spontaneous process in which the change in enthalpy of the system is positive.
Answer: Melting of ice above 0°C,

Chemical Thermodynamics Questions

Question 57. What will be the signs of enthalpy change and entropy change for a process to be spontaneous at all temperatures?
Answer: ΔH < 0 , ΔS > 0

Question 58. What will be the signs of AG for melting ice at 267K and 276K temperature and atm pressure?
Answer: ΔG> 0 ΔG < 0

Question 59. Water and ice remain in equilibrium at 0°C and 1 atm pressure. What will be the value of AG and the sign of the S system at this equilibrium?
Answer: ΔG = 0 , ΔSsys>0,

Question 60. Which one of the following relation or expression is true for a spontaneous process of ΔG = 0, ΔH=TΔS, ΔG > 0, ΔG < 0?
Answer: AG<0

Question 61. What will be the sign of entropy change for the process I2(g)-+I2(s)?
Answer: ΔS = -ve

Question 62. What will be the change in entropy of the surroundings in a spontaneous process occurring in an isolated system?
Answer: Zero

Question 63. Is the entropy of the universe constant?
Answer: No, increases continuously

Chemical Thermodynamics Questions

Question 64. Give an example of a process in which the change in enthalpy of the system is negative.
Answer: Solidification of liquid

Question 65. what sign (or-) of entropy change in an endothermic relation makes the reaction Non-spontaneous at any temperature?
Answer: ΔS < 0,

Question 66. In which case the Asys will lie maximum between die following two processes ice – water I₂(s) – l₂(g)
Answer: I₂(s) – l₂(g)

Question 67. Predict the sign of AS0 for the given reaction. \(2 \mathrm{H}_2 \mathrm{~S}(g)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(g)+2 \mathrm{SO}_2(g)\)
Answer: – ve

Question 68. Does die Gibbs free energy of a substance decrease or increase if the amount of the substance is increased?
Answer: Will increase

Question 69. In a process, the value of the change in entropy of the die system and its surroundings are x and -yj. IC^-1 . If x > y, then will the process be spontaneous?
Answer: No

Question 70. The change in entropy of the system in a process A -4 B -4 C is 25 J K^-1. If the change in entropy of the system in step B → C is 15 J K^-1, then what will be the change in entropy of the system in step B→ A?
Answer: -10J.K^-1

Thermodynamics MCQ’s

Question 1. Which of the following statements is true—

1. Entropy increases when water vaporizes
2. Randomness decreases in the fusion of ice
3. Randomness increases in the condensation of water vapor
4. Randomness remains unchanged during the vaporization of water

Chemical Thermodynamics Questions

Answer: 1. Entropy increases when water vaporizes

Vaporization of water (water → water vapor) involves an increase in the entropy of the system because the molecular randomness in water vapor is greater than that in water.

Question 2. Identify the correct statement in a chemical reaction—

1. The entropy always increases
2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction
3. The enthalpy always decreases
4. Both tire enthalpy and tire entropy remain constant

Answer: 2. The change in entropy along with a suitable change in enthalpy decides the tire rate of reaction.

For a reaction occurring at a constant temperature and pressure, ΔG<0, ie., ΔH-TΔS<0 the change in entropy (Δs) and (flng with the change in enthalpy (ΔH) determines the spontaneity of a reaction.

Question 3. The condition for the spontaneity of a  process is—

1. Lowering of entropy at constant temperature & pressure
2. Lowering of Gibbs free energy of tire system at constant temperature and pressure
3. Increase in entropy of tire system at constant temperature and pressure
4. Increase in Gibbs free energy of the universe at constant temperature and pressure

Answer: 2. Lowering of Gibbs free energy of tire system at constant temperature and pressure

The condition of spontaneity for a reaction to occur at constant t and p is ag < 0.

Question 4. P-v work done by an ideal gaseous system at constant volume is the internal energy of the system)

1. \(-\frac{\delta p}{p}\)
2. Zero
3. -Vδp
4. -Δe

Answer: 2.  -Vδp

As the system’s volume remains constant in the process, the system cannot do any external work.

Question 5. Mixing of two different ideal gases under an isothermal reversible condition wall leads to—

1. Increase in Gibbs free energy of the system
2. No change in the entropy of the system
3. Increase in entropy of the system
4. Increase in enthalpy of the system

Answer: 3. Increase in entropy of the system

The molecular randomness in a gas mixture is larger than that in the individual gases because a gas mixture contains a larger number of molecules than that in the individual gases. Consequently, the mixing of two gases will lead to an increase in the entropy of the system.

Question 6. For an isothermal expansion of an ideal gas, the correct of the thermodynamic parameters will be—

1. ΔU =0,Q = 0,ω≠0 and ΔH≠ 0
2. ΔU ≠0,Q ≠ 0,ω≠0 and ΔH= 0
3. ΔU =0,Q ≠ 0,ω=0 and ΔH≠ 0
4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

Answer: 4. ΔU =0,Q≠ 0,ω≠0 and ΔH= 0

For an ideal gas undergoing an isothermal process, ‘ah = 0. When a gas undergoes expansion it absorbs heat from its surroundings and performs external work. So, in such a process, q≠0 and w≠0.

We know, Δh = Δu+ Δ(pv). For an ideal gas, pv = nrΔt.

∴ Δh = Δu+Δ(nrt) = Δu+nrΔt

In an isothermal process of an ideal gas, aΔu = 0, at- 0, and hence ΔH = 0.

Chemical Thermodynamics Questions

Question 7. The change in entropy (ds) is defined as—

1. \(d s=\frac{\delta q}{t}\)
2. \(d s=\frac{d h}{t}\)
3. \(d s=\frac{\delta q_{r e v}}{t}\)
4. \(d s=\frac{(d h-d g)}{t}\)

Answer: 3. \(d s=\frac{\delta q_{r e v}}{t}\)

If a system undergoing a reversible process at tk absorbs an amount of heat, 6qrev, then the change in entropy of the system in the process, \(d s=\frac{\delta q_{r e v}}{t}.\).

Question 8. Δh for cooling 2 mol ideal monoatomic gas from 225 °c to 125°c at constant pressure will be?

1. 250R
2. -500R
3. 500R
4. -250R

Answer: 2. ΔH=nCp(T2-T1)

Question 9. For a spontaneous process, correct statement(s) is (are)—

1. \(\left(\delta g_{s y s}\right)_{t, p}>0\)
2. \(\delta s_{\text {sys }}+\delta s_{\text {surr }}>0\)
3. \(\left(\delta g_{s y s}\right)_{t, p}<0\)
4. \(\left(\delta u_{s y s}\right)_{t, v}>0\)

Answer: 3. \(\left(\delta g_{s y s}\right)_{t, p}<0\)

For a spontaneous process at constant pressure and temperature, \(\delta s_{\text {univ }}>0 \text { or, } \delta s_{\text {sys }}+\delta s_{\text {surr }}>0\) now, \(\delta g=-t \delta s_{u n I v}\). For a spontaneous process \(\delta s_{u n I v}>0 \text {, and hence } \delta g<0 \text {. }\).

Question 10. Given: C+O₂→CO₂; ΔH0=-xKJ; 2CO+O₂→2CO₂; ΔH0=-yKJ The heat of formation of carbon monoxide will be—

1. Y+2x
2. 2X-y

Answer: C+O₂→CO₂; ΔH0=-xKJ; 2CO+O₂→2CO₂;ΔH0=-yK

Subtracting equation [2] from the equation obtained by multiplying equation [1] by 2, we have,

2CO+O₂→2CO₂;ΔH0=-yK

∴ Heat of formation of co \(=\left(\frac{y-2 x}{2}\right) \mathrm{kj} \cdot \mathrm{mol}^{-1}\)

Question 11. The enthalpy of vaporization of a certain liquid at its boiling point of 35°c is 24.64 kl-mol-1. The value of change in entropy for the process is—

1. 704 J.k-1 mol-1
2. 80J.k-1. Mol-1
3. 24.64j.k-1. Mol-1
4. 7.04 j.k-1. Mol-1

Answer: 2. 80J.k-1. Mol-1

⇒ \(\begin{aligned}
\delta s_{\text {vap }}=\frac{\delta h_{\text {vap }}}{t_b} & =\frac{24.64 \times 10^3 \mathrm{~j} \cdot \mathrm{mol}^{-1}}{(273+35) \mathrm{k}} \\
& =80 \mathrm{~j} \cdot \mathrm{k}^{-1} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Chemical Thermodynamics Questions

Question 12. ΔH and ΔS of a certain reaction are -400 kj. Mol-1 and -20kj.mol-1.k_1 respectively. The temperature below which the reaction is spontaneous is—

