WBCHSE Chemistry Class 11 Equilibrium Question And Answers

Equilibrium Long Answer Type Questions

Equilibrium Constant Calculation Question 1. Equilibria involving physical and chemical changes are dynamic in nature—why?
Answer: In a system, if two processes occur simultaneously at the same rate, then the system is said to be in a state of dynamic equilibrium. When a chemical reaction remains at equilibrium, the forward and the backward reactions occur simultaneously at the same rate.

Due to this, the equilibrium established in a chemical reaction is regarded as a dynamic equilibrium.

Equilibrium Constant Calculation

Similarly, when a physical process remains at equilibrium, the two opposite processes occur simultaneously at the same rate.

For this reason, a physical equilibrium is also dynamic. For example, at equilibrium established during the evaporation of a liquid in a closed container, the evaporation of the liquid and the condensation of its vapor take place simultaneously at the same rate.

Question 2. Give examples of two systems involving solid-vapor equilibrium.
Answer: The equilibrium, solid vapour is found to be established when naphthalene or iodine undergoes sublimation in a closed container.

Question 3. The equilibrium established in the dissolution of a solid in a liquid or a gas in a liquid is dynamic. Explain
Answer: If a solid solute is continuously dissolved in a suitable solvent, eventually a saturated solution of the solute is obtained. In this solution, an equilibrium exists between dissolved solute and undissolved solute.

At this equilibrium, the solute particles from undissolved solid solute get dissolved at the same rate as the solute particles from the solution get deposited on the surface of undissolved solid solute. Since these two processes occur at the same rate, the equilibrium established between dissolved solute and undissolved solute is dynamic.

When a gas is brought in contact with a suitable solvent, the gas keeps on dissolving in the solvent, and eventually, a saturated solution of the gas is formed. At this state, the gas over the liquid surface remains in equilibrium with the dissolved gas in the liquid.

In this state of equilibrium, the gas molecules from the gas phase enter the liquid phase at the same rate as the dissolved gas molecules from the liquid enter the gas phase. Two opposite processes therefore occur at the same rate. For this reason, the equilibrium that forms when a gas remains in equilibrium with its saturated solution is dynamic.

Question 4. The reaction Fe₂O₃(s) + 3CO(gH→2Fe(s) + 3CO₂(g) is carried out separately in a closed vessel and an open vessel. In which case do you expect a higher yield of CO₂(g)?
Answer: If the reaction Fe₂O₃(s) + 3CO(g)→2Fe(s) + 3CO₂(g) is carried out in a closed vessel, an equilibrium is established between the reactants and the products. As a result, the reaction vessel always contains a mixture of reactants and products.

On the other hand, if the reaction is carried out in an open vessel, CO₂(g) formed in the reaction diffuses out from the vessel and mixes with the air and does not have the opportunity to react with Fe(s).

As a result, the reaction becomes irreversible and goes to completion. At the end of the reaction, the vessel only contains Fe(s). Since the reaction reaches completion in an open vessel, the yield of CO₂(g) will be higher in this condition.

Question 5. If a reversible reaction is carried out in a closed vessel, the reactant(s) is/are never used up completely. Explain the reaction with an example.
Answer: When a reversible reaction is carried out in a closed vessel, an equilibrium is established between the reactants and the products.

As a result, the reaction system always contains a mixture of reactants and products. In other words, a reversible reaction occurring in a closed vessel never gets completed. Consequently, the reactants are never used up.

Equilibrium Constant Calculation

Question 6. The following reaction is carried out in a closed vessel at a fixed temperature: A(g) 2B(g). The concentrations of A(g) and B (g) in the course ofthe reaction are as follows When does the reaction attain equilibrium? What
are the equilibrium concentrations of A and B?
Answer: From the time of 60 minutes, the concentrations of reactant and product are found to remain the same with time. Therefore, the reaction has arrived at an equilibrium state at 60 minutes.

At the state of equilibrium, [A] = 0.58 mol- L-1 and [J3] = 0.84 mol- L-1.

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Question 7. At a given temperature, for reaction A B, the rate constant (k) of the forward reaction is greater than that of the backward reaction (kb). Is the value of the equilibrium constant (K) for this reaction greater than, less than, or equal to 1?
Answer: The equilibrium constant (K) of a reaction

⇒ \(=\frac{\text { Rate constant of the forward reaction }}{\text { Rate constant of the backward reaction }}=\frac{K_f}{K_b}\)

Given: Kj→ Kb. Therefore K > 1.

Question 8. Write the expressions of Kc and Kp for the following reactions:

1. 2SO₂(g) + O₂(g) 2SO₃(g)
2. 2BrF₅(g) Br₂(g) + 5F₂(g)
3. 3O₂(g)⇒2O₃(g)
4. 4HCl(g) + O₂(g) 2Cl₂(g) + 2H₂O(g)
5. P4(g) + 3O₂(g) P₄O₆(s)
6. CO(g) + 2H₂(g) CH₃OH(Z)

Answer:

⇒ \(K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{SO}_3}^2}{p_{\mathrm{SO}_2}^2 \times p_{\mathrm{O}_2}}\)

⇒ \(K_c=\frac{\left[\mathrm{Br}_2\right]\left[\mathrm{F}_2\right]^5}{\left[\mathrm{BrF}_5\right]^2} ; K_p=\frac{p_{\mathrm{Br}_2} \times p_{\mathrm{F}_2}^5}{p_{\mathrm{BrF}_5}^2}\)

⇒ \(K_c=\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3} ; K_p=\frac{p_{\mathrm{O}_3}^2}{p_{\mathrm{O}_2}^3}\)

⇒ \(K_c=\frac{\left[\mathrm{Cl}_2\right]^2\left[\mathrm{H}_2 \mathrm{O}\right]^2}{[\mathrm{HCl}]^4\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{Cl}_2}^2 \times p_{\mathrm{H}_2 \mathrm{O}}^2}{p_{\mathrm{HCl}}^4 \times p_{\mathrm{O}_2}}\)

⇒ \(K_c=\frac{\left[\mathrm{P}_4 \mathrm{O}_6(s)\right]}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}=\frac{1}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}\) [since [P4O6(s)]=1]

⇒ \(K_p=\frac{1}{p_{\mathrm{P}_4} \times p_{\mathrm{O}_2}^3}\)

⇒ \(K_c=\frac{\left[\mathrm{CH}_3 \mathrm{OH}(l)\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}=\frac{1}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}\)

[since [CH₃OH(l)]=1

⇒ \(K_p=\frac{1}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2}^2}\)

Question 9. Write the expression for the following reactions:

1. 2Ag+(aq) + Cu(s) Cu2+(aq) + 2Ag(s)
2. NH₄NO₃(s) N₂O(g) + 2H₂O(g)
3. Au+(aq) + 2CN-(aq) [Au(CN)2]-(aq)
4. 3Cu(s) + 2NO₃ {aq) + 8H+(aq)
5. 3Cu2+(aq) + 2NO(g) + 4H₂O(l)
6. Cl₂(g) + 2Br-(a<7) Br₂(l) + 2Cl-(a<7)

Equilibrium Constant Calculation

Answer:

⇒ \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right][\mathrm{Ag}(s)]^2}{\left[\mathrm{Ag}^{+}(a q)\right]^2[\mathrm{Cu}(s)]}=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]}{\left[\mathrm{Ag}^{+}(a q)\right]^2}\)

[since [Ag(s)]=1 and [Cu(s)]=1]

⇒ \(K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2}{\left[\mathrm{NH}_4 \mathrm{NO}_3(s)\right]}=\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2\)

[since [NH₄NO₃(s)]=1]

⇒ \(K_c=\frac{\left[\mathrm{Au}(\mathrm{CN})_2^{-}(a q)\right]}{\left[\mathrm{Au}^{+}(a q)\right]\left[\mathrm{CN}^{-}(a q)\right]^2}\)

⇒ \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2\left[\mathrm{H}_2 \mathrm{O}(l)\right]^4}{[\mathrm{Cu}(s)]^3\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\)

⇒ \(=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2}{\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\) [since (s)=1 and H₂O(l)]=1

⇒ \(K_c=\frac{\left[\mathrm{Br}_2(l)\right]\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}=\frac{\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}\)

[since [Br₂(l)]=1]

Question 10. Establish the relation between Kp and Kc for the following: aA + bB₂=± IL + mM [where the terms have their usualsignificance]
Answer: \(K_c=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b}\) where[A],[B],[l] and [M] are Molar Concentrations Of A, B, [l] and [M] are Molar Concentrations of A, B, L and At M At Equilibrium \(K_p=\frac{p_L^l \times p_M^m}{p_A^a \times p_B^b}; \text { where } p_A, p_B, p_L \text { and } p_M\) partial pressures of A, B, L and M respectively at equilibrium. Now, pA = [A]RT, pB = [B]RT, pL = [L]RT and PM = [M]RT. Putting the values of pA, pB, pL, and PM in the expression of, we have—

⇒ \(\begin{aligned}
& K_p=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b} \times(R T)^{(l+m)-(a+b)} \\
& \text { or, } K_p=K_c \times(R T)^{(l+m)-(a+b)}
\end{aligned}\)

Question 11. What will be the correlation between Kp and Kc for the given equilibrium?CO(g) + H₂O(g)⇒CO₂(g) + H₂(g)
Answer: We know, \(K_p=K_c(R T)^{\Delta n}\)

For the given reaction, An = (1 + 1) — (1 +1) = 0

Therefore, for the given reaction Kp = KC(RT)° – K

Question 12. Give two examples of chemical reactions for each of the following cases: (1)Kp > Kc (2)Kp<Kc&(3) Kp = Kc
Answer: We know, Kp = Kc(RT)An

Kp will be greater than Kc if An > 0. Reactions for

Question 13. Find the relation between Kp &Kc for the given reactions:

1. NH₃(g) +HCl(g) NH₄Cl(s)
2. C(s) + CO₂(g) + 2Cl₂(g) 2COCl₂(g)
3. lN₂(g) + O₂(g) NO₂(g)
4. C(s) + H₂O(g) CO(g) + H₂(g)
5. Fe(s) + H₂O(g) FeO(s) + H₂(g)
6. CH₄(g) + H₂O(l) CO(g) + 3H₂(g)
7. CO(g) + 2H₂(g) CH₃OH(Z)

Answer:

⇒ \(\begin{aligned}
& \Delta n=0-(1+1)=-2 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-2} \\
& \Delta n=2-(1+2)=-1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-1}
\end{aligned}\)

⇒ \(\Delta n=1-\left(\frac{1}{2}+1\right)=\left(-\frac{1}{2}\right) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-\frac{1}{2}}\)

⇒ \(\begin{aligned}
& \Delta n=(1+1)-1=+1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T) \\
& \Delta n=1-1=0 ; K_p=K_c(R T)^{\Delta n}=K_c
\end{aligned}\)

⇒ \(\begin{aligned}
& \Delta n=1-1=0 ; K_p=K_c(R T)^{\Delta n}=K_c \\
& \Delta n=(1+3)-1=+3 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^3 \\
& \Delta n=0-(1+2)=(-3) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-3}
\end{aligned}\)

Question 14. Give an example of a reaction for which = Kc = Kx.
Answer: We know, Kp = Kc(RT)An and Kp = Kx(P)An For the reaction: N₂(g) + O₂(g)2NO(g), An = 2- (1 + 1) = 0 . Therefore, Kp = Kc = Kx.

Equilibrium Constant Calculation

Question 15. At 200°C, the equilibrium N₂O4(g) 2NO₂(g) is achieved through the following two pathways: 0.1 mol N₂O₄ is heated in a closed vessel L volume, A mixture of 0.05 mol N₂O₄(g) and 0.05 mol NO₂(g) is heated at 200°C in a closed vessel of 1 L volume. In these two cases, will the equilibrium concentrations of N₂O₄(g) and NO₂(g) and the values of equilibrium constants be the same?
Answer: The value of the equilibrium constant of a reaction depends only on temperature. It does not depend on the initial concentrations of the reactants. Since the temperature is the same for both experiments, the value of the equilibrium constant will be the same in both cases.

The initial concentrations of the reactant(s) in the two experiments are not the same. As a result, the molar concentrations of N₂O₄(g) and NO₂(g) at equilibrium will be different in the two experiments.

Question 16. At a constant temperature, the equilibrium constant of the reaction N₂(g) + 3H₂(g) 2NH₃(g) is K, and that of the reaction N₂(g) + H₂(g) NH₃(g) is JR. Explain this difference in K values even though the reactants and products in both the reactions are same.
Answer: The value of the equilibrium constant of a reaction depends upon the mode of writing the balanced equation of the reaction.

N₂(g) + 3H₂(g) 2NH₃(g)

⇒ \(\text { Equilibrium constant, } K=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\)

⇒ \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) ;\)

Equilibrium constant \(K_1=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2} \times\left[\mathrm{H}_2\right]^{3 / 2}}\)

Therefore \(K_1^2=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=K \quad \text { or, } K_1=\sqrt{K}\)

In both reactions 1 and 2 reactants and products are the same, but in the equations of these two reactions, the stoichiometric coefficients of reactants and products are different, resulting in different values of the corresponding equilibrium constant.

Question 17. For the reaction 2H₂(g) + O₂(g) +=± 2H₂O(g) — Kp = KC(RT)X. Find the value of.
Answer: For the given reaction, An = 2-(2+l) = -l

We know, Kp = Kc(RT)-n

As An = -1, Kp = Kc(RT)-l

Given, Kf) = KC{RT)X

Comparing equations (1) and [2], we have x = -1

Question 18. If the value of the equilibrium constant of the reaction 2SO₂(g) + O₂(g)-2SO₃(g) is K, then what will be the lues of equilibrium constants of the following reactions?

1. 4SO₂(g) + 2O₂(g) 4SO₃(g)
2. 2SO₃(g) 2SO₂(g) + O₂(g)
3. SO₂(g) + iO₂(g)⇒SO₃(g)
4. SO₃(g)⇒SO₂(g) + iO₂(g)

Answer: \(2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) 2 \mathrm{SO}_3(g) ; K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}\)

⇒ \(\begin{aligned}
\text { (I) } 4 \mathrm{SO}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) & \rightleftharpoons 4 \mathrm{SO}_3(\mathrm{~g}) ; \\
\text { Equilibrium constant } & =\frac{\left[\mathrm{SO}_3\right]^4}{\left[\mathrm{SO}_2\right]^4\left[\mathrm{O}_2\right]^2}=K^2
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { (ii) } 2 \mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \\
& \text { Equilibrium constant }=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}=\frac{1}{K}
\end{aligned}\)

⇒ \(\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)\)

⇒ \(\text { Equilibrium constant }=\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}=\sqrt{K}\)

Question 19. At a particular temperature, the following reaction is carried out with 1 mol of A(g) and 1 mol of B(g) in a closed vessel: A(g) + 4B(g) AB4(g). Will the equilibrium concentration of AB4(g) be higher than that of A(g)?
Answer: According to the given equation of the reaction, 1 mol of AB4 forms due to the reaction between 1 mol of A and 4 mol of B. Suppose, the concentration of AB4 at the equilibrium of the reaction is x mol- L-1.

So, according to the given equation, the concentration of A and B at equilibrium will be (1 – x) mol- L-1 and (1-4) mol. L-l, respectively. At equilibrium, if the concentration of AB, was greater than that of A, then x would be greater than (1 – x), i.e., x > 1 – x or, x > 0.5.

If x was greater than 0.5, then the concentration of B would be negative. This is impossible. Therefore, the concentration of AB4 can never be greater than that of A.