1. 100K
2. 20°C
3. 20K
4. 120°C

Answer: 3. 20K

For a spontaneous reaction, Δg = Δh- tΔs <0

⇒ \(t \delta s>\delta h \text { or, } t>\frac{\delta h}{\delta s} \text {, or, } t>\frac{400}{20} \mathrm{~k} \text {, or, } t>20 \mathrm{~k}\)

Question 13. For the reaction x2y4(/)→2xy(g) at 300 k the values of a u and as are 2kcal and 20 cal.k-1 respectively. The valuve of Δg for the reaction is-

1. -3400 Cal
2. 3400 Cal
3. -2800 Cal
4. 2000 Cal

Answer: 3.  Δh = Δu+angrt or, Δh = 2000 + 2 x 2 x 300

An = 2-0 = 2, r = 2 cal -k-1- mol-1]

Ah = 3200 cal

Ag = ah- tas = 3200- (300 x 20) =-2800 cal

Question 14. For the reaction 2sO₂(g) + O(g)⇌ 2sO₃(g) at 300 k, the value of ac0 is -690.9/1, the equilibrium constant value for the reaction ax that temperature is (r is gas constant)—

1. 10 Atm-1
2. 10 Atm
3. 10
4. 1

Answer: 3. 10

2sO₂(g) + O₂(g)⇌ 2sO₃(g)

Now, ΔG = -RTlnk

Or, -690.9r = -RTlnk

∴K = 101

Question 15. The condition for the reaction to occur spontaneously is

1. Δh must be negative
2. Δs must be negative
3. Δh- tδs) must be negative
4. Δh + tδs) must be negative

Answer: 3. Δh- tδs) must be negative

At constant temperature and pressure, for a spontaneous process of the reaction a g < 0. According to Gibbs’s equation ag = ah-tas. Therefore the condition spontaneity is, ah- tas < 0.

Question 16. During a reversible adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio \(\frac{c_p}{c_v}\) for the gas is-

1. \(\frac{3}{2}\)
2. \(\frac{7}{2}\)
3. \(\frac{5}{3}\)
4. \(\frac{9}{7}\)

Chemical Thermodynamics Questions

Answer: 1. \(\frac{3}{2}\)

For adiabatic reversible process, \(t p^{\frac{(1-\gamma)}{\gamma}}=\text { constant }\)

Or, \(p t^{\frac{\gamma}{(1-\gamma)}}=\text { constant }\)

Again as mentioned, poct3

Or, pt-2 = constant

Comparing equations (1) and (2), it may be written as

⇒ \(\frac{\gamma}{1-\gamma}=-3\)

∴ \(\gamma=\frac{3}{2}\)

Question 17. The heat of neutralization of a strong base and a strong add is 13.7 kcal the heat released when 0.6 mol hc1 solution is added to 0.25 mol of NaOH is-

1. 3.425kcal
2. 8.22kcal
3. 11.645kcal
4. 13.7kcal

Answer: 1. 3.425kcal

⇒ \(\begin{aligned}
& \mathrm{hcl}(a q)+\mathrm{naoh}(a q) \rightarrow \mathrm{nacl}(a q)+\mathrm{h}_2 \mathrm{o}(l) ; \\
& \delta h^0=-13.7 \mathrm{kcal}
\end{aligned}\)

⇒ \(\begin{array}{r}
0.25 \mathrm{~mol} \mathrm{hcl}(a q)+0.25 \mathrm{~mol} \mathrm{naoh}(a q) \\
0.25 \mathrm{~mol} \mathrm{nacl}(a q)+0.25 \mathrm{~mol} \mathrm{h}_2 \mathrm{o}(l) ; \\
\delta h^0=-13.7 \times 0.25=-3.425 \mathrm{kcal}
\end{array}\)

∴ Amount of heat released = 3.425 kcal

Question 18. The entropy change involved in the isothermal reversible expansion of 2 mol of an ideal gas from a volume of 10 cm3 to a volume of 100 cm3 at 27°cis —

1. 38.3j.mol-1.k-1
2. 35.8j.mol-1.k-1
3. 32.3j.mol-1. K-1
4. 42.3j.mol-1.k-1

Answer: 1. 38.3j.mol-1.k-1

For isothermal reversible expansion,

⇒ \(\delta s=n r \ln \frac{v_2}{v_1}=2 \times 8.314 \ln \frac{100}{10}=38.3 \mathrm{~j} \cdot \mathrm{mol}^{-1} \cdot k^{-1}\)

Chemical Thermodynamics Questions

Question 19. Among the following expressions which one is incorrect-

1. \(w_{\text {rev, iso }}=-n r t \ln \frac{v_f}{v_i}\)
2. \(\ln k=-\frac{\delta h^0-t \delta s^0}{r t}\)
3. \(k=e^{-\delta g^0 / r t}\)
4. \(\frac{\delta g_{\text {sys }}}{\delta s_{\text {total }}}=-t\)

Answer: 2. \(\ln k=-\frac{\delta h^0-t \delta s^0}{r t}\)

ΔG° = -RTInk and ΔG° = ΔH0- TΔS0

\(\Delta H^0-T \Delta S^0=-R T \ln K \text { and } \ln K=-\left(\frac{\Delta H^0-T \Delta S^0}{R T}\right)\)

Question 20.  A cylinder filled with 0.04mol of ideal gas expands reversibly from 50ml to 375ml at a constant temperature of 37.0°c. As it does so, it absorbs 208 j heat, q and w for the process will be (a = 8.314 j.mol-1.k-1 )

1. q = +208 J, w = +208 J
2. q = +208 J, w = -208 J
3. q = -208 J , w = -208 J
4. q = -208 J , w = +208 J

Answer: 2. Q = +208 J, w = -208 J

The process is isothermal and the system is an ideal gas. So, in this process, a u = 0. Given that q = +208J.

∴ Au = q + w or, 0 = 208 + w

∴ W = -208J .

Question 21. For complete combustion of ethanol, \(\mathrm{c}_2 \mathrm{h}_5 \mathrm{oh}(l)+3 \mathrm{o}_2(g) \rightarrow 2 \mathrm{co}_2(g)+3 \mathrm{h}_2 \mathrm{o}(l)\)

The amount of host produced ao measured in a bomb calorimeter, u 1364,47kJ mol 1 at 25°c, assuming ideality the enthalpy of combustion, ΔHc(  – mol 1 j for the reaction will be (R= 8.3m jk-1. Mol-1)

1. -1350.50
2. -1366.95
3. -1361.95
4. -1460.50

Answer: 2. -1366.95

In a bomb calorimeter, a reaction occurs under a condition of constant volume. Hence, qv = alt. For the given reaction, an = 2-3 = -1 we know, ah = a u + art

∴ Ah = [- 1364.47-lx 8.314 x 10~3 x 298]kj . Mol-1

=-1366.95 kj- mol-1

Question 22. The following reaction is performed at 298k. 2No(g) + O₂(g/ v= 2nO₂(g) the standard free energy of formation is 86.6 kj/mol at 298k, what is the standard free energy of formation of NO₂(g) at 298k (kp – 1.6 x 10 l2) –

1. \(8660-\frac{\ln \left(1.6 \times 10^{12}\right)}{r(298)}\)
2. 0.5[2 x 86600 -R(298) in(1.6 x 1012)
3. R(298)ln(1.6 x 1012)- 86600
4. 86600 + R(298) In (1.6 x 1012)

Chemical Thermodynamics Questions

Answer: 2. 0.5[2 x 86600 -R(298) in(1.6 x 1012)

Given: \(T=298 \mathrm{~K}, \Delta G_f^0(\mathrm{NO})=86.6 \mathrm{~kJ} / \mathrm{mol},\)

⇒ \(\begin{aligned}
& \Delta G_f^0\left(\mathrm{NO}_2\right)=?, K_p=1.6 \times 10^{12} \\
& 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_0(g) \rightleftharpoons 2 \mathrm{NO}_0(g)
\end{aligned}\)

∴ \(\Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \Delta G_f^0(\mathrm{NO})-\Delta G_f^0\left(\mathrm{O}_2\right)\)

\(\text { or, } \Delta G_r^0=2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600\)

⇒ \(Again,\Delta G_r^0=-R T \ln K_p
or, 2 \Delta G_f^0\left(\mathrm{NO}_2\right)-2 \times 86600=-R(298) \ln \left(1.6 \times 10^{12}\right)\)

⇒ \(\text { or, } \delta g_f^0\left(\mathrm{no}_2\right)=\frac{2 \times 86600-r(298) \ln \left(1.6 \times 10^{12}\right)}{2}\)

= 0.5[2 x 86600- (298)Ln(1.6 x 1012)

Question 23. The standard Gibbs energy change at 300k for the reaction 2az±b + c is 2494.2 j. At a given time, the composition of the reaction mixture is [a] \([a]=\frac{1}{2},[b]=2 \text { and }[c]=\frac{1}{2}\) reaction proceeds in the [a = 8.314)/k/mol, c=2.718] —

1. Forward direction because q<kc
2. Reverse direction because q<kc
3. Forward direction because q > kc
4. Reverse direction because q > kc

Answer: 4. Reverse direction because q > kc

⇒ \(2 A \rightleftharpoons B+C\)

Given: = 300k , ΔG0 =2494.2J

R = 8.314 I .K-1.mol-1

Now, AG° = -2.303RT logKc

Question 24. The heats of carbon and carbon monoxide combustion are -393.5 and -283.5 kj.mol-1 respectively. The heat of formation (in kj) of carbon monoxide per mole is

1. 110.5
2. 676.5
3. -676.5
4. -110.5

Answer: c(s) + O₂(g) →CO₂(g) , Δh0 = -393.5 kj.mol-1

⇒ \(\mathrm{co}(\mathrm{g})+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}_2(\mathrm{~g}), \delta h^0=-283.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Subtracting equation (2) from equation (1) we get,

⇒ \(\mathrm{c}(\mathrm{s})+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}(\mathrm{g}),\)

Ah0 =[- 393.5- (-283.5)] kj.mol-1

=-110.0 kj-mol-1

Therefore, the heat of the formation of co(g)

= -110.0 kj-mol-1

Question 25. Given, c(graphite) + O₂(g) -4 CO₂(g),aÿ = -393.5 kj .mol-1

⇒ \(\begin{aligned}
& \mathrm{h}_2(\mathrm{~g})+\frac{1}{2} \mathrm{o}_2(g) \rightarrow \mathrm{h}_2 \mathrm{o}(l), \delta_r h^0=-285.8 \mathrm{~kj} \cdot \mathrm{mol}^{-1} \\
& \mathrm{co}_2(\mathrm{~g})+2 \mathrm{h}_2 \mathrm{o}(l) \rightarrow \mathrm{ch}_4(g)+2 \mathrm{o}_2(\mathrm{~g})
\end{aligned}\)

Based on the above thermochemical equations, the value of arh° at 298 k for the reaction, c(graphite) + 2h2(g) → Ch4(g) , will be—

1. -748 Kj.mol-1
2. -144.0kj.mol-1
3. +74.8 kj.mol-1
4. +144.0kj.mol-1

Chemical Thermodynamics Questions

Answer:  C(graphite) + O₂(g)→CO₂(g),

= -393.5 kj. Mol-1

⇒ \(\mathrm{h}_2(\mathrm{~g})+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{h}_2 \mathrm{o}(l), \delta_r h^0=-285.8 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

⇒ \(\begin{aligned}
& \mathrm{co}_2(g)+2 \mathrm{h}_2 \mathrm{o}(l) \rightarrow \mathrm{ch}_4(g)+2 \mathrm{o}_2(\mathrm{~g}) \\
& \delta_r h^0=+890.3 \mathrm{~kj} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 26. The combustion of benzene (z) gives CO₂(g) and h₂O(z). Given that the heat of combustion of benzene at constant volume is -3263.9 kj -mol-1 at 25°c, the heat of combustion (in kj .mol-1 ) of benzene at constant pressure will be [r = 8.314 j.k-1.mol-1 )—

1. -4152.6
2. -452.46
3. 3260
4. -3267.6

Answer: 4. -3267.6

⇒ \(\mathrm{c}_6 \mathrm{h}_6(l)+\frac{15}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow 6 \mathrm{co}_2(\mathrm{~g})+3 \mathrm{h}_2 \mathrm{o}(l)\)

⇒ \(\delta n=6-\frac{15}{2}=-\frac{3}{2}\)

Ah = au+anrt

⇒ \(\begin{aligned}
\delta h & =\left[-3263.9-\frac{3}{2} \times 8.314 \times 10^{-3} \times 298\right] \mathrm{kj} \cdot \mathrm{mol}^{-1} \\
& =-3267.6 \mathrm{~kj} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 27. If \(\frac{1}{2} a \rightarrow b, \quad \delta h=+150 \mathrm{~kj} \cdot \mathrm{mol}^{-1} ; \quad 3 b \rightarrow 2 c+d\)   ah=-125 kj mol-1 ; e + a-*2d , ah= + 350 k. Mol-1 then ah of the reaction b + d—>£ + 2c will be

1. 525Kj.mol-1
2. -175Kjmol-1
3. -325Kj.mol-1
4. 352Kj.mol-1

Answer: 2. -175Kjmol-1

⇒ \(\frac{1}{2} a \rightarrow b ; \delta h=+150 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

3B→2c + d; Δh = -125 kj. Mol-1

E + a→2d; Δh = +350 kj . Mol-1

Subtracting equation [3] from (2 x equation [1] + equation [2]), we have—

B + d→e + 2c; ah = 2 x 150- 125- 350 = -175 kj.

Chemical Thermodynamics Questions

Question 28. Which is the correct option for free expansion of an ideal gas under adiabatic conditions—

1. q=0, ΔT≠0, W=0
2. q≠0, ΔT≠0, W=0
3. q=0, ΔT≠0, W=0
4. q=0, ΔT<0, W≠0

Answer: 3. q=0, ΔT≠0, W=0

For an adiabatic process, q = 0, and for free expansion of a gas, w = 0.

∴ Δu=q + w = 0 + 0 = 0. For an ideal gas undergoing an isothermal process, ah = 0. So, the given process is isothermal and hence at = 0.

Question 29. The enthalpy change for the reaction, 4h(g)→2h2(g) is -869.5 kl. The dissociation energy of the fi — h bond is

1. -434.8 kj
2. -869.6kj/mol
3. +434.8 kj
4. +217.4kj

Answer: 3. +434.8 kj

We know, bond formation enthalpy =(-)x bond dissociation enthalpy. Now, bond formation enthalpy for 2 mol h —h bonds = -869 kj.

Bond dissociation enthalpy for h— h bond \(=\frac{1}{2} \times 869=434.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 30. If the enthalpy change for the transition of liquid water to steam is 30 kj.mol-1 at 27°c, the entropy change in ] mol-1 k_1 for the process would be

1. 10
2. 1.0
3. 0.1
4. 100

Answer: 4. 100

Question 31. In which of the given reactions, standard reaction entropy change (as0) is positive and standard gibbs energy change (ag°) decreases sharply with increasing temperature—

1. \(\mathrm{c} \text { (graphite) }+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}(\mathrm{g})\)
2. \(\mathrm{co}(\mathrm{g})+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}_2(\mathrm{~g})\)
3. \(\mathrm{mg}(\mathrm{s})+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{mgo}(\mathrm{s})\)
4. \(\mathrm{c}(\mathrm{s})+\mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}_2(\mathrm{~g})\)

Answer: 1. \(\mathrm{c} \text { (graphite) }+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}(\mathrm{g})\)

In the reaction, c(graphite) \(+\frac{1}{2} \mathrm{o}_2(\mathrm{~g}) \rightarrow \mathrm{co}(\mathrm{g})\) number of gas molecules increases. As a result, the entropy of the system increases (as0 > 0).

We know, ag = a/7- tas for the given reaction a77° < 0, as it is a combustion reaction. Since at/0 < 0 and as0 > 0, according to the reaction [1], ag° decreases with rise in temperature.

Chemical Thermodynamics Questions

Question 32. For the reaction, x2O₄(l)→2xO₂(g), ah =2.1 kcal, as = 20 cal.k-1 at 300 k. Hence, ag is—

1. 2.7 kcal
2. -2.7 kcal
3. 9.3kcal
4. -9.3kcal

Answer: 2. -2.7 kcal

ΔH= ΔU +ΔnRT

For the given reaction, an = 2.