Equilibrium Constant Calculation

Question 20. A few reactions and their equilibrium constants are given:

A⇌B + C⇌Kc = 2; C⇌B +D⇌Kc = 3D ⇌B +E; Kc = 4. Find the equilibrium constant ofthe reaction, A⇌=±3B +E

Answer:

1. \(K_c=2=\frac{|B|[C \mid}{|A|}\)
2. \(K_c=3=\frac{[B \| D]}{[C]}\)
3. \(K_c=4=\frac{[B][E]}{[D]}\)

For the reaction A 36 + E. equilibrium constant \(K_c=\frac{[B]^3[E]}{[A]}\) \(K_c=4=\frac{[B] \mid E]}{[D]} ; \text { For the reaction } A \rightleftharpoons 3 B+E \text {, }\) equilibrium constant
\(K_c=\frac{[B]^3[E]}{[A]}\) Multiplying equilibrium constants of reactions 1; 2 we have \(2 \times 3 \times 4=\frac{[B][C]}{[A]} \times \frac{[B][D]}{[C]} \times \frac{[B][E]}{[D]}\text { or, } 24=\frac{[B\rfloor^3[E]}{[A]}\)

Question 21. What does the sign of AG of a reaction at a given pressure and temperature indicate about the feasibility of the reaction?
Answer: \(\begin{aligned}
& A+2 B \Longrightarrow C ; K_c=2=\frac{[C]}{[A][B]^2} \\
& C \rightleftharpoons 2 D ; K_C=4=\frac{[D]^2}{[C]}
\end{aligned}\)

Adding equations 11 1 and [2), A + 26 s=± 2D

The equilibrium constant for equation [3] = Product of The equilibrium constants for reactions [1] and [2]

⇒ \(=\frac{[D]^2}{[A][B]^2}=2 \times 4=8\)

The equilibrium constant For the reaction 2D⇒ A + 2B, the equilibrium constant = reciprocal of the equilibrium constant for reaction (3) \(=\frac{1}{8}=0.125\)

Question 22. For the reaction A + B C + D, the equilibrium constant is K. What would be the value of the reaction quotient (Q) when the reaction just starts and when it reaches equilibrium?
Answer: For the given reaction, equilibrium constant,

⇒ \(K=\frac{[C]_{e q} \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}} \text { where }[A]_{e q},[B]_{e q},[C]_{e q}\)

[D]eq are the equilibrium concentrations of A, B, C, and D respectively.

For the given reaction, the reaction quotient (Q) at any moment during the reaction.

⇒ \(Q=\frac{[C] \times[D]}{[A] \times[B]} ; \text { where }[A],[B],[C] \text { and }[D]\) are the molar concentrations of.4, B, C, and D respectively at the
moment under consideration.

Initially: [C] = 0, [D] = 0. Therefore, Q = 0.

At The Equilibrium state; \([A]=[A]_{e q},[B]=[B]_{e q}\) \([C]=[C]_{e q} \text { and }[D]=[D]_{e q} \text {. Hence, } Q=K \text {. }\)

Question 23. We know, AG° = -RT\nKc and AG° = -RTlnKp. Therefore in case of a reaction occurring in the gaseous phase at a given temperature, AG° is the same even if the values of Kp and Kc are different. Is the statement true? Give reasons.
Answer: The statement is not true. In the equation, AG° = -RTnKc, the concentration of each of the reactants and products at standard state is taken as l(M).

On the other hand, in the equation, AG° = -RTnKp, the partial pressure of each of the reactants and products at standard state is taken as 1 atm. Therefore, the different values of AG° will be obtained from these two equations.

The values of the equilibrium constant (K) of a reaction at 25°C and 50°C are 2 X 10-‘1 and 2 x 10-2 respectively. Is the reaction an exothermic or endothermic?

Question 24. Consider the reaction, \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\) Heat and answer the following questions: (1) Find the relation among a, b, and c. State whether the equilibrium will be shifted towards right or left if the temperature is Increased.
Answer: According to the equation ofthe reaction, c mol of XY forms when a mol of X₂ reacts with b mol of Y₂. So, the number of X atoms in a mol of X₂ = The number of X atoms in c mol of XY.

since \(2 \times a=c \text { or, } a=\frac{c}{2} \text {. }\)

Similarly, the number of Y atoms in b mol of Y2 = the number of Y atoms in c mol of XY.

∴ \(2 \times b=c \text { or, } b=\frac{c}{2}\)

∴ \(a=b=\frac{c}{2} \text {. }\)

The reaction is exothermic. So, according to Le Chatelier’s principle, a temperature rise will cause the equilibrium ofthe reaction to shift to the left.

Question 25. State Le Chatelier’s principle, explain the effect of (a) pressure and (b) continuous removal at the constant temperature on the position of equilibrium of the following reaction: H₂(g) + I₂(g) ?=± 2111(g)
Answer: Pressure does not have any effect on the position of the equilibrium because the reaction is not associated with any volume change [Total no. of molecules of HI(g) = Total no. of molecules of H₂(g) and I₂(g)].

If HI is removed continuously from the reaction system, then the equilibrium goes on shifting towards the right, and finally, the reaction moves towards completion.

Question 26. Consider thefollowing reaction: 2A(g) + B₂(g) 2AB(g); AH < 0. How can the yield of (g) be Increased?
Answer:

For the reaction AH < 0, so it is an exothermic reaction. According to Le Chatelier’s principle, if the temperature of an exothermic reaction at equilibrium is decreased, the equilibrium of the reaction shifts to the right. So, the decrease in temperature will result in a higher yield of AB(g).

The given reversible reaction is associated with the decrease in number of moles [An = 2- (2 + 1) =-l ] in the forward direction. So, according to Le Chatelier’s principle, if the pressure is increased at the equilibrium of the reaction, the equilibrium shifts to the right, thereby increasing the yield of AB(g)

Equilibrium Constant Calculation

At constant temperature and volume, if the reactant A₂(g) or B₂(g) is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of AB(g) will increase

Question 27. 2BrP₂(g);=± Br₂(g) + 5F₂(g); At constant temperature, how the increase in pressure will affect the following at equilibrium— Equilibrium constant, Position of the equilibrium, and yield of the product?
Answer: Pressure does not affect the magnitude of the equilibrium constant.

The given reversible reaction involves a decrease in several gas molecules in the backward direction. So, increasing pressure at the equilibrium of the reaction will favor the backward reaction Consequently, the yield of the product will decrease.

Question 28. Consider the reaction, A(g) + 2B(g) 2D(g) + 3E(g) + heat, and state how the following changes at equilibrium will affect the yield of the product, D(g)— Temperature is increased, Pressure is increased at a constant temperature, Some amount of E (g) is removed from the reaction system at constant temperature and volume, and Some amount of D(g) is added to the reaction system, keeping temperature and volume constant.
Answer: The reaction is exothermic. So, the increasing temperature at the equilibrium of the reaction will cause the equilibrium to shift to tire left. This decreases the yield of D(g). As a result, the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

The reaction involves an increase in the number of gas molecules [An = (2 + 3)-(l + 2) =+2] in the forward direction. Therefore, at a constant temperature, if the pressure is increased at the equilibrium of the reaction, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease and the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

At constant temperature and volume, if some amount of E(g) [product] is removed from the reaction system, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of D(g) will increase, and the concentration of D(g) will increase. Thus, the concentration of D(g) at the new equilibrium will be more than that at the original equilibrium.

At constant temperature and volume, if some amount of D(g) [product] is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease. However, the concentration of D(g) at the new equilibrium will be greater than that at the original equilibrium.

Question 29. Consider the following reactions; What will be the effect on the following if temperature is increased? (a) the equilibrium constant (b) the position of equilibrium (c) the yield of products.
Answer: The reaction (<) is endothermic as AH > 0.

As the reaction is endothermic, the value of its equilibrium constant will increase with the rise in temperature According to Le Chatelier’s principle, a rise in Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂. In its aqueous solution, Na₂CO₃ dissociates completely and forms Na+ and CO₃ ions. In water, COl- is a stronger base than HaO, and hence it reacts with water, forming OH- and HCO₂ ions.temperature at the equilibrium of an endothermic.

Reaction shifts the equilibrium to the right. As a result, the yields of the products increase.

The reaction (ft) is exothermic as AH < 0.

As the reaction is exothermic, the value of its equilibrium constant will decrease with the temperature rise. According to Le Chatelier’s principle, increasing temperature at the equilibrium of an exothermic reaction shifts the equilibrium to the left. Consequently, the yields of the products decrease.

Question 30. For these reactions, predict the effect on the following if pressure is increased at equilibrium Position of equilibrium, Yield of the products.
Answer: The reaction involves no change in volume. Therefore, pressure has no effect either on the position of the equilibrium or on the yield of the product.

The reaction involves a decrease in volume in the forward direction. So, at a constant temperature increasing pressure at the equilibrium of the reaction favors the shifting of equilibrium to the right, thereby increasing the yield of the product.

The reaction involves a decrease in volume in the reverse direction. So, at constant temperature, increasing pressure at the equilibrium of the reaction shifts the equilibrium to the left, thereby decreasing the yield ofthe product

Question 31. At constant temperature, in a closed vessel, the following equilibrium is established during the decomposition of NH₄C1(S); NH₄Cl(s)⇌NH₃(g) + HCl(g). If pressure of. the equilibrium mixture =P and value of equilibrium =K , then show that \(P=2 \sqrt{K_p} \text {. }\)
Answer ∴ Total number of moles at equilibrium = x + x = 2x

∴ \(p_{\mathrm{NH}_3}=\left(\frac{x}{2 x}\right) P ; p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P\)

∴ \(K_p=p_{\mathrm{NH}_3} \times p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P \times\left(\frac{x}{2 x}\right) P=\frac{p^2}{4}\)

∴ \(K_p=\frac{P^2}{4} \quad \text { or, } P^2=4 K_p\)

∴ \(P=2 \sqrt{K_p}\)

Question 32. Write the conjugate bases of the following acids and give the reason: HN₃, CH₃OH, [A1(H₂O)₆]₃+, NH+, HPO2-, H₂O₂, OH-
Answer: According to the Bronsted-Lowry concept, an acid is a substance that can donate protons. The species formed when an acid donates a proton is called the conjugate base of the acid. The conjugate base of an acid has one fewer H-atom than the acid. Therefore, the conjugate bases of HN₃, CH₃OH, [Al(H₂O)g]3+, NH+, HPO2-, H₂O₂ and OH- are N₂, CH₆CT, [A1(H₂O)₅OH]2+, NH₃, PO4-, HO₂ and O2-, respectively.

Question 33. Write the conjugate acids of the following bases and give
the reason: OH-, H₂PO₄, O2-, H₂-, SO₂T, HCO₂, NH₂, NH₃, H-, C₆H₅NH₂ S₂O₂-8 CO₂–₃
Answer: According to the Bronsted-Lowry concept, a base is a proton acceptor. When a base accepts a proton, the conjugate acid of the die base is formed. The conjugate acid of a base contains one more H-atom than the base.

Equilibrium Constant Calculation

Therefore, the conjugate acids of OH-, H₂PO₄ , O2-, HS-, SO₂-, H₂0, HCO₂, NH₂ , NH₃, H-, C₆H₅NH₂ , S₂Og- and CO3- are H₂O , H₃PO₄ , OH- , H₂S, HS0₃, H₃0+, H₂CO₃, NH₃, NH+, H₂, C₆H₅NH₂, HS₂Og and HCO₃ respectively.

Question 34. Which of the following is the strongest Bronsted base? CIO-, C1O₂, C1O₃, C10₄.
Answer: CIO- is the strongest Bronsted base

Question 35. Identify the Lewis acids and Lewis bases in the following reactions H+ + OH- H₂
Answer: According to Lewis’s concept, an acid is a substance that can accept a pair of electrons. Generally, cations (e.g, Ag+, H+, K+, etc.), molecules with the central atom having incomplete octet (e.g., SiF₄, A1F₃, RMgX, BF₂, etc.), and molecules in which the central atom is linked to an electronegative atom through double bonds (e.g. CO2 ) can act as Lewis acid. In the given reactions, Lewis acids are H+, Co3+, BF3, CO₂, and A1F₃.

According to Lewis’s concept, a base is a substance that can donate a pair of electrons. Anions (for example; F-, OH-, etc.) and molecules having unshared electron pairs act as Lewis bases. In the given reactions, Lewis bases are OH-, NH₃, F-

Question 36. All Lewis bases are fact bases—explain. Each of HCO₂ and HPO- can act both as Bronsted acid and hose—why? Write the formula of conjugate base and conjugate acid in each case.
Answer: A substance that can accept a proton is called a Bronsted base, while a substance that can donate a pair of electrons is called a Lewis base. NH₃ molecule can accept a proton. So, it is a Bronsted base. Again, the NH₃ molecule donates a pair of electrons to form a coordinate bond with a proton. So it also acts as a Lewis base. Therefore, it can be inferred that a Lewis base is a Bronsted base.

Question 37. Each of HCO-3 and HPO- can act both as Bronsted acid and hose—why? Write the formula conjugate base and conjugate acid in each case.
Answer: According to the Bronsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. Since the given species are capable of accepting and donating a proton, they can act as an acid as well as a base

Question 38. Of the two solutions of acetic acid with concentrations 0.1(N) and 0.01(N), in which one does acetic acid have a higher degree of dissociation?
Answer: Acetic acid is a weak acid. Such an acid undergoes partial ionization in water. The degree of ionization of a weak acid in its solution depends upon the concentration of the solution. The higher the concentration of a solution of a weak acid, the smaller the degree of ionization of the acid in that solution. So, the degree of ionization of acetic acid will be higher in 0.01(N) solution.

Question 39. At a certain temperature, the ionization constants of three weak acids HA, HB, and HC are 4.0 X 10-5, 5.2 X 10-4, and 8.6 x 10-3, respectively. If the molar concentrations of their solutions are the same, then arrange them in order of their increasing strength.
Answer: The larger the ionization constant (Ka) of an acid, the greater the extent to which it undergoes ionization in its aqueous solution, and hence the higher the concentration of H30+ ions produced by it in the solution. Alternatively, the larger the value of Ka of an acid, the greater its strength.

The increasing order ofthe given acids concerning their Ka: HA < HB < HC. Consequently, the order of the given acids in terms of increasing strength in their aqueous solutions will be HA < HB < HC.

Equilibrium Constant Calculation

Question 40. A weak tribasic acid, H3A, in its aqueous solution, ionizes in the following three steps: If the ionization constants of the steps are K₁, K₂, and K₃ respectively, then determine the overall ionization constant of HgA.