∴ ΔH = 2.1 + 2 x 1.987 x 10-3 x 300 =3.2922 kcal and

Δg = ΔH- TΔS= 3.2922 -300 x 20 x 10-3 =-2.7 kcal

Question 33. The heat of combustion of carbon to CO₂ is -393.5 kj/mol. The heat released upon the formation of 35.2g of CO₂ from carbon and oxygen gas is

1. -315 Kj
2. +315 Kj
3. -630 Kj
4. -3.15kj

Answer: c(s) + O₂(g)→CO₂(g); afh = -393.5 kj.mol-1

The heat released on formation of 44g CO₂ =-395.5 kj-mol-1

The heat released by the formation of 35.2g of CO₂

⇒ \(=-\frac{393.5 \mathrm{~kj} \cdot \mathrm{mol}^{-1}}{44 \mathrm{~g}} \times 35.2 \mathrm{~g}=-315 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

Question 34. For the expansion of n mol of an ideal gas, the change in entropy of reversible isothermal process is \(\delta s=n r \ln \frac{v_f}{v_l}\)

At constant temeratrue, for n mol ofideal gas \(\frac{v_f}{v_i}=\frac{p_i}{p_f}\)

Therefore \(\delta s=n r \ln \left(\frac{p_i}{p_j}\right)\)

Question 35. Consider the following liquid-vapor equilibrium. Liquid vapour which of the following relations is correct

1. \(\frac{d \ln p}{d \mathrm{~t}^2}=-\frac{\delta h_v}{t^2}\)
2. \(\frac{d \ln p}{d \mathrm{~t}}=\frac{\delta h_v}{r t^2}\)
3. \(\frac{d \ln g}{d t^2}=\frac{\delta h_v}{r t^2}\)
4. \(\frac{d \ln p}{d t}=-\frac{\delta h_\nu}{r t}\)

Answer: 1. At constant temperature and pressure, the reaction is to be spontaneous if ag < 0. According to the gibbs free energy ag = a/7- tas, ag will be negative at any temperature if a/7 < 0 and as > 0

Question 37. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 l to a final volume of 4.50 l. The change in internal energy a u of the gas in joules will be—

1. -500 Jk-1
2. -505 J
3. +505 J
4. 1136.25j

Answer: 2. -505 J

Work done of irreversible process, w =-pgxav =-2.5(4.5-2.5) =-5 l- atm =-(5 x 101.3)j =-506.5j since, it is an insulated system, q = 0. From the first law ofthermodynamics a u = q+w or, a 7/ = 0- 506.5j = -506.5j.

Chemical Thermodynamics Questions

Question 38. For a given reaction, ah = 35.5 kj. mol 1 and as = 83.6 j.mol-1. The reaction is spontaneous at (assume that ah and as do not vary with temperature)-

1. T > 425 k
2. All temperatures
3. T> 298 k
4. T< 425 k

Answer: 1. T > 425 k

Ag = ah-tas = 35.5 x 103- tx 83.6 reaction is to be spontaneous if a g < 0. Thus, 35.5 x 103- tx 83.6 < 0 therefore, t x 83.6 > 35.5 x 103 or, t> 424.64 k

Question 39. The bond dissociation energies of x2, y2, and xy are in the ratio of 1 : 0.5: 1 . Ah for the formation of xy is -200 kj.mol-1. The bond dissociation energy of x2 will be

1. 200Kj.mo1-1
2. 100Kj.mol-1
3. 800 Kj.mol-1
4. 400Kj.mol-1

Answer: 3. 800 Kj.mol-1

⇒ \(\frac{1}{2} \mathrm{x}_2+\frac{1}{2} \mathrm{y}_2 \rightarrow \mathrm{xy}\)

⇒ \(\begin{aligned}
& \delta \mathrm{h}_{\text {reaction }}=\sum(\mathrm{be})_{\text {reactant }}-\sum(\mathrm{be})_{\text {product }} \\
& {[\mathrm{be}=\text { bond energy] }}
\end{aligned}\)

If the bond energy of x2 is a kj.mol-1 then the bond energy of y2 and xy are 0.5a and a kj.mol-1 respectively.

⇒ \(-200=\frac{a}{2}+\frac{0.5}{2} a-a=-0.25 a \quad \text { or, } a=800\) ”

Therefore, bond energy of x2 = 800kj.mol-1

Question 40. Which of the following is an intensive property

1. Enthalpy
2. Entropy
3. Specific heat
4. Volume

Answer: 3. Specific heat

Intensive property: specific heat; extensive property-: enthalpy, entropy, volume.

Question 41. Which of the tires following is not a thermodynamic function—

1. Internal energy
2. Work done
3. Enthalpy
4. Entropy

Answer: 2. Workdone

Thermodynamic functions are internal energy, enthalpy, entropy, pressure, volume, temperature, free energy, and number of moles.

Chemical Thermodynamics Questions

Question 42. For the adiabatic process, which is correct-

1. At = 0
2. As = 0
3. Q = 0
4. Qp = 0

Answer: 3. Q = 0

For the adiabatic process, no exchange of heat takes place between the system and surroundings. i.e., Q = 0.

Question 43. The enthalpy of formation of CO(g), CO₂(g), N₂0(g) and N₂O₄(g) is -110, -393, +811′ and lokj/mol respectively for the reaction N₂O₄(g) + 3C0(g)→N₂O(g) + 3CO₂(g) ahr (kj/mol) is –

1. -212
2. +212
3. +48
4. -48

Answer: 4. -48

N₂O₄(g) + 3CO(g)→ N₂O(g) + 3CO₂(g)

ΔH_reaction =∑ Heat of formation of products heat of formation of reactants

-∑ Heat of formation of reactants

Question 44. The bond dissociation energy of ch4 is 360 kj/mol and c2h6 is 620 kj/mol. Then bond dissociation energy of the c- c bond is—

1. 170 Kj/mol
2. 50Kj/mol
3. 80Kj/mol
4. 220Kj/mol

Answer: 3. 80Kj/mol

Dissociation energy of-methane = 360 kj.mol-1

∴ Bond energy of c—h bond \(=\frac{360}{4}=90 \mathrm{~kj} \cdot \mathrm{mol}^{-1}\)

The bond energy of ethane,

• 1 X B.E. (C-C) + 6 X B.E. (C-H) = 620 KJ.mol-1
• Or, B.E.(C—C) + 6 X 90 = 620
• Or, B.E. (C—C) + 540 = 620
• Or, B.E. (C—C) = 620-540
• Or, B.E. (C—C) = 80 KJ-mol-1
• Bond dissociation of c —C bond = 80 kj. Mol-1

Chemical Thermodynamics Questions

Question 45. Which thermodynamic parameter is not a state function-

1. Q at constant pressure
2. Q at constant volume
3. W at adiabatic
4. W at isothermal

Answer: 4. W at isothermal

H and u are state functions but w and q are not state functions. From the equation, Δh = Δu+Δpv at constant pressure, Δh = Δu+pΔv at constant volume, Δh = Δu+ vΔp at constant pressure, Δp = 0, Δh = qp so, it is a state function.

At constant volume, Δv = 0, Δu = qΔ so, it is a state function. Work done in any adiabatic process is a state function. Δu = q- w (Δq – 0) Δu = -w work done in the isothermal process is not a state function. W = -q (since Δt = 0, q=0)

Question 46. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Alt < 0 , w < 0
3. Al/> 0, w = 0
4. A17 > 0 , w> 0

Answer: 1. Au< 0, w = 0

For adiabatic conditions, PV^ϒ  = constant

⇒ \(p_1 v_1^\gamma=p_2 v_2^\gamma ; v_2=\frac{1}{2} v_1\)

⇒ \(p_2=p_1\left(\frac{v_1}{v_2}\right)^\gamma \text { [for diatomic gas, } \gamma=1.4 \text { ] }\)

⇒ \(p_2=p_1\left(\frac{v_1 \times 2}{v_1}\right)^{1.4} p_2=p_1(2)^{1.4}=(2)^{1.4} p\)

Question 47. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, a u and w correspond to—

1. Au< 0, w = 0
2. Au< 0, w <0
3. Au> 0, w = 0
4. Au> 0, w>0

Answer: 1. Au< 0, w = 0

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb.

If the process is rim in a sealed container then no expansion or compression is allowed, so w = 0 and au = q ΔU< 0, w = 0.

Chemical Thermodynamics Questions

Question 48. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25° c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g-1 °c-1, and the value of the heat of combustion of benzoic acid is—

1. 881.1kcal
2. 771.124kcal
3. 981.1kcal
4. 871.2kcal

Answer: 2. 771.124kcal

Given: weight of benzoic acid = 1.89 g; the temperature of bomb calorimeter =25°c=298k; the mass of water (m) = 18.94 kg = 18940 g; increase in temperature (at) = 0.632°c and specific heat of water (s) = 0.998 cal- gÿc-1 we know that heat gained by water or heat liberated by benzoic acid (q) = meat

= 18940 X 0.998 x 0.632 = 11946.14 cal

Since 1.89 g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122 g of acid \(=\frac{11946.14 \times 122}{1.89}\)

= 771126.5 cal =771.12 kcal

(Where 122 g is the molecular weight of benzoic acid)

Question 49. The heat is liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25°c and it increases the temperature of 18.94 kg of water by 0.632°c. The specific heat of water at 25°c is 0.998 cal.g_1 .°C-1, the value of the heat of combustion of benzoic acid is—

1. 881.1kcal
2. 771.124kcal
3. 981. kcal
4. 871.2kcal

Answer: 2. 771.124kcal

Chemical Thermodynamics Questions

Question 50. What is the entropy change in 2 mol n2, when its temperature is taken from 400 k to 800 k, at constant pressure

1. 30J/k
2. 60J/k
3. 40J/k
4. 20J/k

Answer: 3. 40J/k

⇒ \(\delta s=n c_p \ln \frac{t_2}{t_1}=2 \times \frac{7}{2} r \times \ln \frac{800}{400}=40 \mathrm{~j} \cdot \mathrm{k}^{-1}\)

Question 51. 1 Mole of an ideal gas expands isothermally reversible from 2 liters to 4 liters and 3 moles of the same gas expand from 2 liters to x liter and do the same work, what is ‘x’-

1. (8)^1/3
2. (4)^2/3
3. 2
4. 4

Answer: 2. (4)^2/3

⇒ \(w=n r t \ln \frac{v_2}{v_1} \text { or, } r t \ln \frac{4}{2}=3 r t \ln \frac{x}{2}\)

⇒ \(\text { or, } \ln 2=\ln \left(\frac{x}{2}\right)^3 \text { or, }\left(x^3=16\right) \text { or, } x=(16)^{\frac{1}{3}}=4^{\frac{2}{3}}\)

Question 52. Which of the following are extensive properties

1. Volume and enthalpy
2. Volume and temperature
3. Volume and specific heat
4. Pressure and temperature

Answer: 1. Extensive properties depend upon the quantity of the matter contained in the system, e.g., Volume and enthalpy, etc. Intensive properties depend only upon the nature of the substance and are independent of the amount of the substance present in the system e.g., Temperature, pressure-specific heat, etc.

Chemical Thermodynamics Questions

Question 53. If one monoatomic gas is expanded adiabatically from 2 l to 10 l at1 atm external pressure then the value of a u (in atm. l )is-

1. -8
2. 0
3. -66.7
4. 58.2

Answer: 1. -8

q = 0 (since process is adiabatic.) ∴ Δu =w = -pav

=-1(10 -3) atm.L=-8 atm.L

Question 54. The equation of state for ‘ n’ mol of an ideal gas is PV – nRT. fn this equation, the respective number of intensive and extensive properties are—

1. 2,3
2. 3,2
3. 1,4
4. 4, 1

Answer: 2. 3,2

Question 55. \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) ; \Delta H^0=57.32 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

\(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \quad \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) 25°C. If \(\Delta H_f^0\left[\mathrm{H}^{+}(a q)\right]=0,\),then the standard heat of formation (kJ.mol-1) for OH-(aq) at 25°Cls—

1. -142.9
2. -228.48
3. -343.12
4. -253.71

Answer: 2. -228.48

Chemical Thermodynamics Questions

Question 56. For a reaction at T K, ΔH> 0 and ΔS > 0. If the reaction attains equilibrium at a temperature of T1K, (assume ΔH and ΔS are independent of temperature) then—

1. T<T₁
2. T>T₁
3. T=T₁
4. T>T₁

Answer: 2. T>T₁

Question 57. The change in entropy for 2 mol ideal gas in an isothermal reversible expansion from 10 mL to 100 mL at 27°C is—

1. 26.79 J. K^-1
2. 38.29 J.K^-1
3. 59.07 J-K^-1
4. 46.26 J-K^-1

Answer: 2. 38.29 J.K^-1

Question 58. Which of the statements is true—

1. A reaction, in which a 77 < 0 is always spontaneous;
2. A reaction, in which ah > 0 can never occur spontaneously
3. For a spontaneous process in an isolated system,
4. For a spontaneous process in an isolated system,

Answer: 3. For a spontaneous process in an isolated system,

Question 59. For the reaction, CaCO₃(s)→CaO(s) + CO₂(g), ΔH⁰= +179.1 kl-mol-1 and ΔSO = 160.2 If ΔH⁰ and ΔS⁰ are temperature independent, then the temperature above which the reaction will be spontaneous is equal to—

1. 1008 K
2. 1200 K
3. 845 K
4. 1118 K

Answer: 4. 1118 K

Chemical Thermodynamics Questions

Question 60. When a gas (molar mass =28 g-mol^-1) of mass 3.5g is burnt completely in the presence of excess oxygen in a bomb calorimeter) the temperature of the calorimeter increases from 208 K to 298.45 K. The heat of combustion at constant volume for the gas (Given; the heat capacity of the calorimeter 2.5 k K4)

1. 4.5 kJ .mol^-1
2. 8.0 kJ. mol^-1
3. 9.0 KJ .mol^-1
4. 9.5 kJ mol^-1

Answer: 3. 9.0 KJ .mol^-1

Question 61. For the reaction, 2NH₃(g)→ N₂(g)+3H₂(g)-

1. ΔH< 0, ΔS> 0
2. ΔH> 0, ΔS > 0
3. ΔH > 0, ΔV <0
4. ΔH < 0, Δ5 < 0

Answer: 2. AH> 0, AS > 0

Question 62. Which of the following pairs is true for the process C₆H₆(g)[1atm, 80.1°C]→C₆H₆(l)[1 atm, 80.1°C]

1. ΔG< 0, ΔS> 0
2. ΔG< 0, ΔS< 0
3. ΔG = 0, ΔS < 0
4. ΔG = 0, ΔS > 0

Answer: 3. ΔG = 0, ΔS < 0

Chemical Thermodynamics Questions

Question 63. The internal energy change when a system goes from state P to Q is 30kJ> mol-1. If the system goes from P to Q by a reversible path and returns to state P by an irreversible path, what would be the net change in internal energy—

1. 30kj
2. <30kJ
3. zero
4. >30kJ

Answer: 3. Zero

Question 64. If at normal pressure and 100°C the changes in enthalpy and entropy for the process, H₂O(l)→H₂O(g), are ΔH and ΔS respectively, then ΔH-ΔU is—

1. 5.6 kj. mol-1
2. 6.2 kj – mol-1
3. 3.1 kj-mol-1
4. 4.8 kj-mol-1

Answer: 3. 3.1 kj-mol-1

Question 65. At 25°C, the standard heat of formation for Br₂(g) is 30.9 kj.mol-1. At this temperature, the heat of vaporization for Br₂(l) is—

1. <30.9 kj. mol^-1
2. 30.9 kj .mol^-1
3. >30.9 kj. mol^-1
4. Cannot Be Predicted

Answer: 2. 3.1 kj-mol^-1

Question 66. At 25°C, when 0.5 mol of HC1 reacts completely with 0.5 mol of NaOH in a dilute solution, 28.65 kj of heat is liberated. If at 25°C ΔH0f[H₂O(7)]=  then AH0fOH- (aq) is—

1. -314.45 kj. mol^-1
2. -285.8 kj.mol^-1
3. -228.5 kj. mol^-1
4. -343.1 kJ. mol^-1

Answer: 2.  -285.8 kj.mol^-1

Chemical Thermodynamics Questions

Question 67. On combustion, C_xH_Y(l) forms CO₂(g) and H₂O(l). At a given temperature and pressure, the value of \(\left(\frac{\Delta H-\Delta U}{R T}\right)\) in this combustion reaction is—

1. \(\frac{x}{5}\)
2. \(\frac{x+y}{3}\)
3. \(\frac{y}{4}\)
4. \(\frac{x-y}{4}\)

Answer: 3. \(\frac{y}{4}\)

Question 68. An ideal gas is compressed isothermally at 25°C from a volume of 10 L to a volume of 6 L. Which of the following is not true for this process—

1. q<0
2. w>0
3. AU = 0
4. AH > 0.

Answer: 4. AH > 0.