Answer:

⇒ \(\text { Step 1: } K_1=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}{\left[\mathrm{H}_3 \mathrm{~A}\right]} \text {; }\)

⇒ \(\text { Step 2: } K_2=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{HA}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}\)

⇒ \(\text { Step 3: } K_3=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{3-}\right]}{\left[\mathrm{HA}^{2-}\right]}\)

The equation for the overall ionisation of H₃A is: \(\mathrm{H}_3 \mathrm{~A}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{A}^{3-}(a q)+3 \mathrm{H}_3 \mathrm{O}^{+}(a q)\)

If the overall ionization constant of H₃A is K, then

Equilibrium Constant Calculation

⇒ \(K=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Multiplying ionization constants of steps 1,2 and 3

we have, \(K_1 \times K_2 \times K_3=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Therefore, K = K₁ x K₂x K₃

Question 41. At a certain temperature, the ratio of ionization constants of weak acids HA and HB is 100: 1. The molarity ofthe solution is the same as that of HB, and the degrees of ionization of HA and HB in their respective solutions are Q’1 and a₂ respectively, then show that Q₂ = 10a2
Answer: Suppose, the ionization constants of HA and MB arc K1 and K2, respectively. If the concentration of each of the solutions of HA and HB is c mol-L-1, then

⇒ \(\alpha_1=\sqrt{\frac{K_1}{c}} \text { and } \alpha_2=\sqrt{\frac{K_2}{c}} \text {; }\) where QT1 and a2 are the degrees of ionization of HA and HB in their respective solutions. It is given that Kl: = 100: 1

⇒ \(\text { So, } \frac{\alpha_1}{\alpha_2}=\sqrt{\frac{K_1}{K_2}}=\sqrt{100}=10 \text {, i.e., } \alpha_1=10 \alpha_2\)

Question 42. Find [OH-] in pure water if [H-,0+] in it is x mol L~1. Also, find the relation between x Kw.
Answer: In pure water, \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)

Therefore, [OH-] in pure water \(=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

We know, Kw = [H₃O+] x [OH-]

∴ Kw = xxx

or, Kw = x2

∴ \(x=\sqrt{K_w}\)

Question 43. Why is the ionicproduct of water at 50 greater than that at 25°C?
Answer: Ionisation of water \(\left[2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right]\)

Is an endothermic process. The equilibrium constant referring to the ionization equilibrium of water is called the ionic product of water (Kw). As the process is endothermic, Kw increases with the temperature rise. Thus, Kw of water at 50 °C is greater than that at 25 °C.

Question 44. Why does the concentration of OH“ ions in pure water increase with temperature rise? Does this increase make pure water alkaline? Explain.
Answer: In pure water, [OH-] = JKW. KW increases with temperature, and so does [OH-]. This increase in [OH-] does not however mean that pure water becomes alkaline at a higher temperature as pure water always contains an equal number of H30+ and OH- ions at any temperature.

Question 45. At 25°C what is the concentration of H₃O+ ions in an aqueous solution in which the concentration of OH- tons is 2 x 10-5(M)?
Answer: At 25 °C, Kw = 10-14. Now, for an aqueous solution, [H₃O+] X [OH-] = Kw At 25 °C, [H₃O+][OH-] = 10-14

Given: [OH-] = 2 x 10-5(M)

Therefore, [H₃O+] \(=\frac{10^{-14}}{2 \times 10^{-5}}=5 \times 10^{-10}(\mathrm{M})\)

Question 46. By what factor will the concentration ofH30+ ions in an aqueous solution be increased or decreased if its pH is increased by one unit?
Answer: If the pH of an aqueous solution be x, then [H₃O+] in the solution = 10-PH = 10-x mol.L-1

Question 47. The pH of solution A is twice that of solution B. If the concentrations of H30+ ions in A and B are x (M) and y (M), respectively, then what is the relation between candy?
Answer: Given: pH of the solution A = 2 X pH of the solution B,

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=x \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B=y \mathrm{~mol} \cdot \mathrm{L}^{-1} \text {. }\) Therefore \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=2 \times-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B\) or, \(x=y^2 \text { or, } y=\sqrt{x}\)

Equilibrium Constant Calculation

Question 48. In the case of pure water or an aqueous solution, show that pH + pOH =14. Comment on whether this value will be greater than or less than 14 at 0°C and 100°C.
Answer: Second part: We know, pH+pOH = pKw and pKw = -log Rise in temperature increases the value of Kw. Therefore, < KJ25-C) < ATIOO-C) Since pKw = -log10KW, pKJl00°C) < pKw(25 -C) < pKJ0°C).

Now, at 25 °C, = 14. Therefore, p/Cf100 °C) < 14 < pKw(0°C) and pH + pOH(100°C) < pH + pOH(25°C) < pH+ pOH(0°C) Hence, the value of (pH+ pOH) will be higher at 0°C than at 25 °C , but it will be lower at 100 °C than at 25 °C.

Question 49. The values of pure water at 0°C and 25°C are x and y respectively. Is it greater than or less than y?
Answer: For pure water \(p H=\frac{1}{2} p K_w\) Rise in temperature increases the value of Kw. So, Kw(0°C) < KW(25°C) & hence pKJ0°C) > pKJ100-C) as pKw = -log10. As in the case of pure water, \(p H=\frac{1}{2} p K_w\) pH of pure at 0 °C will be greater than that at 25 °C. Hence, x > y.

Question 50. pKw = 12.26 at 100°C. What is the range of pH -scale at this temperature? What will be the pH of a neutral solution at this temperature?
Answer: At 100 °C, pKw = 12.26. So, at this temperature, the pH scale ranges from 0 to 12.26. At this temperature, \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{K_w}\) in a neutral aqueous solution. Therefore, pH of this solution \(=\frac{1}{2} p K_w=\frac{1}{2} \times 12.26=6.13\)

Question 51. Between 0.1(M) and 0.01(M) acetic acid solutions which one will have a higher pH, and why? the pH of 10-8 (M) HC1 is 8— is it true? Give reasons.
Answer: In an aqueous solution of a weak monoprotic add (example CH3COOH ), \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{c \times K_a}\) where c = molar concentration of the solution and Ka = ionization constant of the weak acid.

Therefore, in 0.1(M) CH3COOH solution, \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{0.1 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and in 0.01(M) CHgCOOII solution, [H30+] \(\sqrt{0.01 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

Hence, the concentration of H₃O+ ions in 0.1(M) CH₆COO₂ solution will be more than that in 0.01(M) CH₆COOH solution. Therefore, the pH of 0.01 (M) CHgCOOH solution will be greater than that of O.l (M) CH₃COO₂ solution.

Question 52. The concentration ofH30+ ions in solution is 1000 times that in solution B. What is the difference between the values of pH of these two solutions?
Answer: Given: [HgO+]a = 1000 x [H30+]fl; where [H₂O+]a and [H₃O+]A are the concentrations of H30+ ions in solutions A and B respectively. Therefore, pH (solution A) = -log10[H₃O+]/1 = -log(103x[H₃O+]B) =- 3- log₁₀[H₃O+]B =- 3 + pH of solution B

Question 53. An aqueous solution o/(NH4)₂SO₄ is acidic. Explain
Answer: (NH4)₂SO₄ is a salt of a weak base (NH3) and a strong acid (H₂SO₄). In its aqueous solution, (NH₄)₂SO₄ dissociates almost completely forming NH- and SO- ions. SO- ion is a conjugate base of strong acid H₂SO₄ and hence in aqueous solution, it is a very weak base in comparison to H₂O.

As a result, the SO- ion does not react with water in aqueous solution. On the other hand, NH₄ is a conjugate acid of weak base NH₃. In an aqueous solution, the NH+ ion shows a higher acidic character than H₂O. As a result, NH₄ ions react with water [NH₄(O-7) + H₂O(l);=NH₃(aq) + H₃O+(aq)] , causing an increase in concentration of H₃O+(aq) ions in the solution. This makes the solution of (NH4)2S04 acidic.

Question 54. What color change does a blue or red litmus paper exhibit when it is put separately in each of the following aqueous solutions? CH₃COONa₂CH₃COONH₄, NH₄C1
Answer: An aqueous solution of CH3COONa is alkaline because in solution CH₃COO- ions resulting from the complete dissociation of CH₃COONa undergo hydrolysis, thereby increasing the concentration of OH-. So, if a red litmus paper is dipped into this solution, it will turn blue

In its aqueous solution, CH₃COONH₄ dissociates completely forming NH+(aqr) and CH₃COO-(aq) ions. Both these ions undergo hydrolysis in water. The important fact is that in water the strength of CH₃COO- ion as a base and the strength of NH₂ ion as an acid are the same. Consequently, the concentration of OH- ions produced due to hydrolysis of CH₃COO- ions:

Equilibrium Constant Calculation

⇒ \(\left[\mathrm{CH}_3 \mathrm{COO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q)\right]\) is almost the same as the concentration of H30+ ions produced due to hydrolysis of NH- ions:

⇒ \(\left[\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)\right]\)

As a result, an aqueous solution of CH₃COONH₄ is neutral. So, if a litmus paper is dipped into this solution, it will not exhibit any color charge.

In its aqueous solution, NH₄C1 dissociates completely forming NH₂(aq) and Cl-{aq). NH₂ ions so formed undergo hydrolysis and thereby increase the concentration of H₃O+ ions in the solution. As a result, an aqueous solution of NH₄C1 is acidic. Therefore, if a blue litmus paper is dipped into this solution, its color will change to red.

Question 55. Why is an aqueous solution of NaNO₃ neutral?
Answer: NaNO₃ is a salt of strong acid HNO₃ and strong base NaOH. In its aqueous solution, NaNO₃ dissociates completely, forming Na+ and NO₃ ions. In an aqueous solution, Na+(aq) is a weaker acid than H₂O and NO₃ is a weaker base than H₂O.

So, neither Na+(a₂) nor NO₂(aq) reacts with water. As a result, there is no change in the concentration of either H₃O+ ions or OH- ions. Due to this, an aqueous solution of NaN03 is neutral

Equilibrium Constant Calculation

Question 56. Arrange the following aqueous solutions in order of their increasing pH values: Na₂CO₃, CH₃COONH₄, and CuSO₄.
Answer: Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂. In its aqueous solution, Na₂CO₃ dissociates completely and forms Na+ and CO₃ ions. In water, COl- is a stronger base than HaO, and hence it reacts with water, forming OH- and HCO₂ ions.

This results in an increase in the concentration of OH ions in the solution of Na₂CO₃ and makes the solution alkaline.

In its aqueous solution, CuSO₄ dissociates completely, forming [Cu(H₂O)₆]2+ and SO- ions. SO- ion, being a conjugate base of strong acid H₂SO₄, is a very weak base and cannot react with water.

In [Cu(H₂O)g]2+, the charge density of the Cu2+ ion is very high, and consequently, H₂0 molecules bonded to the Cu2+ ion get polarized. This causes the weakening of the O — H bonds in the H₂O molecule. Consequently, the O — H bond gets ionized easily, thereby forming an H+ ion. This H+ ion is then accepted by H₂O, which forms the H30+ ion.

⇒ \(\begin{aligned}
& {\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}(a q)+} \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \\
& {\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{OH}\right]^{+}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) }
\end{aligned}\)

Therefore, the hydrolysis of CuSO₄ in its solution increases the concentration of H30+ ions, thereby making the solution acidic. So, the increasing order of pH values of the given aqueous solutions: CuSO₄ < CH₃COONH₄ < Na₂CO₃.

Question 57. Give an example of a salt solution whose pH is independent of salt concentration.
Answer: In the case of a solution of a salt formed from a weak acid and a weak base, the pH of the solution does not depend upon the concentration of the salt. One such salt is CH₃COONH₄.

Question 58. Which of the given salts will undergo cationic or anionic or both cationic and anionic hydrolysis? NH4F, NaCN, AIC1₃, Na₂CO₃, NH₄C1
Answer: Both NH4C1 and A1C13 are the salts of strong acids and weak bases. In aqueous solution of such salts, cationic hydrolysis takes place. NaCN and Na2CO₃ are the salts of weak acids and strong bases. In an aqueous solution of such salts, anionic hydrolysis takes place. NH₄F is a salt of weak acid and weak base. In aqueous solution of such salt, both cationic and anionic hydrolysis take place

Question 59. Which of the following are buffer solutions? Give reasons:

1. 100 mL 0.1 (M)NH₃ + 50 mL 0.1(M) HCI
2. 100 mL 0.1 (M)CH₃COOH + 50 mL0.2 (M) NaOH
3. 100 mL 0.1 (M) CH₃COOH +100 mL 0.15 (M) NaOH
4. 100 mL 0.1 (M) CH₃COONa +25mL0.1(M) HCI
5. 50 mL 0.2 (M) NH₄C1 + 50 mL 0.1 (M) NaOH

Answer:

100 mL of 0.1(M) NH3 = O.Olmol of NH₃ and 50mL of 0.1(M) HCI = 0.005mol of HC1. In the mixed solution, 0.005 mol of HCl reacts completely with the same amount of NH₃, forming 0.005 mol of NH₄C1, and 0.005 mol of NH₃ remains in the solution. Therefore, the mixed solution contains weak base NH3 and its salt NH₄C1. Hence, it is a buffer solution.

Equilibrium Constant Calculation

100mL of 0.1(M) CH₃COOH = 0.01 mol of CH₃COOH and 50mL of 0.2 (M) NaOH = 0.01 mol of NaOH. In the mixed solution, 0.01 mol of CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa. Therefore, the mixed solution is a solution of CH₃COONa. Hence, it is not a buffer solution

100mL of 0.1 (M) CH₃COOH = 0.01 mol of CH₃COOH and 100mL of 0.15(M) NaOH = 0.015 mol of NaOH. In the mixed solution, 0.01 mol CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa, and 0.005 mol of NaOH remains in the solution. Therefore, the mixed solution is a solution of NaOH. Hence, it cannot be a buffer solution.

100mL of 0.1(M) CH₃COONa = O.Olmol of CH₃COONa and 25mL of 0.1(M) HCI = 0.0025 mol of HCI. In the mixed solution, 0.0025 mol of HCl reacts completely with the same amount of CH₃COONa, forming CH₃COOH and NaCl. The number of moles of CH₃COONa remaining in the solution is (0.01- 0.0025) = 7.5 x 10-3 mol. Therefore, the mixed solution contains CH3COOH and its salt CH₃COONa. Hence, it is a buffer solution.

50mL of 0.2 (M) NH₄C1 EE 0.01 mol of NH₄C1 and 50mL of 0.1 (M) NaOH = 0.005 mol of NaOH . In the mixed solution, 0.005mol of NaOH reacts completely with the same amount of NH₄C1, thereby forming NH₃ and NaCl. The number of moles of NH₄C1 left behind in the solution is (0.01-0.005) = 0.005 mol. Therefore, the mixed solution contains NH₃ and NH₄C1. Hence, it is a buffer solution.

Question 60. You are supplied with HCOOH (pKa =3.74), CH₃COOH (pKa=4.74), and NaOH solutions. To prepare a buffer solution of pH =3.8, which acid will you select? Give reason
Answer: The buffer capacity of a buffer solution consisting of a weak acid and its salt becomes maximum when the pH ofthe buffer solution is equal to the pKa of the weak acid. Among the given acids, the pKa of HCOOH is very close to the desired pH of the buffer solution. Hence, one should use HCOOH for preparing the buffer

Question 61. pH of a buffer solution composed of NH₃ and NH₄C1 is 9.26. Will there be any change in pH if 100 mL of distilled water is added to 100 mL of this buffer solution?
Answer: For the buffer solution made up of NH₃ and NH₄C1, the pH of the solutions is given by \(p H=14-p K_b-\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_3\right]}\)

If 100 mL of distilled water is added to 100 mL of this buffer solution, no change occurs in the ratio of [NH₄C1] to [NH₃] and the pH ofthe solution remains the same.