Question 69. At 27°C, for the reaction aA(g) + B(g)→2C(g) PΔV = -2.5 kj . The value of ‘a’ is

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 70. When 1 mol of an ideal gas is compressed in a reversible isothermal process at T K, the pressure of the gas changes from 1 atm to 10 atm. In the process, if the work done by the gas is 5.744 kJ, then T is—

1. 400k
2. 300k
3. 420k
4. 520k

Answer: 2. 300k

Chemical Thermodynamics Questions

Question 71. For 0.5 mol of an ideal gas, 15 cal of heat is required to raise its temperature by 10 K at constant volume. The molar heat capacity for the gas at constant pressure is-

1. 3cal.K^-1 . mol^-1
2. 4 cal.k^-1mol^-1
3. 5cal.k^-1 .mol^-1
4. 4.5cal.k^-1.mol^-1

Answer: 3. 5cal.k-1 .mol^-1

Question 72. According to the enthalpy diagram given below, the standard heat of formation (kJ mol^-1) of CO(g) at 25°C is—

1. -283.0
2. -110.5
3. +283.0
4. +110.5

Answer: 2. -110.5

Question 73. 1 mol of an ideal gas is enclosed in a cylinder fitted with a frictionless and weightless piston. The gas absorbs x kJ heat and undergoes expansion. If the amount of expansion work done by the gas is x kJ, then the expansion is—

1. Adiabatic
2. Cyclic
3. Isothermal
4. It cannot Be Predicted

Answer: 3. Isothermal

Chemical Thermodynamics Questions

Question 74. Given (at 25°C)

⇒ \(\mathrm{Ca}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CaO}(\mathrm{s}) ; \Delta H^0=-635.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\begin{aligned}
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s}) ; \\
& \Delta H^0=-65.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

The heat of formation (in kj.mol-1 ) for Ca(OH)2(s) is –

1. -855.4
2. -673.9
3. -986.6
4. -731.7

Answer: 3. -986.6

Question 75. Given (at 25°C)

⇒ \(\begin{aligned}
& 2 \mathrm{FeSO}_4(s) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(s)+\mathrm{SO}_2(\mathrm{~g})+\mathrm{SO}_3(g) ; \Delta H^0=340.1 \mathrm{~kJ} \\
& 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) ; \Delta H^0=-824.2 \mathrm{~kJ}
\end{aligned}\)

1. \(8660-\frac{\ln \left(1.6 \times 10^{12}\right)}{R(298)}\)
2. 0.5[2 X 86600 -R(298)In(1.6 X 1012)
3. R(298) lnf1.6 x 1012)- 86600
4. 86600 + R(298) In (1.6 X 1012)

Answer: 4. 86600 + R(298) In (1.6 X 1012)

Question 76. At 25°C, the standard heats of formation of H₂O(g) , H₂O₂(g), H(g) and O(g) are -241.8, -135.66, 218 and 249.17 k).mol-1 respectively. The bond energy (in kj.mol-1 ) of O —O bond in H₂O₂(g) molecule is—

1. 179.23
2. 160.19
3. 142.60
4. 157.16

Answer: 3. 142.60

Chemical Thermodynamics Questions

Question 77. Given (at 25°C):

⇒ \(\begin{aligned}
\mathrm{C}(\mathrm{s} \text {, graphite })+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) & ; \\
& \Delta H^0=-110.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{O}(\mathrm{g}) ; \Delta H^0=+249.1 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{O}(\mathrm{g}) ; \Delta H^0=1073.24 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

The standard enthalpy change for the process, C(s, graphite)→C(g) at 25°C is—

1. +934.64 KJ.Mol-1
2. 713.64 KJ.Mol-1
3. 962.64 KJ.Mol-1
4. 652.64 KJ.Mol-1

Answer: 2. 713.64 KJ.Mol-1

Question 78. At 25°C, for the reaction, H+{aq) + OU-(aq)yH₂O(l) , AHO = -57.3 kj . mol-1 . If the ionization enthalpy of HCN in water is 45.2 kj.mol-1, then the standard heat of reaction (in kj mol-1 ) for the reaction, HCN(aq) + NaOH(at jr)-NaCN(ag) + H₂O(Z), in dilute aqueous solution is

1. -113.5
2. -12.1
3. -102.5
4. -35.7

Answer: 2. -12.1

Question 79. The temperature of a bomb calorimeter changes from 25°C to 32.7°C when wg of naphthalene mass (molar = 128 g. mol-1 ) is burnt completely in the calorimeter. If the heat of combustion at constant volume for naphthalene is -5152 kj mol-1, then w is (heat capacity of the calorimeter = 8.19 kj K-1)

1. 0.87 g
2. 1.91 g
3. 2.37 g
4. 1.57g

Answer: 4. 1.57g

Chemical Thermodynamics Questions

Question 80. In which of the following processes the change in entropy for the system is zero

1. Irreversible adiabatic processes
2. Reversible adiabatic process
3. A spontaneous process occurring in an isolated system
4. Isothermal expansion of an ideal gas

Answer: 2. Reversible adiabatic process

Question 81. A system undergoes the process: A→ B → C→ D. In this process, the change in a state function (.X) of the system is x. In steps A→B and B→C of the process, if the changes in X are y and z respectively, then the change in X in step D→C is—

1. x-y-z
2. x-z+y
3. y+z-x
4. y-z-x

Answer: 3. y+z-x

Question 82. At 27°C, AH = + 6 kJ for the reaction A + 2B-3C. In the reaction, if ΔASuniv = 2 J . K^-1 , then ASys (in J . K^-1 ) is

1. +2
2. +3
3. +20
4. +22

Answer: 4. +22

Question 83. An LPG cylinder contains 14 kg of butane. A family requires 2 X 104 kj of heat for their cooking purpose every day. By how many days will the butane in the cylinder be used up (Given: heat of combustion for butane = -2658 kj-mol-1 )—

1. 15 days
2. 20 days
3. 32 days
4. 40 days

Answer: 3. 32 days

Question 84. For a reaction involving 1 mol of Zn and 1 mol of H₂SO₄ in a bomb calorimeter

1. ΔH> 0 , w> 0
2. ΔU> 0 , w> 0
3. ΔU< 0 , w> 0
4. ΔU< 0 , w> 0

Answer: 4. ΔU< 0 , w> 0

Chemical Thermodynamics Questions

Question 85. Assuming that water vapor is an ideal gas, the internal energy change (ALT) when mol of water is vaporized at bar pressure and 100°C, will be (Given: at bar and 373K, molar enthalpy of vaporization of water is 41 kj. mol-1, R = 8.3 J.mol-1 . K-1 )

1. 4.100 KJ. mol
2. 3.7904 KJ. mol-1
3. 37.904 kj. mol-1
4. 41.00 kj. mol-1

Answer: 3. 37.904 kj. mol-1

Question 86. At 25°C and 1 atm pressure, ΔH and pressure-volume work for the reaction, 2H₂(g) + O₂(g)y₂H₂O(g) are —483.7 kj and 2.47 kj respectively. In this reaction the value U is-

1. -483.7 kJ
2. -481.23 KJ
3. -400.23 Kj
4. -492.6 KJ

Answer: 2. -481.23 KJ

Question 87. An ideal gas’s initial state of 1 mol is (P₁, V₂, T₁ ). The gas is expanded by a reversible isothermal process and also by a reversible adiabatic process separately. If the final volume of the gas is the same in both of the processes, and changes in internal energy in the isothermal and adiabatic processes are ΔU₁ and ΔU₂ respectively, then

1. ΔU₁=ΔU₂
2. ΔU₁<ΔU₂
3. ΔU₁>ΔU₂
4. Cannot Be Predicated

Answer: 3. ΔU₁>ΔU₂

Question 88. At constant pressure, the amount of heat required to raise the temperature of 1 mol of an ideal gas by 10-C is x kj. If the same increase in temperature were carried out at constant volume, then the heat required would be

1. > xKJ
2. <x Kj
3. = x KJ
4. > xKJ

Answer: 1. > xKJ

Chemical Thermodynamics Questions

Question 89. The enthalpy of fusion of ice at 0- C and 1 atm is 6.02 kj – mol^-1. The change in enthalpy (J K^-1)of the surroundings when 9 g of water is frozen at 0°C and 1 atm pressure is

1. +11.02
2. -11.02
3. -20.27
4. +23.09

Answer: 2. -11.02

Question 90. At a given pressure and a temperature of 300 K, Δ_Surr and Δ_sys for a reaction are 8.0 J. K-1 and 4.0 J . K-1 surr respectively. ΔG for this reaction is-—

1. -3.0 KJ
2. -3.6 KJ
3. 3.0 kJ
4. -4.2 kJ

Answer: 2. -3.6 KJ

Question 91. In a reversible process, if changes in the entropy of the system and its surroundings are ΔS₁ and ΔS₂ respectively, then

1. ΔS₁ +ΔS₂ >0
2. ΔS₁ +ΔS₂<0
3. ΔS₁+ΔS₂= 0
4. ΔS₁ +ΔS₂>0

Answer: 3. ΔS₁+ΔS₂= 0

Question 92. A flask of volume 1 L contains 1 mol of an ideal gas. The flask is connected to an evacuated flask, and as a result, the volume of the gas becomes 10 L. The change in entropy (J. K-1) of the gas in this process is

1. 9.56
2. 19.14
3. 11.37
4. 14.29

Answer: 2. 19.14

Chemical Thermodynamics Questions

Question 93. The heats of neutralization of four acids A, B, C, and D are 13.7, 9.4, 11.2, and 12.4kcal respectively when they are neutralized against a common base. The weakest add among A, B, C, and D is

1. A
2. B
3. C
4. D

Answer: 2. B

Question 94. In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which is true-

1. ΔU= w≠0, q=0
2. ΔU=0, w=0, q≠0
3. ΔU=0 w=0, q≠0
4. w=0 w≠0, q=0

Answer: 1. ΔU= w≠0, q=0

Question 95. How many calories are required to increase the temperature of 40g of Ar from 40-C to 100°C at a constant volume (R = 2 cal.mol^-1.K^-1)