Question 62. The buffer capacity of a buffer solution consisting of a weak acid, HA, and its salt, NaA becomes maximum when its pH is 5.0. At this pH, what is the relation between the molar concentrations of HA and NaA? Abo finds the value of a of HA.
Answer: In the case of a buffer solution consisting of a weak acid and its salt, the buffer capacity of the solution is maximum when pH = pKa. It is given that the buffer capacity of the given buffer is maximum when its pH = 5. Therefore, the pKa ofthe weak acid present in the buffer is 5. Now, the pH of a buffer solution consisting of a weak acid and its salt is given by \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\)

Equilibrium Constant Calculation

Since, \(p H=p K_a=5, \log \frac{[\text { salt }]}{[\text { acid }]}=0\) , [salt] = [acid]

∴ [NaA]=[HA]

Question 63. A, B, and C are three buffer solutions, each of which is composed of a weak acid and its salt. For increasing the pH by 0.02 units, it is found that 1.0, 1.4, and 1.2 millimol of NaOH are required for A, B, and C, respectively. Arrange the solutions in the increasing order of their buffer capacities.
Answer: The higher the buffer capacity of a buffer solution, the greater the amount of a strong acid or a strong base to be required for increasing the pH of the buffer. It is given that increasing the pH of the buffer by the same amount requires a minimum amount of NaOH for buffer A and a maximum amount of NaOH for buffer B. Therefore, the increasing order of buffer capacity of the given buffers is A < C < B.

Question 64. Of the two bottles, one contains HC1 solution and the other a buffer solution. Each of the bottles b labelled as pH = 5. How can you identify the solutions?
Answer: If the pH of the solutions is measured after adding equal drops of NaOH solution to each ofthe solutions, it will be found that pH has increased markedly for one solution, while it remains almost the same for the other solution. The solution for which pH remains almost the same as a buffer solution and the solution for which pH increases is a solution of HC1.

Question 65. At a given temperature, if the solubility product and the
solubility of a sparingly soluble salt M₂X₃ are Ksp and S, respectively, then prove that S \(S=\left(\frac{K_{s p}}{108}\right)^{1 / 5}.\)
Answer: In its saturated solution, M₂X₃ ionizes partially, forming M₃ (aq) and X₂ (aq) ions. Eventually, the following equilibrium is established.

⇒ \(\mathrm{M}_2 \mathrm{X}_3(s) \rightleftharpoons 2 \mathrm{M}^{3+}(a q)+3 \mathrm{X}^{2-}(a q)\)

If the solubility of M₂X₃ in its saturated solution is S mol -L-1 , then in the solution [M3+] = 2S mol -L_1 and [X2-] = 3S mol-L-1 Therefore, the solubility product of M₂X₃

⇒ \(K_{s p}=\left[\mathrm{M}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3=(2 S)^2(3 S)^3=108 S^5\)

∴ \(S=\left(\frac{K_{s p}}{108}\right)^{\frac{1}{5}}\)

Question 66. At a certain temperature, the Ksp of AgCl in water is 1.8 X 10-10. What will be its Ksp is a 0.1M solution of AgNO₃ at some temperature.
Answer: At a certain temperature, the solubility of AgCl decreases in the presence of a common ion (Ag+), but the solubility product of AgCl remains the same. Therefore, Ksp for AgCl in 0.1(M) aqueous solution of AgNO₃ will be the same as that in water.

Equilibrium Very Short Answer Type Questions

Question 1. To find out the equilibrium constant (K) of a reaction, it is compulsory to mention the balanced equation of the reaction—why?
Answer: The value of the equilibrium constant (K) depends on how the balanced equation of the reaction is written, example the formation of H₂ (g) from H₂ (g) and I₂ (g) can be written in two different ways

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

Both 1 and 2 express the same reaction but the coefficients of reactants and products are different. As a result, the value of the equilibrium constant (K) for the reaction will not be the same in the above two cases.

Question 2. For what kind of solids, is solid vapor equilibrium achieved easily?
Answer: Solid substances that can be easily converted to vapors on heating under normal pressure (i.e., sublimable substances). For example, solid CO₂, camphor, ammonium chloride, naphthalene, etc.

Equilibrium Constant Calculation

Question 3. At a fixed temperature, a liquid is in equilibrium with its vapors in a closed vessel. Which measurable tint quantity for the liquid gets fixed at equilibrium?
Answer: When a liquid remains in equilibrium with its vapor in a closed vessel at a particular temperature, the vapor pressure of the liquid is found to acquire a fixed value.

Question 4. The equilibrium established in the evaporation of a liquid at a given temperature is due to the same rate of two processes. What are these two processes?
Answer: The rate of evaporation of the liquid will be equal to the rate of condensation of its vapors at the equilibrium.

Question 5. In a soda water bottle, CO₂ gas remains dissolved in water under high pressure. Write down the equilibrium established in this case.
Answer: CO₂(gas) CO₂ (solution)

Question 6. At 0°C and 1 atm pressure, why is the equilibrium established between water and ice regarded as dynamic?
Answer: At 0°C and under 1 atm pressure, ‘water ice’ equilibrium is said to be dynamic since at equilibrium the rate of melting of ice is equal to the rate of freezing of water.

Question 7. Find Out Kp/Kc for the solutions \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g)\)
Answer: We know that \(K_p=K_c(R T)^{\Delta n}.\)

For the given reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Thus \(K_p=K_c(R T)^{-\frac{1}{2}}\)

⇒ \(\frac{K_p}{K_c}=\frac{1}{\sqrt{R T}}\)

Question 8. Under experimental conditions/ for the reaction, N₂(g) + 3H₂(g).  2NH₃(g), tithe equilibrium mixture contains 17% of NH₃ and 83% of (N₂ + H₂). Under the same conditions, what will be the percentages of NH₃ and (N₂ + H₂) in the reaction 2NH₃(g) N₂(g) + 3H₂(g) at equilibrium?
Answer: Under similar experimental conditions, the same equilibrium will be established for a reversible reaction irrespective of whether the reaction is initiated with the reactants or the products. thus for reaction, 2NH₃(g)N₂(g)+3H₂(g), The same percentages of NH₃ and (N₂ + H₂) will be present at equilibrium as in the case of the reaction N₂(g) +3H₂(g) 2NH₂(g)

Question 9. The values of the equilibrium constant (if) of a reaction at 25°C & 50°C are 2 X 10-4 & 2 X 10-2, respectively. Is the reaction exothermic or endothermic?
Answer: With the increase in temperature, the value equilibrium constant (K) increases for an endothermic reaction, while it decreases for an exothermic reaction. For the reaction, JC(50°C) > K(25°C), indicating it is an exothermic reaction.

Question 10. For a gaseous reaction, Kp > Kc. What will be the effect on equilibrium if pressure is increased at a constant temperature? Will it affect the yields of the products?
Answer: According to the relation Kp = Kc(RT)An, if Kp> Kc, then An > 0. The positive value of An implies that the reaction occurs with an increase in volume in the forward direction. For such a reaction, if pressure is increased at equilibrium, then according to Le Chatelier’s principle the equilibrium will shift to the left and thus the yield ofthe product will decrease.

Question 11. In the case of the thermal decomposition of H2(g) to H(g), which conditions of pressure and temperature will be favourable for an increase in the yield of H(g)?
Answer: Since the formation of H(g) from H₂(g) [H₂(g) 2H(g)] occurs through decomposition, it is an endothermic reaction. Because 2 moles of H(g) are formed from 1 mole of H₂(g), the reaction is associated with a volume increase. So, according to Le Chateliehs principle, the yield of H (g) will increase if the reaction is carried out at a high temperature and low pressure.

Equilibrium Constant Calculation

Question 12. In the case of the reaction A₂(g) + 4B₂(g), i 2AB₄(g), the change in enthalpy (AH) is negative. Mention the conditions of pressure and temperature at which the yield of the product, AB₄(g) will decrease.
Answer: Since AH < 0, it is an endothermic reaction. The volume of the reaction system decreases in the forward direction [1 molecule of A₂(g) combines with 4 molecules of B₂(g) to form 2 molecules of AB₄(g)]. Thus; according to Le Chatelier’s principle, under the conditions of high temperature and low pressure, the yield of AB₄(g) will decrease

Question 13. How will the equilibrium of the reaction, H₂(g) + I₂(g) ?=± 2HI(g) be affected if the volume of the reaction system at equilibrium is doubled, keeping the temperature constant?
Answer: Doubling the volume of the reaction system at equilibrium will reduce the total pressure of the system by half. But for the reaction An = 0. So, according to Le Chatelier’s principle, the equilibrium ofthe reaction will not be affected by a change in pressure.

Question 14. EacB HS- NH3 can act as Bronsted acid and Bronsted base—why? Write the formula of conjugate base and conjugate acid in each case.
Answer: According to the Bronsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. The given species can accept as well as donate protons, so they can act as both acid and base.

Question 15. In the reaction, I2+I- I3, which one acts as a Lewis base?
Answer: In the reaction between and I2, the I- ion donates an electron pair to the I2 molecule, resulting in the formation of the I3 ion [I2 +1- I-]. Therefore, I- ion acts as a Lewis base in this reaction.

Question 16. The pKa values of the three weak acids HA, HB and HC are 4.74, 3.75 and 4.20, respectively. Arrange them in order their of increasing acid strengths.
Answer: As pKa = -log10kTa, the smaller the value of Ka the larger the value of pKa. So, an acid with a larger pKa will have a smaller Ka. As, pJCa(HB) < pKa(HC) < pKa(HA) , pCa(HB) > Kf₂(HC) > Ka(HA). At a certain temperature, a higher value of Ka for an acid indicates a higher strength of the acid. Therefore, the increasing order of acid strengths of the given acids will be— HA < HC < HB.

Question 17. X and Y are two aqueous solutions of added HA with concentrations of 0.1 M & 0.01M, respectively. In which solution will the degree of ionisation of U A be higher
Answer: According to Ostwald’s dilution law, the degree of ionisation of a weak electrolyte increases with the increase in dilution of its aqueous solution. Since, the concentration of solution Y is less than that of X, the degree of ionisation of HA will be higher in solution Y.

Question 18. Which one of the following two acids will have a higher concentration of H30+ ions in their 0.1(M) aqueous solutions—O HC1 and 0 CH₃COOH?
Answer: HC1 is a strong acid, while CH₆COOH is a weak acid. Thus, HC1 ionises almost completely in aqueous solution, whereas CH₃COOH undergoes partial ionisation. As a result, the concentration of H₃0+ ions in 0.1(M) HC1 solution is higher than that in 0.1(M) CH₃COOH solution.

Equilibrium Constant Calculation

Question 19. Show7 that [OH-]>\(\sqrt{K_w}\) in an alkaline solution.
Answer: We know, [H30+] x [OH ] = Kw. In pure water, [H30+] = [OH-]. This gives \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{K_w}\) an alkaline solution, the concentration of OH- ions is higher than that in pure water. Therefore, in an alkaline solution, \(\left[\mathrm{OH}^{-}\right]>\sqrt{K_w} \text {. }\).

Question 20. Will the concentration of HgO+ ions in pure water at 0°C be more than or less than that at 4°C?
Answer: Ionisation of water is an endothermic t process: [2H₂O(1) H₃O+(aq) + OH-(aq)]. Hence, with, a temperature rise, the ionic product of water (Kw) increases. Therefore, KM,(4°C) In pure water, [H₃O+] = Jÿw- Since,(4°C) > Kw(0°C) , the concentration of H₂O+ ions in pure water at 4 °C will be higher than that at 0°C.

Question 21. At a certain temperature, what is the value for the die sum of pH and pOH for an aqueous solution? What will be its value at 25°C?
Answer: In case of any aqueous solution at a certain temperature pH+POH=pk_w. At 25C, Pk_w= 14. Hence at 25C PH+POH=14.

Question 22. An acid bottle labeled pH – 5 Is this add a weak acid
Answer: The cell may be a weak add or a very dilute strong acid. the pH of an added solution depends upon the die concentration of ll30+ Ions in the solution. So, from the value of pH it Is not possible to predict whether the acid IN is weak or strong.

Question 23. At a certain temperature, Kw of pure water =10-12. VVluit will be the pll -range, and pll of the neutral solution at that temperature.
Answer: At the given temperature, Kw = 10-12. Hence, at that temperature, pKw= 12. We know that at a particular temperature, pll -scale ranges from 0 to pKw and pH = \(=\frac{1}{2} p K_w\) for a neutral solution at that temperature. Therefore, at the given temperature, the range of pH -scale will be from 0 to 12 and the pH of a neutral solution at that temperature will be \(\frac {1}{2} p K_w=\frac{1}{2} \times 12=6\)

Question 24. pH of an aqueous 0.1 (M) CH₃COOH solution is 2.87. State whether the pH of the solution will decrease or increase if CH₃COONa is added to this solution.
Answer: pH will increase.

Question 25. At 25°C, the ionization constant (Ka) of weak acid HA is 10-6. What will the value of the ionization constant (Kb) of its conjugate base (A-) be at that temperature?
Answer: We know \(K_a \times K_b=10^{-14} \text { [at } 25^{\circ} \mathrm{C} \text { ]. } K_a \text { of } \mathrm{HA}=10^{-6} \text {. }\)

Therefore \(K_b \text { of } \mathrm{A}^{-}=\frac{10^{-14}}{K_a}=\frac{10^{-14}}{10^{-6}}=10^{-8} .\)

Question 26. Which of the following mixtures will act as buffer solution(s)? O 50 mL 0.1 (M) NH₃ + 100 mL 0.025(M) HC1 0 100mL 0.05(M) CH₆COOH + 50mL 0.1 (M) NaOH
Answer: Will act as a buffer solution.

Question 27. Here are a few salts. Whose aqueous solution(s) at 25°C has (have) a pH greater than 7, less than 7 or equal to 7? (NH₄)2SO₄, CH₃COONH₄, K₂CO₃, NaNO₃
Answer: pH < 7 : (NH₄)2SO₄,pH> 7 : K2CO₂₃ pH = 7 : CH₃COONH₄, NaNO₃

Question 28. A liquid is in equilibrium with its vapor at its boiling point. On average which property of the molecules is equal in two phases?
Answer: At the boiling point of a liquid in equilibrium with its vapor, the average kinetic energy of the molecules in the two phases is equal.

Equilibrium Constant Calculation

Question 29. According to Le Chatelier’s principle, what is the effect of adding heat to a solid and liquid in equilibrium?
Answer: In the equilibrium system solid-liquid, the forward process is endothermic. Therefore, if temperature is increased at equilibrium, then, according to Le Chatelier’s principle, equilibrium will shift to the right, thereby increasing the amount of liquid.