1. 120
2. 2400
3. 1200
4. 180

Answer: 4. 180

Question 96. Water is supercooled to -4°C. The enthalpy (H) of the supercooled water is

1. Same as ice at -4°c
2. More than ice at -4°c
3. Same as ice at 0°c
4. Less than ice at -4°c

Answer: 4. Less than ice at -4°c

Question 97. The standard entropy of X₂, Y₂, and XY₃ are 60, 40, and 50 J.K-1.mol-1 respectively. For the reaction\(\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3,(\Delta H=-30 \mathrm{~kJ})\) (AH = -30 kl ) to be at equilibrium, the temperature will be

1. 1250 k
2. 750k
3. 500 k
4. 1000 k

Answer: 2. 750k

Question 98. Two moles of gas of volume 50L and pressure 1 atm are compressed adiabatically and reversibly to 10atm. What is the atomicity of the gas (T₁/T₂= 0.4)

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Chemical Thermodynamics Questions

Question 99. Given (at 25°C): C(s, graphite)-C(g) ; ΔH0 = +713.64 kj. mol-1

\(\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g}); \Delta H^0=+218 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) C(s, graphite)+3H2(g)→C6H6(g); AH0-+82.93kj – mol-1 At 25°C, if the bond energy of C—H and C— C bonds are 418 and 347 kj .mol-1 respectively, then the C=C bond energy is—

1. +679.81 KJ. mol^-1
2. +652.63 kj. mol^-1
3. +808.75 KJ. mol^-1
4. +763.39 kJ. mol^-1

Answer: 2. +652.63 kj. mol^-1

Question 100. Given (at 25 °C):

⇒ \(\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_6(g)+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta H^0=-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
& \Delta H^0=-890 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \\
& \qquad \Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25 °C , if the standaÿenthalpy of formation of C3H8(g) is -103.8 kj. mol-1 , then the standard heat of reaction for the reaction; C3H8(g) + H2(g)yC2H6(g) + CH4(g) is

1. +98.45 kJ
2. -55.70KJ
3. 62.37 KJ
4. -47.25 KJ

Answer: 2. -55.70KJ

Question 101. At 0°C and normal pressure, the enthalpy of fusion of ice is 334.7 J . g^-1. At this temperature and pressure, if 1 mol of water is converted into 1 mol ice, then the change in entropy of the system will be

1. 16.7 J.K^-1
2. -16.7 J.K^-1
3. 22.06 J.K^-1
4. -22.06 J.K^-1

Answer: 4. -22.06 J.K^-1

Question 102. 5 mol of gas is put through a series of changes as shown graphically in a cyclic process. The process X→Y, Y→Z and Z→X respectively are

1. Isochoric, isobaric, isothermal
2. Isobaric, isochoric, isothermal
3. Isothermal, isobaric, isochoric
4. Isochoric, isothermal, isobaric

Answer: 1. Isochoric, isobaric, isothermal

Question 103. Given \(\mathrm{NH}_3(g)+3 \mathrm{Cl}_2(g) \rightarrow \mathrm{NCl}_3(g)+3 \mathrm{HCl}(g) ;-\Delta H_1\) N2(g) + 3H2(g)→2NH3(g) ; AH2; H2(g) + Cl2(g→2HCl(g); AH3 Heat of formation (ΔHf) of NCl3(g) in terms of ΔH1 , ΔH2 and ΔH3 is-

1. \(-\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)
2. \(\Delta H_1+\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)
3. \(\Delta H_1-\frac{\Delta H_2}{2}-\frac{3}{2} \Delta H_3\)
4. None of the above

Answer: 4. None of the above

Chemical Thermodynamics Questions

Question 104. When lg of graphite is completely burnt in a bomb calorimeter, the temperature of the bomb and water rises from 25°C to 30.5°C. The heat capacity of the calorimeter is 5.96 kj. °C-1, then the heat of combustion per mole of graphite at constant volume is

1. -357.13 kj.mol-1
2. -289.71 kj.mol-1
3. -393.36 kj .mol-1
4. -307.94 kj-mol-1

Answer: 3. -393.36 kj .mol-1

Question 105. The volume of a gas is reduced to half of its original volume. The specific heat will—

1. Reduce to half
2. Double
3. Remain Constant
4. Increase Four times

Answer: 3. Remain Constant

Question 106. For which molar-specific heat is temperature-independent

1. ARgn
2. Hydrogen
3. Nitrogen
4. Carbon dioxide

Answer: 1. ARgn

Question 107. Which of the following quantities are state functions

1. q
2. q+w
3. w
4. U+pv

Answer: 1. q

Question 108. A monoatomic ideal gas undergoes the cyclic process, Which of the comments are true for this process

1. For the whole process q = +1.134J
2. For the whole process ASys > 0
3. For the whole process ASys = 0
4. For the whole process q = -2.310 J

Answer: 1. For the whole process q = +1.134J

Question 109. Which of the following comments is true—

1. Only for an ideal gas, cp m > cv m
2. For any gas, cp m > cv m
3. For a solid substance, cp m – cv> m
4. For ‘ n ’ mol ofideal gas, cp m- cv m = nr

Answer: 2. For any gas, cp m > cv m

Question 110. When 3g of ethane gas is brunt at 25°C, 156 kl of heat is liberated. If the standard enthalpies of formation for CO₂(g) and H₂O(l) are -393.5 and -285.8 kj.mol-1 respectively, then for ethane gas-

1. Standard heat of combustion = -1560 kJ mol-1
2. Standard heat of formation =-67.9 kJ. Mol-1
3. Standard heat of combustion =-832 kJ .mol-1
4. Standard heat of formation = -84.4 kJ. Mol-1

Answer: 1. Standard heat of combustion = -1560 kJ. mol-1

Chemical Thermodynamics Questions

Question 111. A reaction is spontaneous at a temperature of 300K, but it is non-spontaneous at a temperature of 400 K. If ΔH and ΔS for the reaction do not depend on temperature, then

1. ΔH> 0
2. ΔH < 0
3. ΔS > 0
4. ΔS<0

Answer: 2. AH < 0

Question 112. The reaction, 3O₂(g)→2O₃(g), is non-spontaneous at any temperature. Hence—

1. The reverse reaction is spontaneous at any temperature
2. ΔH < 0 and ΔS < 0 for the reverse reaction
3. ΔH > 0 , ΔS > 0 for the reverse reaction
4. ΔH < 0 and ΔS > 0 for the reverse reaction

Answer: 1. The reverse reaction is spontaneous at any temperature

Question 113. For the isothermal free expansion of ideal gas

1. ΔH =0
2. ΔS < 0
3. ΔS > 0
4. ΔH > 0

Answer: 1. ΔH =0

Question 114. The changes in which of the following quantities are for a cyclic process

1. Enthalpy
2. Work
3. Entropy
4. Internal energy

Answer: 1. Enthalpy

Question 115. Which of the following relations are true for the reaction, PCl₅(g)→PCl₃(g) + Cl₂(g)

1. ΔH< 0
2. ΔH >0
3. ΔS <0
4. ΔS>0

Answer: 2. ΔH >0

Chemical Thermodynamics Questions

Question 116. An ideal gas performs only pressure-volume work in the given cyclic process. In the diagram, AB, BC, and CA are the reversible isothermal, isobaric, and isochoric processes respectively. Identify the correct statements regarding this cycle

1. Total Work Done In This Process ( W) = WA→B + WB→C
2. Changes In Internal Energy In The Step Ab = 0
3. ΔSA→b = ASB→C+ΔSC→A
4. If The Total Heat And Work Involved In The Process Are Q And W Respectively, Then q + W = 0

Answer: 2. Changes In Internal Energy In The Step Ab = 0

Question 117. Identify the correct statements—

1. Standard state of bromine (25°C, 1 atm) is Br2(g)
2. C (graphite, s)-C (diamond, s); here AH=0
3. Standard enthalpy change for the reaction N₂(g) + O₂(g)→2NO(g) at 25°C and 1 atm is standard enthalpy offormation of NO(g)
4. At a particular temperature and pressure, if ΔH = xkj for the reaction A + 3.B→2C then AH \(-\frac{x}{2} \mathrm{~kJ}\) for the reaction \(C \rightarrow \frac{1}{2} A+\frac{3}{2} B\)

Answer: 2. C (graphite, s)-C (diamond, s); here ΔH=0

Question 118. Which of the following statements is correct—

1. In any adiabatic process, ΔSsys = 0
2. In the isothermal expansion of ideal gas, ah = 0
3. An endothermic reaction will be spontaneous if in this reaction ΔSsys > 0
4. Heat capacity is a padi-dependent quantity