Question 30. At a constant temperature and pressure, what is the value of AG for a reaction at equilibrium?
Answer: Zero

Equilibrium Constant Calculation

Question 31. For a chemical reaction, Kc> 1. Will the value of AG° for this reaction be negative or positive?
Answer: Negative,

Question 32. Find the relation between K and Kc for the following system: 2NaHCO₃(s) Na₂CO₃(s) + CO₂(g) + H₂O(g)
Answer: Kp=Kc(RT)₂

Question 33. The values of the equilibrium constant for a particular reaction at 25°C and 50°C are 0.08 and 0.12 respectively. State whether it is an exothermic or an endothermic reaction.
Answer: Exothermic;

Question 34. If ‘mol-L-1 ‘ and ‘atm’ are the units of concentration and pressure respectively, then what will be the value of Kp/Kc for the reaction, N₂O₄(g) 2NO₂(g) at 300K?
Answer: 24.63L.atmmol-1

Question 35. Give an example of a reaction for which, Kp = Kc = Kx.
Answer: N₂(g) + O₂(g) ⇌ 2NO(g)

Question 36. Consider the reaction: A(s) + 2B₂(g) AB4(g); A₂<O. At equilibrium, what would be the effect of increasing temperature on the concentration of B₂(g)? For tile reaction, N₂O₄(g) ⇌2NO₂(g) A₂> 0
Answer: Increases

Question 37. Mention two factors whose change at the equilibrium of the reaction will increase the yield of NO₂(g).
Answer: Decrease in temperature, increase in pressure at constant temperature

Equilibrium Constant Calculation

Question 38. Give an example of a reaction for which, K /Kc Is independent of temperature.
Answer: H₂(g) + I₂(g)⇌2HI(g)

Question 39. At a given temperature, if the total pressure of the equilibrium A(s) B(g) + C(g) is P, then what is the value of K?
Answer: Kp = P2/2

Equilibrium Constant Calculation

Question 40. What are the conjugate acid and conjugate base of HPO-?
Answer: Conjugate acid: H₂PO₄ , conjugate base: PO3-4

Question 41. What are the conjugate acid and conjugate base of H.,0?
Answer: Conjugate acid: H₃O+

Question 42. Kb for NH3 = 1.8 x 10-5 at 25°C. What is the value of Ka for NH₄  ions at the same temperature?
Answer: 5.55 x 10-10

Question 43. For pure water, pKW=12 at a certain temperature. At this temperature, what is the molar concentration of H₂O+ In a neutral aqueous solution?
Answer: 10-6(M)

Question 44. At a certain temperature, pKw for pure water is 12. At the same temperature, the concentration of H30+ in aqueous solution is 10-8 mol L-l. Is this solution acidic or basic?
Answer: Basic

Question 45. pH of a sample of pure water is x at a certain temperature. Find the value of pKw at that temperature.
Answer: 2X

Equilibrium Constant Calculation

Question 46. pKa values for two acids, HA and HB are 4 and 6 respectively. Which one is the stronger acid?
Answer: HA

Question 47. pKa values for two acids, HA, and HB are 4 and 5 respectively. If each of the acid solutions has a concentration of 0.1(M), then in which solution the molar concentration of H₃0+ ions is greater?
Answer: HA

Question 49. At 25 °C, pKa for weak acid, HA = x, and pKb for A-, the conjugate base of HA = y. Find the value of (x + y) at this temperature.
Answer: (x+y)=14

Question 50. Give examples of two salts in case of which the pH value of their aqueous solutions is independent of their concentrations.
Answer: CH₃

Equilibrium Constant Calculation

Question 51. For a weak acid, HA, pKa = 5. What is the effective range of pH for a buffer comprised of HA and its salt, NaA?
Answer: 4 to 6

Question 52. Give examples of two buffer solutions prepared by mixing two salt solutions of polybasic acid.
Answer: (NaH₂PO₄ ,Na₂HPO₄ )and (NaH₆P₃ , Na₂CO₃)

Question 53. In a buffer solution comprising a weak base (B) and its ion (BH+), [B] = [BH+]. If Kb for the weak base =10-5, then find the pH value ofthe buffer solution.
Answer: 9

Question 54. 100 mL of 0.05(M) NaOH solution is added to a 100 mL of 0.1(M) NaH₂PO₄ solution. Will the mixed solution act as a buffer?
Answer: Yes

Equilibrium Short Answer Type Questions

Question 1. The unit of the equilibrium constant of the reaction, A + 3B nC is L₂.mol-2. What is the value of
Answer: For the given \(K_c=\frac{[C]^n}{[A][B]^3}\)

Thus, the unit of kc \(=\frac{\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^n}{\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right) \times\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)^3}\)

= (mol .L-1)n-4

Hence, L₂.mol-2 = (mol . L-1)n-4 or, n = 2

Question 2. Find out the value of Kp/Kc for the reaction \(\mathrm{PCl}_5(g) \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\) at 298K, consider the unit of concentration is mol L-1 and the unit of pressure is atm.
Answer: We know. Kp = Kc(RT)An For the given reaction, An = (1 +1-1) = 1

Thus, \(K_p=K_c \times R T \quad \text { or, } \frac{K_p}{K_c}=R T\)

⇒ \(\begin{aligned}
& =0.0821 \mathrm{~L} \cdot \operatorname{atm} \cdot \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K} \\
& =24.465 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Equilibrium Constant Calculation

Question 3. Mention two ways by which the equilibrium of the L-given reaction can be shifted to the right?
Answer: According to Le Chatelier’s principle, the equilibrium of the above reaction can be shifted to the right by the addition of excess reactants [i.e., CH₃COOH(Z) or C₂H₅OH(Z) ] or by the removal of the products [i.e., CH₃COOC₂H₅(Z) or H₂O(Z)] from the reaction system at a given temperature, keeping the volume ofthe reaction system constant.

Question 4. When steam is passed over a red-hot iron, H₂ gas is produced. In this reaction, the yield of H₂(g) is found to increase when the partial pressure of steam is increased. Explain.
Answer: \(\text { Reaction: } \mathrm{Fe}(s)+4 \mathrm{H}_{-} \mathrm{O}(\rho) \rightleftharpoons \mathrm{Fe}_{-} \mathrm{O} \cdot(\mathrm{s})+4 \mathrm{H}_{-}(g)\) A steam is one of the reactants in the above reaction, increasing its partial pressure at equilibrium will shift the equilibrium position to the right. As a result, the yield of the product i.e., H₂(g) will increase.

Question 5. Why does not the equilibrium constant expression for a reaction involving pure solids or liquids contain the concentration terms of the solids or liquids?
Answer: For a pure solid or liquid, the molar concentration is directly proportional to density. As the density is constant at a given temperature, therefore, for a pure solid or liquid at a constant temperature, the molar concentration is a constant quantity/ which is usually taken as unity (1). For this reason, The equilibrium constant expression for a reaction involving pure solids or liquids does not contain concentration terms of the solids or liquids.

Question 6. Find out the equilibrium constant for the reaction, XeO₄(g) + 2HF(g)iXeO₃F₂(g) + HaO(g) Consider K₁ as the equUibrium constant for the reaction, XeF₆(g) + HaO(g) XeOF₄(g) + 2HF(g) and K₂ as the equilibrium constant for the reaction, XeO₄(g) + XeF₆(g) XeOF₄(g) + XeO₃F₂(g).
Answer: According to the given condition

⇒ \(K_1=\frac{\left[\mathrm{XeOF}_4\right] \times[\mathrm{HF}]^2}{\left[\mathrm{XeF}_6\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]} \text { and } K_2=\frac{\left[\mathrm{XeOF}_4\right] \times\left[\mathrm{XeO}_3 \mathrm{~F}_2\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{XeF}_6\right]}\)

∴ \(\text { For } \mathrm{XeO}_4(g)+2 \mathrm{HF}(g) \rightleftharpoons \mathrm{XeO}_3 \mathrm{~F}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)

Equilibrium constant ,\(K=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{HF}^2\right.}\)

⇒ \(\text { Now, } \frac{K_2}{K_1}=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times[\mathrm{HF}]^2}=K\)

∴ K= \(\frac{K_2}{K_1}\)

Question 7. How will the following reaction equilibrium be affected if the volume of each reaction system is increased at a constant temperature?
Answer: At constant temperature, if the volume of the reaction system is increased, then the total pressure at equilibrium will be decreased. Thus, the equilibrium of the system will be disturbed. According to Le Chatelier’s principle, equilibrium will be shifted, in a direction that increases the total number of molecules.

In the case ofthe first reaction, the equilibrium will shift to the left side. As a result, the yield of the product, [SO₃] will decrease. On the other hand, in the case ofthe second reaction, the equilibrium will shift to the right side “thereby increasing the yield of the product [CO(g)].

Equilibrium Constant Calculation

Question 8. The equilibrium of any -reversible reaction may be shifted to the left or reality changing the conditions. Will this change cause any alteration in the value of the equilibrium constant?
Answer: At ascertain temperature The equilibrium constant of a reversible reaction has a definite value.

If temperature remains fixed, then the equilibrium can be shifted to the left or right by changing the conditions of pressure, temperature, etc: on which the equilibrium of a reversible reaction depends. Consequently, the respective amount of both reactants and products will change, but the value of the equilibrium constant remains unchanged since the temperature remains fixed.

If the equilibrium is shifted due to temperature change, then the amounts of both reactants and products as well as the value of the equilibrium constant will change. With the increase in temperature, the value of the equilibrium constant will increase for an endothermic reaction and decrease for an exothermic reaction.

Question 9. At constant temperature, the following reaction is at equilibrium in a closed container: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})\) At constant temperature, if the amount of the solid carbon is reduced to half at equilibrium, then what will be the change in the concentration of CO(g)?
Answer: At a particular temperature, the concentration of any pure solid is independent of its amount. Thus, keeping the temperature constant, if the amount of solid carbon is reduced to half at equilibrium, then its concentration will remain unchanged. So, the concentration of CO(g) will also remain unaffected.

Question 10. At constant temperature, if the pressure is changed at the equilibrium of a gaseous reaction, then will the values of Kp, Kc, and Kx change?
Answer: we known \(\Delta G^0=-R T \ln K_p \quad \text { or, } K_p=e^{-\frac{\Delta G^0}{R T}}\) where G° = standard free energy change ofthe reaction.

The value of AG° depends only on temperature. Its value is independent of pressure. So, the value of Kp is independent of pressures.

We know, Kp = K(RT)An. Since the value of Kp does not depend on pressure, according to this relationship, the value of Kc is also independent of pressure.

Again, we know, Kp=Kx(p)n or \(K_x=\frac{K_p}{(P)^{\Delta n}}\)

As Kp does not depend on pressure, according to this relation, the value of Kx depends on pressure. However if An = 0, then pressure will not affect Kx.

Question 11. How can the yield of the products be increased by changing the volume of the reaction system in the given reactions at constant temperature?

⇒ \(\begin{aligned}
& \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \\
& 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)
Answer: In this case, the volume increases in the reaction as written (since, An = +1 ). Thus, if the volume of the reaction system is increased at a constant temperature, then equilibrium will shift to the right, and consequently, the yields ofthe products will increase.

In this case, the volume decreases in the reaction as written (since, An = -3 ). Thus, if the volume of the reaction system is decreased at constant temperature then equilibrium will shift to the right. As a result, the yield of products will increase.

Equilibrium Constant Calculation

Question 12. Mention two factors for which die yields of the products in the given reaction increase.

⇒ \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})  \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \text {-heat }\)

Answer: Keeping both temperature and volume fixed, if we add some reactants [CO(g) or H₂O(g)] to the reaction system or remove some products [CO₂(g) or H₂(g) ] from the reaction system, equilibrium will shift to the right, which will result in higher yields ofthe products.

The reaction is endothermic. Thus, on increasing the temperature at equilibrium, the equilibrium of the reaction will shift to the right. This will cause higher yields of the products

Question 13. At a particular temperature, for the reaction, aA + bB dC + cD, the equilibrium constant is K. Find out the equilibrium constants for the following reactions at the same temperature.\(m a A+m b B m d C+m c D\)\(\frac{1}{m} a A+\frac{1}{m} b B\frac{1}{m} d C+\frac{1}{m} c D\)
Answer: For the reaction, aA + bB dC+cD the equilibrium constant \(K=\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\)

Now, 1 For the reaction maA + mbB mdC + mcD equilibrium constant, \(K_1=\frac{[C]^{m d} \times[D]^{m c}}{[A]^{m a} \times[B]^{m b_3}}\)

\(=\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^m=K^m\)

For the reaction \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d C+\frac{1}{m} c D\) equilibrium constant, \(K_2=\frac{[C]^{d / m} \times[D]^{c / m}}{[A]^{a / m} \times[B]^{b / m}}\) \(=\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^{1 / m}=(K)^{1 / m}\)

Question 14. What would the effect on the yield of products be If the temperature of the following reaction systems Is changed at equilibrium?
Answer: Reaction is endothermic (as AH > 0 ). If the temperature is increased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On the other hand, if the temperature is reduced at the equilibrium of the reaction, then the equilibrium will shift to the left, and the die yield of products will be reduced.

The reaction is exothermic (as AH < 0 ). If the temperature is decreased at the equilibrium of this reaction, then the equilibrium will shift to the right, and the yield of products will increase. On die other hand, if the temperature is increased at the equilibrium of the reaction, then equilibrium will shift to the left, and the yield of products will be reduced.

Equilibrium Constant Calculation

Question 15. Identify Lewis acids and Lewis bases in the following reactions and give reasons: SiF₄ + 2F-S₁F₂-6 \(\mathrm{RMgX}+2\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \ddot{\mathrm{O}}: \rightarrow \mathrm{RMg}\left[\mathrm{O}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2\right]_2 \mathrm{X}\)
Answer: According to Lewis’s concept, an acid is a substance that can accept one or more electron pair(s). Generally, cations (such as Ag+, H+, K+ ), compounds having a central atom with an incomplete octet (such as SiF₄, A1F₃, RMgX, BF₃), and compounds whose central atom is linked to an electronegative atom by a double bond (such as GO₂ ) can act as Lewis acids. In the given reactions, Lewis acids are SiF₄, RMgX, Ag+, and H+.

According to Lewis’s concept, a base is a substance that can donate one or more electron pair(s). Anions (such as F-, OH₂Si₂) and compounds with lone pairs of electrons can act as Lewis bases. Therefore, in the given reactions, Lewis bases are F-, NH₃, (C₂H₅)₂O, and NH₃.

Question 16. State the nature of aqueous solutions containing the following ions with reason: NH+4, F-, Cl-.
Answer: NH+: NH₂ ion is the conjugate acid of a weak base, NH₃. In an aqueous solution, NH₂ is stronger than H₂O [weak Bronsted acid]. Hence, in an aqueous solution, NH₂ ion reacts with water to produce H₃0+ ions, leaving the die solution acidic.

Equilibrium Constant Calculation

NH+4(aq) + H₂O(l) =± NH₃(a<7) + H₃O+(aq)

F-: F- ion is the conjugate base of a weak acid, HF. F- is stronger titan H₂O [weak Bronsted base] in aqueous solution. For this reason, in an aqueous solution, the F- ion reacts with water to give OH- ions, and as a result, the solution becomes basic. \(\mathrm{F}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{OH}^{-}(a q)\)

Cl-: Cl- is the die conjugate base of strong acid, HCl. Hence, it is very weak and cannot react with water. Consequently, the die aqueous solution of Cl- is neutral.

Question 17. The solubility of zinc phosphate in water is S mol. L-1 . Derive the mathematical expression of the solubility product of the compound.
Answer: The following equilibrium is established by zinc phosphate, [Zn3(P04)2] in its saturated aqueous solution.

⇒ \(\mathrm{Zn}_3\left(\mathrm{PO}_4\right)_2(s) \rightleftharpoons 3 \mathrm{Zn}^{2+}(a q)+2 \mathrm{PO}_4^{3-}(a q)\)

∴ \(K_{s p}=\left[\mathrm{Zn}^{2+}\right]^3 \times\left[\mathrm{PO}_4^{3-}\right]^2\)

As given, the solubility of Zn3(P04)2 is S mol. L-1 . Hence, \(\left[\mathrm{Zn}^{2+}\right]=3 S \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{PO}_4^{3-}\right]=2 S \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ Ksp = (3S)₃ X (2S)₂ = 108S5; this is the mathematical expression of the solubility product of zinc phosphate.

Equilibrium Constant Calculation

Question 18. At a certain temperature, the ionization constants of two weak acids, HA and HB are Ka₁ and’ ka₂, respectively. If Ka₁> Ka₂ and the concentration of the aqueous solutions of both acids is 0.1(M), then which solution will have a higher pH?
Answer: As the ionization constant of HA is higher than that of HB, the degree of ionization ofHA in its 0.1 M solution is higher than that of HB in its 0.1 M solution. As a result, the molar concentration of H₃0+ions in 0.1(M) HA solution will be more than that in 0.1(M) HB solution. Therefore, the pH of the HB solution will be greater than that of the HA solution.

Equilibrium Constant Calculation

Question 19. What will be the change in concentrations of H₃O+ & OH- and the ionic product of water (Kw) if NaOH is added to pure water at a certain temperature?
Answer: Since Kw is fixed at a certain temperature, it will not undergo any change due to the addition of NaOH in pure water. However, the concentration of OH- ions increases due to the addition of NaOH, causing the dissociation equilibrium of H2O to shift to the left

⇒ \(\left[\mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right]\)

As a result, the concentration of H3O+ ions in the solution is reduced

Question 20. 20 mL of 0.15(M) HC1 solution is mixed with 50 mL of 0.1(M) CH₃COONa solution. State whether the mixed solution will act as a buffer or not.
Answer: Number of millimoles of CH₃COONa₃ in 50 mL 0.1(M) CH₆COONa₃ =0.1 X 50 = 5 and that of HC1 in20mL 0.15(M)HC1= 0.15X20 = 3.

The reaction between CH6COONa2 and HCl is:

CH₃COONa(O + HCl(aq) -4 CH₃COOH(aq) + NaCl(aq) Hence, 3 millimol of HC1 + 3 millimol of CH₆COONa → 3 millimol of CH₃COOH + 3 millimol of NaCl . Therefore, at the end of the reaction, there remains 3 millimol of CH₃COOH and (5-3) = 2 millimol of CH₃COONa.

∴ The resulting solution consists of weak acid (CH₃COOH) and its salt (CH₃COONa). So, it acts as a buffer.

Question 21. The pH of a buffer solution remains almost unchanged even after dilution—explain.
Answer: Let a buffer solution that consists of a weak acid (HX) and its salt (MX). For this buffer, \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\) At a certain temperature, the value of pKa is constant. Hence, at a given temperature, the pH of the buffer solution depends upon the ratio ofthe molar concentrations ofthe salt to the acid in the solution. As the solution is diluted, this ratio remains the same. As a result, the pH ofthe solution remains unaltered.

Question 22. Consider the salts given below. For which salt(s) will the pH of the aqueous solution(s) be independent of the concentration of the salt? CH₃NH₃1, (NH4)₃PO₄, KCN and (NH4)₂CO₃.
Answer: The pH of an aqueous solution of the salt of a weak acid and a weak base does not depend upon the concentration of salt.

Among the given salts, (NH₄)₃PO₄ and (NH₄)2CO₃ are produced from a weak acid and weak base. Hence, the pH of their aqueous solutions does not depend on the concentration of the salt.

Equilibrium Constant Calculation

Question 23. When H₂S gas Is passed through an acidified solution of Cu2+ and Zn2+, only CuS is precipitated—why?
Answer: H₂S is a very weak acid. In its aqueous solution, only a small fraction of it ionizes to produce H₃O+ and S2- ions

⇒\(\left[\mathrm{H}_2 \mathrm{~S}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q)\right]\)

In an acidified solution of H₂S, due to the common ion (H₃O+) effect, the ionization of H₂S is further reduced, and as a result, the concentration of S2- ions becomes extremely low.

When H₂S gas is passed through an acidified solution of Cu2+ and Zn2+ ions, the concentration of S2- ions becomes so low that only the product of the concentrations of Cu2+ and S2- ions exceeds the solubility product of CuS. However, the product of the concentrations of Zn2+ and S2- ions lies well below the value of the solubility product of ZnS. This is why only CuS is precipitated in preference to ZnS.

Question 24. What will happen when a solution of potassium chloride is added to a saturated solution of lead chloride? Give reason.
Answer: When potassium chloride solution is added to a saturated lead chloride solution, then the solubility of lead chloride decreases due to the common ion (Cl-) effect.

Explanation: The following equilibrium is established in an aqueous PbCl2 solution: PbCl2(s) Pb2+(ag) + 2Cl-(ag) The addition of KC1 to the saturated solution of PbCl2 increases the concentration of common ion Cl- the above equilibrium to get disturbed. To re-establish the equilibrium, some of the Cl- ions will combine with an equivalent amount of Pb2+ ions to form solid PbCl2. Therefore, as an overall effect, the equilibrium is shifted to the left. Hence, the solubility of PbCl2 decreases.

Question 25. Why does not MgS04 form any precipitate when it reacts with NH₃ in the presence of NH₄C1?
Answer: NH₃ is a weak base. In an aqueous solution, it ionizes partially to produce NH+ and OH- ions.

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

In the presence of NH4C1, owing to the common ion effect of NH4, the degree of ionization of NH3 is further suppressed. Thus, the concentration of OH- ions decreases to a large extent.

At this low concentration of OH- ions, the product of the concentration of Mg2+ ions and square of the concentration of OH- ions (as ATsp[Mg(OH)₂] = [Mg2+][OH-]₂) cannot exceed the solubility product of Mg(OH)₂, i.e., [Mg2+] x [OH-]2 < Ksp (solubility product). As a result, Mg(OH)₂ does not get precipitated.

Question 26. Why is the aqueous solution of Cu(N03)2 acidic?
Answer: Being a strong electrolyte, Cu(NO₃)₂ dissociates almost completely in its aqueous solution to form [Cu(H₂O)g]2+ and NO₂ ions. In the solution, H₂O also ionizes slightly to form H₃O+ and OH- ions

⇒ \(\begin{gathered}
\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}(a q)+2 \mathrm{NO}_3^{-}(a q) \\
2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)
\end{gathered}\)

NO-3 is the conjugate base of a strong acid, HNO3. Hence it is a very weak base. So, NO₂ ions cannot react with water. On the other hand, since the charge density of Cu2+ ion in the cationf[Cu(H₂O)₆]2+ is very high, the H₂O molecules attached to it get polarised, and their O—H bonds become weaker.; These O —H bonds dissociate to produce H+ ions, which are accepted by H20 to give H30+ ions.

⇒ \(\begin{aligned}
{\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}(a q)+} & \mathrm{H}_2 \mathrm{O}(l) \\
& \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{OH}\right]^{+}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)
\end{aligned}\)

As a result, the concentration of H₃O+ ions in the aqueous solution of Cu(NO₃)2 becomes higher than that of OH- ions (from H₂O), and consequently, the solution becomes acidic.

Equilibrium Constant Calculation

Question 27. Despite being a neutral salt, the aqueous solution of Na2CO₃ is alkaline—why?
Answer: Na₂CO₃, being a strong electrolyte, dissociates almost completely in aqueous solution, forming Na+ and CO₃ ions:

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3(a q) \rightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_3^{2-}(a q)\)

Water, being a very poor electrolyte, ionizes slightly to form H3O+ and OH- ions: \(2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Hydrated Na+ ion (Na+(aq) ] is a very weak acid and cannot react with water.

Equilibrium Constant Calculation

CO₂3-, the conjugate base of weak acid HCO-3, is a stronger base than H₂O (which is a Bronsted base). Therefore, it reacts with water in aqueous solution, forming the following equilibrium.

⇒ \(\mathrm{CO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(a q) \rightleftharpoons \mathrm{HCO}_3^{-}(a q)+\mathrm{OH}^{-}(a q)\)

HCO-3 so formed undergoes ionization

⇒ \(\mathrm{HCO}_3^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_3^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)]\)

But, due to the common ion (CO2-3) effect, this ionization occurs to a very small extent.

As a result, [OH-] in solution is much higher than [H+]. So, aqueous solution of Na2C03 is alkaline.

Question 28. HPO₂-4 can act both as a Bronsted base and as a Bronsted acid. Write the equation of equilibrium established by HPO₂– as an acid and a base in an aqueous solution. Also, write the expressions of Ka & Kb in two cases. 0 What are the conjugate acid and base of HS-?
Answer: HPO₂-4 can denate and accept protons in aqueous solution. Thus it can serve both as an acid and base.

⇒ \(\begin{aligned}
& \mathrm{HPO}_4^{2-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{PO}_4^{3-}(a q)+\mathrm{H}_3 \mathrm{O}_2^{+}(a q) \\
& \mathrm{HPO}_4^{2-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

⇒ \(K_a=\frac{\left[\mathrm{PO}_4^{3-}\right] \times\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HPO}_4^{2-}\right]} ; K_b=\frac{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}\)

HS-(aq) + H₂O(l) H₂S(aq) + OH-(aq)

Hence, the conjugate acid of HS- is H₂S.

HS-(aq) + H₂O(l) S2-{aq) + H₃O+(aq)

Therefore, the conjugate base of HS- is S₂-.

Question 29. Will the pH of pure water at 20°C be lower or higher than that at 50°C?
Answer: Ionic product of water (Kw) increases with, a temperature rise Hence \(K_w\left(50^{\circ} \mathrm{C}\right)>K_w\left(20^{\circ} \mathrm{C}\right)\) or, \(p K_w\left(50^{\circ} \mathrm{C}\right)<p K_w\left(20^{\circ} \mathrm{C}\right).\) (since pKw=-logl0Kw) Now, for pure
water \(p H=\frac{1}{2} p K_w\)

Equilibrium Constant Calculation

Question 30. An aqueous solution of sodium bisulfate is acidic, whereas an aqueous solution of sodium bicarbonate is basic—Explain.
Answer: Since sodium bisulfate (NaHSO₄) is a salt of strong acid (H₂SO₄) and strong base (NaOH), it is not hydrolyzed in an aqueous solution. NaHS04 in its solution dissociates completely to form Na+ and HSO₄ ions and HSO₄ ions so formed get ionized to form H₃O+ and SO2-4 ions. As a result, the aqueous solution of NaHSO₄ becomes acidic.

⇒ \(\begin{gathered}
\mathrm{NaHSO}_4(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HSO}_4^{-}(a q) \\
\mathrm{HSO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)
\end{gathered}\)

NaHCO₃ in its solution dissociates completely to form Na+ and HCO₃ ions [NaHCO₃(aq)-Na+(aq) + HCO₃ (aq) ]. HCO₃ can act both as an acid and a base in aqueous solution.

⇒ \(\begin{aligned}
& \mathrm{HCO}_3^{-}(a q)[\text { Acid }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{CO}_3^{2-}(a q) \\
& \mathrm{HCO}_3^{-}(a q)[\text { Base }]+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Question 31. Both CuS and ZnS are precipitated if H₂S gas is passed through an alkaline solution of Cu2+ and Zn2+. Explain.
Answer: In an aqueous solution, H₂S ionizes to establish the following equilibrium, H₂S(aq) + 2HzO(Z) 2H₃O+(aq) + S2-(aq) The degree of ionization of H₂S increases in alkaline solution because OH- ions present in the solution react with H₃O+ ignite) form unionized water molecules. This shifts the equilibrium to the right, thereby increasing the concentration of S2- ions? ‘in the presence of a high concentration of S2- ions, [Cu2+] (s2-] > and [Zn2+][S2-] > Ksp(ZnS). As a result, both CuS and ZnS are precipitous.

Question 32. Calculate the formation constant of [Ag(NH₃)₂]+
Answer: \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; K_1=3.5 \times 10^3\)

⇒ \(\begin{aligned}
{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; } \\
K_1=1.7 \times 10^3 \ldots
\end{aligned}\)

Equation (3) represents the formation reaction of [Ag(NH₃)₂]2+. Therefore, the equilibrium constant for the reaction represented by equation (3) will be the formation constant for [Ag(NH₃)₂]2+. As equation (3) is obtained by adding equations (1) and (2), the formation constant (fcy) for [Ag(NH₃)₂]2+ equals ky x k2.

Thus, k_y = k_y X k₂ = (3.5 X 103) X (1.7 X 103) = 5.95 X 106

Question 33. Write the correct order of increasing acid strength among HCO₃< HSO₃ < H₃O+ < HSO₃F.
Answer: HSO₃F is a super acid. So, its strength is the maximum among all the given acids. Comparing the dissociation constants of the remaining acids gives the order in terms of dissociation constant:

Now, the larger the value of Ka for an acid, the higher the strength ofthe acid, So, the increasing order of acid strengths of the given acids will be HCO₃< HSO₃ < H₃O+ < HSO₃F.

Question 34. The first and second dissociation constants of an acid H2A are 1 x 10-5 and 5 X 10-1° respectively. Calculate the value of the overall dissociation constant.
Answer: \(\begin{aligned}
\mathrm{H}_2 \mathrm{~A}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HA}^{-}(a q) ; \\
K_1=1 \times 10^{-5} \ldots(1)
\end{aligned}\)

⇒ \(\begin{array}{r}
\mathrm{HA}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}^{2-}(a q) ; \\
K_2=5 \times 10^{-10} \ldots
\end{array}\)

The overall dissociation reaction is the sum of the. reactions (1) and (2). H₂A(aq) + 2H₂O(Z) 2H30+(aq) + A3-(aq) Thus, the overall disputation constant K = K₁ x K₂ =  x 10-5) x (5 X 10-1°) = 5 X 10-15

Equilibrium Constant Calculation

Question 35. a -D-glucose Beta -D glucose, the equilibrium constant for this is 1.8. Calculate the percentage of a -D glucose at equilibrium.
Answer: Suppose, the initial concentration of a -D glucose is a mol L-1 and its degree of conversion to p -D glucose at equilibrium is x mol-L-1 -1

Therefore, the concentration of a -D glucose and D glucose at equilibrium will be as follows—

⇒ \(\begin{array}{ccc}
\begin{array}{c}
\text { Initial concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a & 0 \\
\begin{array}{c}
\text { Equilibrium concentration } \\
\left(\text { in mol } \cdot \mathrm{L}^{-1}\right. \text { ) }
\end{array} & a-a x & a x
\end{array}\)

Equilibrium Constant Calculation

Equilibrium constant for this process, \(K=\frac{[\beta-\mathrm{D}-\text { glucose }]}{[\alpha-\mathrm{D}-\text { glucose }]}\)

⇒ \(1.8=\frac{a x}{a(1-x)}=\frac{x}{1-x}\) or, \(x=\frac{1.8}{2.8}=0.6428\)

Thus, the percentage of a -D-glucose at equilibrium is \(\frac{a(1-0.6428)}{a} \times 100=35.72 \%\)

Question 36. Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 x 10-4 M Na₂CO₃ solution. At what concentration of Ba2+ will a precipitate begin to form? ( Ksp for BaCO₃ = 5.1 X 10-9)
Answer: Ba(NO₃)₂ reacts with Na₂CO₃ to form BaCO₃. BaCO₃ is a sparingly soluble compound that forms the following equilibrium in its saturated solution.

⇒ \(\mathrm{BaCO}_3(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{CO}^{2-}(a q)\)

For \(\mathrm{BaCO}_3, K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]\)

In the solution \(\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M}\)

BaCO₃ will precipitate when [Ba2+][C02-] → Ksp, i.e., when \(\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_3^{2-}\right]>5.1 \times 10^{-9}\left(\text { as } K_{s p}\left[\mathrm{BaCO}_3\right]=5.1 \times 10^{-9}\right)\) \(\text { If }\left[\mathrm{CO}_3^{2-}\right]=1 \times 10^{-4} \mathrm{M} \text {, then }\left[\mathrm{Ba}^{2+}\right]>\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}\) or, [Ba2+] > 5.1 X 10-5 M Thus, the precipitation of BaC03 will start when [Ba2+] in the solution is grater than 5.1 x 10-5 M.