Answer: 2. In the isothermal expansion of ideal gas, ah = 0

Question 119. Correct statements are—

1. A + B→D, AH = x kj.This reaction is performed in the following two steps: A + B→C C→D. If in step AH = ykj.then AH = (x-y)J in step
2. for a spontaneous process occurring in an isolated system ASy > 0 at equilibrium
3. In a spontaneous chemical reaction at constant temperature and pressure, A G = ~TASsurr
4. In a chemical reaction AH > 0 and AS > 0. The reaction attains equilibrium at temperature, Tg. At constant pressure and constant temperature TK the reaction will be spontaneous, if T > T

Answer: 1. A + B-D, AH = x kj.This reaction is performed in the following two steps: A + B—>CC-D. If in step AH = ykj.then AH = (x-y)J in step

Question 120. Some reactions and their ΔH° values are given below: C(graphite,s)
\(+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta H^0=a \mathrm{~kJ}\)

1. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=b \mathrm{~kJ}\)
2. \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=c \mathrm{~kJ}\)
3. \(\begin{aligned}
   & \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=d \mathrm{~kJ} \\
   & 2 \mathrm{C} \text { (graphite,s) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; \Delta H^0=m \mathrm{~kJ}
   \end{aligned}\)

Which of the following statements is correct—

1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1
2. Standard heat of combustion of C₂Hg = (2a + 2b + 3c- m) kj.mol^-1
3. Standard heat of combustion of carbon = a kj.mol^-1
4. Standard of formation of CO₂(g)=(a+fo) kj.mol^-1

Answer: 1. Standard heat of formation of CH₄(g) = (a + b + 2c→d) kj-mol^-1

Question 121. At 25°C, in which of the given reactions do standard enthalpies of reactions indicate standard enthalpies of formation of the products in the respective reactions

1. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{3} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{HI}(\mathrm{g})\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Question 122.

1. CaCO₃(s)→CaO(s)+CO₂(g)
2. \(\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g)\)
3. PCl₅(g)⇒ Pcl₃(g) + C1₂(g)
4. N₂(g) + O₂(g)→2NO(g)

For which of these reactions, P-V work is negative—

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Chemical Thermodynamics Questions

Question 123. Which of the given reactions are endothermic—

1. Combustion of methane
2. Decomposition of water
3. Dehydrogenation of ethane to ethene
4. Conversion of graphite to diamond

Answer: 2. Decomposition of water

Question 124. At constant volume and 298K, mol of gas is heated and the final temperature is 308 K. Ifheat supplied to the gas is 500 J, then for the overall process—

1. w = 0
2. w = -500 J
3. AU = 500 J
4. AU = 0

Answer: 1. w = 0

Question 125. True for spontaneous dissolution of KC₁ in water are

1. ΔG<0
2. ΔH > 0
3. ΔS_surr < 0
4. ΔH<0

Answer: 1. ΔG<0

Question 126. When a bottle of perfume is opened, odorous molecules mix with air and diffuse gradually throughout the room. The correct facts about the process are-

1. ΔS = 0
2. ΔG < 0
3. ΔS> 0
4. ΔS < 0

Answer: 2. ΔG < 0

Chemical Thermodynamics Questions

Question 127. mol of an ideal gas undergoes a cyclic process ABC A represented by the following diagram:

Which of the given statements is correct for the process —

1. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)
2. Work done by the gas in the overall process is P₀ V₀
3. Heat absorbed by the gas in path AB is 2P₀ V₀
4. Heat absorbed by the gas in path BC is \(\frac{1}{2} P_0 V_C\)

Answer: 2. Work done by the gas in the overall process is \(\frac{P_0 V_0^2}{2}\)

Question 128. At 0°C and 10 atm pressure 14g of oxygen is subjected to undergo a reversible. adiabatic expansion to a pressure of 1 atm. Hence in this process—

1. The final temperature of the gas is 141.4 K
2. The final temperature of the gas is 217.3 K
3. Work done = 293.2 cal
4. Work done = -286 cal

Answer: 1. Final temperature of the gas is 141.4 K

Question 129. Choose the reactions in which the standard reaction enthalpy (at 25°C) represents the standard formation enthalpy of the product

1. \(\mathrm{H}_2(g)+\frac{1}{3} \mathrm{O}_3(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)
3. \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
4. \(\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(s) \rightarrow \mathrm{HI}(g)\)

Answer: 2. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s)\)

Thermodynamics Fill In The Blanks

Question 1. The sign of A U in the adiabatic expansion of a gas is ____________.
Answer: -ve

Question 2. In a process q > 0, but w = 0, So, the internal energy of the system will (Increase/decrease) ____________
Answer: Increase

Question 3. According to first Jaw, for a cyclic process q+w =____________
Answer: 0

Question 4. The volume of a substance is a property while the molar volume of a substance is an____________property.
Answer: Extensive, intensive

Chemical Thermodynamics Questions

Question 5. The specific heat capacity of a substance is a; cal. g-1.°C-1. The specific heat capacity of 100 g of that substance will be____________
Answer: x cal. g-1 °C-1

Question 6. The molar heat capacity of a substance at constant pressure is ____________than that at constant volume.
Answer: Greater;

Question 7. In the vaporization of water, the signs of q and w are ____________and respectively.
Answer: q = + ve , w = -ve;

Question 8. For a process occurring at constant volume, U= + 10 kj. In this process, q = ____________
Answer: q = +10 kJ;

Question 9. In isothermal expansion of 1 mol of an ideal gas, q = + 12 kJ. The value of w = and A U =____________
Answer: w = -12 kJ , AU = 0

Question 10. In an alchemical reaction, if the total enthalpies of reactants and products are HR and Hp respectively then Hp > HR for____________ reaction and HR > Hp for ____________reaction.
Answer: Endothermic, exothermic

Question 11. At 25°C, the standard state of liquid ethanol means ____________ethanol at 25 °C and____________ pressure.
Answer: Pure liquid, 1 atm;

Question 12. Given: C₃H₈(g) + 5O₂(g)→3CO₂(g) + 4H₂O(l) and C₃H₂(g) + 5O₂(S)→3CO₂(g) + 4H₂O(g). At a fixed temperature and pressure between these two reactions the heat evolved in the second reaction is more than that in the first one.
Answer: Less

Chemical Thermodynamics Questions

Question 13. At 25°C and 1 atm, the standard reaction enthalpy for the reaction C(graphite, s) + O₂(g)→CO₂(g) is the ____________ of CO₂(g).
Answer: Standard enthalpy of formation;

Question 14. At 25°C and 1 atm, standard reaction enthalpy for the reaction, CH4(g) + 2O₂(g)→ CO₂(g) + 2H₂O(Z) is ____________of CH₄(g).
Answer: Standard enthalpy of combustion

Question 15. The change in enthalpy for die reaction HCN(aq) + NaOH(aq)—>NaCN(aq) + H₂O(Z) is ____________than the change ih! ^enthalpy for the reacdon HCI {aq) + NaOH(aq)→NaCl(aq) + H₂O(/).
Answer: Less

Question 16. The change in enthalpy for the process KCl(s) + 100H₂0(l)-KCl(100H₂O) is called____________.
Answer: Integral heat of solution

Question 17. At 25°C and 1 atm if the bond energy of Cl₂(g) is 242 kj.mol-1, then the standard enthalpy of atomization of chlorine at the same temperature = ____________.
Answer: 121KJ.mol-1

Question 18. In a reaction ΔSys = xJ-K-1 and ΔSsurr = -yJ-K-1. The reaction will be spontaneous if ____________
Answer: x>y

Question 19. In the cyclic process ΔB Δ the changes in entropy of the system are ASX, ΔS_2, and ΔS₃. ΔS ^ + ΔS₂ + ΔS₃ =____________.
Answer: Zero

Question 20. NH₄C1(s) + H₂O(Z) -NH+4(aq) + Cl-(aq). It is an endothermic process. In this process the signs (+ or -) of____________ Δsys and ASsun are and respectively.
Answer: ASys > 0 and ASsun < 0

Chemical Thermodynamics Questions

Question 21. A reaction is non-spontaneous at all temperatures but the____________
reverse reaction is spontaneous at all temperatures. The signs (+ or – ) of ΔH and ΔS for the reverse reaction are____________and respectively.
Answer:  AH < 0 and ASys

Question 22. In an exothermic process ASsys < 0. In this reaction the sys sign (+ or – ) of ASsurr is____________. This reaction will be spontaneous if the numerical value of Ssys is higher than that of ASsurr.
Answer: +ve, less

Question 23. In a spontaneous process occurring at constant pressure and at temperature T K, ΔH > 0, ΔS > 0. In this process, the numerical value of ΔH is ____________ than that of____________.
Answer:  less, TΔS