Question 37. 2.5 m l of \(\frac{2}{5} \mathrm{M}\) weak monoacdic base Kb = 1 x 10-125at 25°C ) is titrated with \(\frac{2}{15}\) in water at 25°C. Calculate the concentration of H + at the equivalence point. (kw = 1 x 10-14)
Answer: Ana. 2.5 mL of \(\frac{2}{15}\) M monobasic acid \(\equiv \frac{2}{5} \times 2.5 \equiv 1\) mmol of the base.

Suppose, VmL of \(\frac{2}{15} \mathrm{M}\) HC1 is required for the neutralisation. \(V \mathrm{~mL} \text { of } \frac{2}{15} \mathrm{M} \mathrm{HCl} \equiv \frac{2}{15} \times V \mathrm{mmol} \mathrm{HCl} .\)

Therefore \(\) mmol HCL will neutralise 1mmol of the base Hence \(\frac{2 \times V}{15}=1 \text { or, } V=7.5 \mathrm{~mL}\)

The total volume of the solution after neutralization = (2.5 + 7.5) mL = 10 mL.

The number of mmol of the salt formed in the neutralization =1 mmol.

So, the concentration of the salt in the final solution \((C)=\frac{1}{10}=0.1 \mathrm{M}\) As the resulting salt is formed from a weak base and a strong acid, it undergoes hydrolysis. The pH of the solution of such a salt is given by the pH
\(=7-\frac{1}{2} p K_b-\frac{1}{2} \log C\) \(\text { As } K_b=10^{-12}, p K_b=-\log _{10}\left(10^{-12}\right)=12\)

Therefore \(p H=7-\frac{1}{2} \times 12-\frac{1}{2} \log (0.1)=7-6+0.5=1.5\) and the concentration of H+(aq) ions at the equivalence point is [H+] = 10-PH M = 10-1-5 M = 0.316 M.

Equilibrium Constant Calculation

Equilibrium Multiple Choice Questions

Question 1. Some reactions, their equilibrium constants are as follows:

\(\begin{aligned}
& \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) ; K_1 \\
& \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) ; K_2 \\
& \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) ; K_3
\end{aligned}\)

The relation among K₁, K₂ and K₃ is—

1. \(K_3 \times K_2^3=K_1^2\)
2. \(K_1 \sqrt{K_2}=K_3\)
3. K₂ x K₃ =K₁
4. K₃ = k₁xk₂

Answer: 4. K₃ = k₁xk₂

Question 2. At a given temperature, the reaction, A(g) 20(g), is In equilibrium In a closed flask. At the same temperature, the reaction, C(g) D(g) + O(g) Is in equilibrium in another closed flask. The values of equilibrium constants of these two reactions are Kp and Kp respectively and the total pressures of the equilibrium mixtures are P₁ and P₂ respectively. If: Kp₂ =1:4 and P₁ P₂ = then the ratio of the degree of dissociation of (g) and C(g) are (assume the degrees of dissociation of bodi care very small compared to —

1. 0.15
2. 0.5
3. 1.0
4. 1.5

Answer: 2. 0.5

Question 3. At a given temperature, the reaction, A₂(g) + B₂(g) 2A₂(g) was started with 0.4 mol of A₂(g) and 0.0 mol of H₂(g) in a flask of volume 2L. When the reaction achieved equilibrium, it was found that the reaction mixture contained 0.5 mol of AO. (The equilibrium constant ( Kp) for the reaction is—

1. 8.30
2. 4.76
3. 10.27
4. 6.49

Answer: 2. 4.76

Equilibrium Constant Calculation

Question 4. At a given temperature, if the degree of dissociation of N₂O₄ in the following reaction N₂O₄(g) 2NO₂(g) is a, and the total pressure of the equilibrium mixture is P, then it can be shown that the equilibrium constant for the reaction, Kp = a₂P (assuming a is very small compared to 1). Which ofthe following comments is true for this relation-

1. Kp increases as a increases
2. Kp increases as P increases.
3. value of Kp does not depend on P but depends on a
4. The value of Kp depends neither on P nor on a

Answer: 4. The value of Kp depends neither on P nor on a

Question 5. For the reactions \( \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g),\) \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {, }\) If the equilibrium constants are Kx, K2 and K3 respectively, then the equilibrium constant for the reaction \(4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) is

1. \(\frac{K_2 K_3^2}{K_1}\)
2. \(\frac{K_2^2 K_3^2}{K_1}\)
3. \(\frac{K_1^3 K_2^2}{K_3}\)
4. \(\frac{K_2^2 K_3^6}{K_1^2}\)

Answer: 4. \(\frac{K_2^2 K_3^6}{K_1^2}\)

Question 6. For a hypothetical reaction, Kc = 0.9 and Kp = 538. Which of the following equations pan represents the reaction properly at 25°C-

1. \(A(g) \rightleftharpoons 2 C(s)+D(g)\)
2. \(B(g) \rightleftharpoons C(l)+D(l)\)
3. \(A(l)+2 B(g) \rightleftharpoons 2 C(g)\)
4. \(A(g)+B(s) \rightleftharpoons 3 C(g)\)

Answer: 4. \(A(l)+2 B(g) \rightleftharpoons 2 C(g)\)

Equilibrium Constant Calculation

Question 7. When a mixture containing N₂ and H₂ in the molar ratio 1:3 heated in presence of a catalyst in a closed vessel, die following equilibrium is established: \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\) At equilibrium, if the mole fraction of NH₃ is 0.6 and the total pressure ofthe equilibrium mixture is 10 atm then Kp for the reaction, \(2 \mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g)\)

1. 1.33 atm^-2
2. 0.75 atm^-2
3. 1.333 atm^-2
4. 0.75 atm^-2

Answer: 1. 1.33 atm^-2

Question 8. The reaction, \(A(g)+4 B(g) \rightleftharpoons 2 C(g)+3 D(g)\) is carried out in a closed vessel of volume 2L by taking 3 mol of A(g) and 4 mol of 5(g). At equilibrium, if the amount of C(g) be1 mol, then Kc for the reaction is-

1. 0.056
2. 0.038
3. 0.084
4. 1.24

Answer: 3. 0.084

Equilibrium Constant Calculation

Question 9. The total pressure at the equilibrium of the reaction, XY(g) X(g) + Y(g) is P. If the equilibrium constant P for the reaction is Kp and \(K_p=\frac{P}{8}\), then the percent dissociation of XY is—

1. 30.49%
2. 33.33%
3. 41.90%
4. 19.26%

Answer: 2. 33.33%

Equilibrium Constant Calculation

Question 10. At 500K, for the reaction, PCl₅(g) PCl₃(g) + Cl₂(g) , the equilibrium constant, Kp = 0.52 . In a closed container, these three gases are mixed. If the partial pressure of each of these gases is 1 atm, then in the reaction system—

1. The number of moles of PCl₅ will increase
2. The number of moles of PCl₃ will increase
3. The reaction will attain equilibrium when 50% of the reaction gets completed
4. The reaction will attain equilibrium when 75% of the reaction is completed

Answer: 1. The number of moles of pc1₅ will increase

Equilibrium Constant Calculation

Question 11. For the reaction: \(2 A(g)+B(g) \rightleftharpoons 3 C(g)+D(g)\), two moles of each A and B were taken into a flask which of the following relation between the concentration terms is true when the system attains equilibrium

1. [A]=[B]
2. [A]<[B]
3. [A]=[B]
4. [A]>[B]

Answer: 2. [A]<[B]

Question 12. The reaction, 2A(g) + 5(g) C(s); AH < 0, is in equilibrium in a closed vessel. Which of the following changes at equilibrium’ will increase the yield of C (s)—

1. Temperature is increased
2. At constant volume and temperature, some amount of 5(g) is added to the reaction system
3. At constant volume and temperature, some amount of c (s) is removed from the reaction system noriw
4. Pressure is decreased at a constant temperature

Answer: 2. At constant volume and temperature, some amount of 5(g) is added to the reaction system.

Question 13. At 300 K, the reaction A(g) + 5(g) C(s), 1 is in equilibrium in a closed, vessel. At the beginning of the reaction, the partial pressures of A and B gases are 0.2 and 0.3atm respectively and the total pressure of the equilibrium mixture is 0.3atm. Kc, for the reaction, is—

1. 6.06×104 L₂.mol^-2
2. 2.59 x 103 L₂.mol^-2
3. 3.03 x 104 L₂.mol^-2
4. 8.2 x 10-2 L₂.mol^-2

Answer: 3. 3.03 x 104 L₂-mol^-2

Question 14. At 300K, for the reaction, \(\mathrm{AB}_3(\mathrm{~g}) \rightleftharpoons \mathrm{AB}_2(\mathrm{~g})+\frac{1}{2} \mathrm{~B}_2(\mathrm{~g}),\) Kp = 1.66. At the same temperature, AG° for the reaction, 2AB₂(g) + B₂(g) 2AB₃(g) is-

1. +2.19kJ
2. -2.52KJ
3. +3.85KJ
4. -3.26kJ

Answer: 2. -2.52KJ

Equilibrium Constant Calculation

Question 15. At a given temperature, when a reversible reaction is carried out in the absence of a catalyst, the ratio of the rate constants for the forward and reverse reactions is found to be 8.0. At the same temperature, if the reaction is carried out in presence  of a catalyst, then the ratio will be—

1. >8.0
2. <8.0
3. =8.0
4. <8.0

Answer: 3. =8.0

Equilibrium Constant Calculation

Question 16. At a given temperature, a closed vessel contains NH₃ gas and solid NH₄HS . The pressure of NH₃ gas in the vessel is 0.50atm. On dissociation, NH₄HS produces NH₃ and H₂S gases. The total pressure in the flask at equilibrium is 0.84 atm. The equilibrium constant for the dissociation reaction (Kp) of NH₄HS is—

1. 0.30 atm^-2
2. 0.16 atm^-2
3. 0.11 atm^-2
4. 0.22 atm^-2

Answer: 3. 0.11 atm^-2

Question 17. N₂O₄ is dissociated to 33% and 40% at total pressures P₁ and P₂ atm respectively. Hence, the ratio of P₁ to P₂ is—

1. \(\frac{7}{3}\)
2. \(\frac{8}{3}\)
3. \(\frac{8}{5}\)
4. \(\frac{7}{4}\)

Answer: 3. \(\frac{8}{5}\)

Question 18. The reaction, A(g) + 25(g) 2C(g) + D(g) was studied using an initial concentration of 5 which was 1.5 times that of. The equilibrium concentrations of A and C were found to be equal. So, Kc for the equilibrium is—

1. 4
2. 0.32
3. 2.73
4. 8.17

Answer: 2. 0.32

Question 19. The equilibrium constants Kp1 and Kp2 for the reactions x ⇌2y and x⇌ p + q, respectively are in the ratio of 1: 9 . If the degree of dissociation and Z is equal then the ratio of total pressure at this equilibrium is

1. 1:36
2. 1:1
3. 1:3
4. 1:9

Answer: 1. 1:36

Question 20. If the equilibrium constant for Mutrotion or-D-glucose .-D-glucose is the percentage ofthe a -form in the equilibrium mixture is

1. 64.5
2. 35.7
3. 53.7
4. 44.8

Answer: 2. 35.7

Question 21. The reaction, C(s) + CO₂(g)⇌2CO(g), is at equilibrium in a closed vessel under a given set of conditions. If the degree of dissociation of CO₂ at equilibrium is a and the total pressure of the equilibrium mixture and the value of equilibrium constant are P and Kp respectively, then a

1. \(\frac{K_p}{\sqrt{2 P}}\)
2. \(\frac{1}{2} \sqrt{\frac{K_p}{p}}\)
3. \(\frac{\sqrt{K_p}}{P}\)
4. \(\sqrt{\frac{P}{K_p}}\)

Answer: 4. \(\sqrt{\frac{P}{K_p}}\)

Equilibrium Constant Calculation

Question 22. At a given temperature, the equilibrium constant, Kc, for the reaction, A + B C is 10. At the same temperature, the reaction is allowed to occur in a closed vessel of volume 1L. At a particular moment during the reaction, if the amounts of A, B, and C in the reaction system are 0.1, 0.4, and 0.3 mol respectively, then—

1. The reaction is in equilibrium at that moment
2. The reaction will occur to a greater extent towards the left to attain equilibrium
3. The reaction will occur to a greater extent towards the right to attain equilibrium
4. The reaction will occur to a greater extent towards the left for achieving equilibrium and concentrations of reactants and product will be the same at a new equilibrium

Answer: 3. The reaction will occur to a greater extent towards the right to attain equilibrium

Question 23. A mixture containing N₂ and H₂ in a mole ratio of 1: 3 is allowed to attain equilibrium when 50% ofthe mixture has reacted. If P is the pressure at equilibrium, then the partial pressure of NH₃ formed is-

1. p/3
2. p/2
3. p/9
4. p/5

Answer: 1. p/3

Question 24. If the concentration of OH- ions in the reaction Fe(OH)₃(s) ⇌ Fe3+(aq) + 30H-(aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+ will be increased by—

1. 64 times
2. 4 times
3. 8 times
4. 16 times

Answer: 1. 64 times

Question 25. The equilibrium constant (Kp) for the decomposition of gaseous H₂O, \(\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g)\) is related to degree of dissociation (a ) at a total pressure (P) as-

1. \(\kappa_p=\frac{\alpha^3 p^{1 / 2}}{(1+\alpha)(2+\alpha)^{1 / 2}}\)
2. \(K_p=\frac{\alpha^3 p^{3 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)
3. \(K_p=\frac{\alpha^{3 / 2} p^2}{(1-\alpha)(2+\alpha)^{1 / 2}}\)
4. \(K_P=\frac{\alpha^{3 / 2} p^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)

Answer: 4. \(K_P=\frac{\alpha^{3 / 2} p^{1 / 2}}{(1-\alpha)(2+\alpha)^{1 / 2}}\)

Question 26. 2 mol of PCl₅(g) is heated at a given temperature in a closed vessel of volume 2L. As a result, PCl₅(g) dissociates and forms PCl₃(g) and Cl₂(g). When the dissociation reaction reaches equilibrium, it Is found that 50% of PClg(g) has dissociated. Kc for the reaction is

1. 0.15
2. 0.30
3. 0.25
4. 0.5

Answer: 4. 0.5

Equilibrium Constant Calculation

Question 27. At a given temperature, the reaction, SO₂Cl₂(g) SO₂(g) + CI₂(g), is in a state of equilibrium in a closed vessel. At constant temperature and volume, if some amount of He gas is added to the reaction system, then —

1. The concentration of SO₂(g) will increase
2. The concentration of SO₂Cl₂(g) will increase
3. The concentrations of SO₂(g) , Cl₂(g) , SO₂cl₂(g) will remain the same
4. The value of the equilibrium constant will decrease

Answer: 3. The concentrations of SO₂(g) , Cl₂(g) , SO₂cl₂(g) will remain the same

Question 28. The reaction, C(s) + CO₂(g);⇌2CO(g), is in a state of equilibrium in a closed vessel at a constant temperature. The equilibrium of the reaction will shift towards the left and get re-established if at constant temperature and volume some amount of—

1. C(s) is removed from the reaction system
2. CO₂(g) is added to the reaction system
3. CO₂(g) is removed from the reaction system
4. CO(g) is removed from the reaction system

Answer: 3. CO₂(g) is removed from the reaction system

Equilibrium Constant Calculation

Question 29. The pair of compounds which cannot exist together in solution is

1. NaHCO₃ and NaOH
2. Na₂CO₃ and NaHCO₃
3. Na₂CO3 and NaOH
4. NaHCO₃ and NaCl

Answer: 1. NaHCO₃ and NaOH

Equilibrium Constant Calculation

Question 30. Equimolar solutions of the following were prepared in water separately. Which of the solutions will have the highest pH—

1. Srcl₂
2. Bacl₂
3. MgC₂
4. Cacl₂

Answer: 2. Bacl₂

Question 31. A student wants to prepare a saturated solution of Ag+ ion. He has only three samples of Ag— AgCl (Ksp = 1.8 X 10-18), AgBr (Ksp = 5 x 10-13), and Ag₂CrO₄ (Ksp = 2.4 x 10-12). Which compound should he take’ to obtain maximum [Ag+] —

1. Agcl
2. AgBr
3. Ag₂CrO₂₄
4. None of these

Answer: 3. Ag₂CrO₄

Question 32. The correct relationship between the pH of picomolar solutions of sodium oxide (pH₂), sodium sulfide (pH₂), sodium selenide (pH₃), and sodium telluride (pH₄) is

1. pH₁>pH₂>pH₃>pH₄
2. pH₁ <pH₂<pH₃ < pH₄
3. pH₁<pH₂<pH₃<pH₄
4. pH₁>pH₂=pH₃>pH₄

Answer: 1. pH₁>pH₂>pH₃>pH₄

Question 33. Solubility product constant (Ksp) of salts of types MX₁, MX₂ & M₃X at temperature TK are 4.0 x10-8, 3.2 x 10-14 & 2.7 x 10-13 respectively. Solubilities (mol dm-3) of the salts at temperature T are in the order—

1. M₃>MX₂>M₃X
2. M₃X>MX₂>MX
3. MX₂>M₃X>MX
4. MX>M₃X>MX₂

Answer: 4. MX>M₃X>MX₂

Equilibrium Constant Calculation

Question 34. If the solubilities of AgCl in H₂O,0.01(M) CaCl₂ , 0.01(M) NaCl and 0.05(M) AgNO₃ are S₁, S₂, S₂ Ansd S₂ respectively, then—

1. S₁>S₂>S₃>S₄
2. S₁ > S₂ = S₃ > S₄
3. S₁ > s₃ > S₂ > S₄
4. S₄>S₂>S₃>S₁

Answer: 3. S₁ > s₃ > S₂ > S₄

Question 35. The degree of hydrolysis of a salt of weak acid and weak base in its 0.01(M) solution is found to be 50%. If the molarity of the solution is 0.2(M), the percentage hydrolysis of the salt should be—

1. 100%
2. 50%
3. 25%
4. 10%

Answer: 2. 50%

Question 36. The first and second dissociation constants of an acid H₂A are 1.0 x 10-5 and 5.0 x 10-I° respectively. The overall dissociation constant of the acid will be

1. 5.0×10-5
2. 5.0×1015
3. 5.0×10-15
4. 0.2×105

Answer: 3. 5.0×10-15

Equilibrium Constant Calculation

Question 37. If three salts P₂X, QY₂, and RZ₃ have the same solubilities in water then the correct relation among their Ksp values is—

1. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)<K_{s p}\left(\mathrm{RZ}_3\right)\)
2. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)>K_{s p}\left(\mathrm{QY}_2\right)=K_{s p}\left(\mathrm{RZ}_3\right)\)
3. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)=K_{s p}\left(\mathrm{RZ}_3\right)\)
4. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)>K_{s p}\left(\mathrm{QY}_2\right)>K_{s p}\left(\mathrm{RZ}_3\right)\)

Answer: 1. \(K_{s p}\left(\mathrm{P}_2 \mathrm{X}\right)=K_{s p}\left(\mathrm{QY}_2\right)<K_{s p}\left(\mathrm{RZ}_3\right)\)

Question 38. The pH of the solution was obtained by mixing 20 mL of 0.01 (M) Ca(OH)₂ and 30 of 0.1 (M) HC1 solution is

1. 6.32
2. 9.85
3. 11.3
4. 4.74

Answer: 3. 11.3

Question 39. The pH of an aqueous solution of MCI is 3,0 and that of an aqueous solution of NaOH is 12. The pH of the solution obtained by mixing 100mL of NaOH solution with 500 ml, of HC1 solution’ is

1. 6.71
2. 10.92
3. 12.05
4. 3.08

Answer: 2. 10.92

Question 40. At 25°C, the pH of 0.1(M) aqueous solution of NH₃ is 11.13. At the same temperature, the pH of a ‘solution containing 0.1 (M) of N1₄C1 and 0.01 (M) of NH₃ is-

1. 4.74
2. 6.25
3. 8.26
4. 9.34

Answer: 3. 8.26

Equilibrium Constant Calculation

Question 41. 800ml. 0.1 (M) HC1 solution is mixed with 200ml. 0.5(M) CH₃NH₂ solution. In the resulting solution, traction of H₃O+ ions is = fix 10 is

1. 3×10-5(M)
2. 1.25 x 10-4(M)
3. 8 x 10-11 (M)
4. 7.2 X 10-10(M)

Answer: 3. 8 x 10-11 (M)

Question 42. A solution of a weak acid ( Kn = 10-5) has a molarity of (M/5). lol of this solution is neutralized completely with a NaOH solution of molarity (M/20). At the neutra¬ lisation point, concentration of H(Of ions (mol-L-1 ) is

1. 4.39x 10-5
2. 1.25 x10-6
3. 7.02 x 10-8
4. 1.58 x10-9

Answer: 4. 1.58 x10-9

Equilibrium Constant Calculation

Question 43. At 25°C, K for PbCI, is 1.6 x 10-5 in water. At the same temperature, the amount of PbCl₂ (molar mass = 278.19 g-mol-1 ) that remains dissolved in 100 mL of a saturated solution of PbCL, is—

1. 0.28g
2. 0.44g
3. 0.17g
4. 0.35g

Answer: 2. 0.44g

Equilibrium Constant Calculation

Question 44. At 25°C, the solubility product for Gd(OH)₂ in water is 1.2 X 1O-14. What would be the pH of tut aqueous solution of 0.01 (M) Cd2ÿ ions when Cd(OH)₂ starts precipitating—

1. 4.29
2. 5.60
3. 8.04
4. 7.56

Answer: 3. 8.04

Question 45. At 25°C the solubility product of a salt AB0 in water is 4.0 x 10- 15. If 0.1 mol of A2- ions are added to 1 1. of a saturated solution of the salt (assuming the volume of the solution does not change on the addition of A2- ions), then—

1. The solubility product of ab2 will increase
2. The solubility product of ab2 will decrease
3. The cone, of b- ions in the solution will be 2 x it)-7 mol l.-1
4. The solubility of ab2 in solution will be 4×10- 10. Mol-l-1

Answer: 3. Cone, of b- ions in solution will be 2 x it)-7 mol l.-1

Question 46. At 25°C, Ksp for Al(OH)₃.( in water is 2 x 10-33 aqueous solution of ph = 13, the solubility of AI(OH)₃ is 2 x 1 0-x. The value of x is

1. 10
2. 15
3. 22
4. 30

Answer: 2. 15

Question 47. At 25°C, K(l for a weak acid, HA in water, is 10- V₂ ml. of 0.1 (M) NaOH solution is added to K, mL of 0.1 (M) solution of HA. How many times would Vl he of K, so that the pH of the solution is 6

1. 2 times
2. 1.5 times
3. 1.1 times
4. 1.4 times

Answer: 3. 1.1 times

Equilibrium Constant Calculation

Question 48. Which of the relations are correct for the given physical change; \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) 

1. \(K_p=p_{\mathrm{H}_2 \mathrm{O}}^2\)
2. Kc = [H₂O(g)]₂
3. Kp – KC(RT)₂
4. Kc = Kp(RT)₂

Answer: 1. \(K_p=p_{\mathrm{H}_2 \mathrm{O}}^2\)

Question 49. At a given temperature, Kc = 6.3 x 10-e for the reaction S-(g) 4S₂(g). At the same temperature, if the lead Ion is started with 2 moles of S0(g) and 0.2 mol of S₂(g) In a closed vessel of volume 1 L, then which of the following comments are true regarding this reaction—

1. At the beginning of the reaction, Qc = 8.0 x 10-4
2. The reaction will occur to a greater extent towards the right To attain equilibrium
3. The reaction will occur to a greater extent towards the left to attain equilibrium
4. The concentration of S₈ at equilibrium is greater than 2 mol. L-l.

Answer: 1. At the beginning of the reaction, Qc = 8.0 x 10-4

Question 50. 4NH₃(g) + 5O₂(g)⇌4NO(g) + 6H₂O(g), is in equilibrlum In a closed container of volume 1L at a given temperature. If the reaction is started with 1 mol NH₃(g) and 1 mol of O₂(g) and the number of mol of H20(g) at equilibrium is 0.6 mol, then at equilibrium-

1. [NH₃] = [NO]
2. [NO]<[O₂]
3. [NO] > [NH₃]
4. [O2] < [H₂O]

Answer: 2. [NO]<[O₂]

Equilibrium Constant Calculation

Question 51. The reaction, fe(s) + H₂O(g) ⇌ CO(g) + H₂(g); ΔH > 0, is in equilibrium. At equilibrium—

1. If the temperature is increased, the partial pressure of H20(g) will decrease
2. The concentration of H₂(g) will decrease if an inert gas is added at constant temperature and volume
3. the concentration of CO(g) will increase if pressure is increased at a constant temperature
4. The equilibrium will move towards the right if an inert gas is added at constant temperature and pressure.

Answer: 1. If the temperature is increased, the partial pressure of H₂O(g) will decrease

Question 52. The reaction, 2NOCl(g)⇌ 2NO(g) + Cl₂(g) ; ΔH>0, Is In equilibrium. Winch of the following changes at equilibrium will decrease The yield of NO(g) — 

1. At constant temperature and volume, some amount of nocl(g) is lidded to the reaction system u
2. At constant temperature and volume, some amount of Cl₂(f> ) is added to the reaction system
3. Temperature is decreased at equilibrium
4. At constant temperature and pressure, some airfoil of the gas is added to the reaction system

Answer: 2. At constant temperature and volume, some amount of Cl₂(f> ) is added to the reaction system

Question 53. The vapor density of the equilibrium mixture of NO₂ and N₂O₄ is found to be 40 for the given equilibrium N₂O₄(g) 2NO₂(g). For the given equilibrium—

1. 1 Mole percent of nO₂ present in the mixture is 59%
2. 1 Mole percent of nO₂ present in the mixture is 26%
3. The degree of dissociation of n₂O₄ is 0.45
4. The degree of dissociation of n₂O₄ is 0.15

Answer: 2. 1 Mole percent of nO₂ present in the mixture is 26%

Question 54. The vapor pressure of liquid methanol at 50°C is 55.5 kPa. These are correct for the equilibrium reaction attained in a closed vessel of 5 L at 50°C for the following equilibrium CH₃OH(Z) CH₃OH(g)

1. Kp = 55.5kPa
2. Kc = 0.021 mol- L-1
3. K = 0.555
4. K = 0.555kPa

Answer: 1. Kp = 55.5kPa

Equilibrium Constant Calculation

Question 55. The reactions in which the yield of the products cannot be increased by the application of high pressure are—

1. 2SO₂(g) + O₂(g) ⇌2SO₃(g)
2. NH₄HS(s)⇌ NH₃(g) + H₂S(g)
3. N₂O₄(g)⇌2NO₂(g)
4. N₂(g) + 3H₂(g)⇌2NH₃(g)

Answer: 2. NH₄HS(s) NH₃(g) + H₂S(g)

Question 56. Aqueous solutions of which of the following compounds on dilution do not suffer any change in pH value—

1. PhCOONH₄
2. NH₄CN
3. HCOONa

Answer: 1. PhCOONH₄

Question 57. Which can act as an acid as well as a base—

1. SO₄^2-
2. HS-
3. HCO^-3
4. HSO^-4

Answer: 2. HS-

Equilibrium Constant Calculation

Question 58. Which mixtures (in molar ratio) can act as buffer—

1. H₂CO₃ + NaOH (3:2)
2. H₂CO₃ + NaOH (3: 4)
3. NH₃ + HC1 (5: 4)
4. NH₃ + HC1 (4: 5)

Answer: 1. H₂CO₃ + NaOH (3:2)

Question 59. If equal volumes of the given solutions are mixed, precipitation of AgCl (Ksp = 1.8 x 10-11) will occur only with—

1. 10-4(M)Ag+ and 10-4(M)Cl-
2. 10-5(M)Ag+ and 10-5(M)C1-
3. 10-6(M)Ag+ and 10-6(M)Cr
4. 10-10(M)Ag+ and l0-10(M)Cl-

Answer: 1. 10-4(M)Ag+ and 10-4(M)Cl-

Question 60. Which ofthe following is true regarding H₃PO₄—

1. Ka = Ka₁ X Ka₂ X Ka₃
2. Ka₁<Ka₂<Ka₃
3. ka₁ >ka₂ > ka₃
4. ka₁=ka₂=ka₂

Answer: 1. Ka = Ka₁ X Ka₂ X Ka₃

Equilibrium Constant Calculation

Question 61. Select the buffer solutions-

1. 0.8(M) H₂S+0.8(M)KHS
2. 2(M) C₆H₅NH₂+2(M)C₆H₅Nh₃Br
3. 3(M)H₂Co₃+3(M)KHCO3
4. 0.05(M) KClO₄+0.05(M) HClO₄

Answer: 1. 0.8(M) H₂S+0.8(M)KHS

Question 62. The solubility Of BaSO₆ will be almost the same in

1. 0.1(M) H2O₄
2. 0.1(M) Ba(OH)₂
3. 0.1(M)BA(NO3)₂
4. 0.2(M) HCl

Answer: 1. 0.1(M) H₂SO₄

Equilibrium Constant Calculation

Question 63. 7he special that can act both and as Bronsted adds and as Bronsted base in water are-

1. H₂PO-4
2. PO3-4
3. HCO-3
4. [Fe(H2O)₆]3+

Answer: 1. H2PO-4

Question 64. Among the following salts, whose aqueous solutions will turn blue litmus paper red-

1. NaHCO₃
2. FeCl₃
3. KCN
4. C₆H₅NH₃Cl

Answer: 2. NaHCO₃

Question 65. At a given temperature, the first and the second ionization constants of the acid, H2A, are 10×10-5 and 5,0 x 10 10 respectively. Which of the following comments are true regarding this add—

1. The concentration of A2+ Ions In 0.01 (M) aqueous solution of 1 f2A Js 0,01 (Ml
2. The overall Ionisation constant for H2A is 5.0 x. 10″15
3. In 0,01 aqueous solution of H2A, the molar concentration of H30* Ions Is twice that of A2″ ions.
4. In 0,01 (M) aqueous solution of H2A, (H3O+) x [HA-)

Answer: 2. The concentration of A2- Ions In 0.01 (M) aqueous solution of if 2A Js 0,01 (Ml)

Equilibrium Constant Calculation

Question 66. At 25C, Kb for CN- (the conjugate base of HCN) is 2.5 x 10-5, If 25rnf, of 0,01 (M) aqueous NaOH solution Is added to 50ml, of0,0 1 (M) HCN solution, then-

1. The pH of the solution is 1 1.2
2. The pH of the respiting solution is 9.4
3. At 20*c, the ionisation constant for him is 4 x 10-18
4. At 23*c, the ionisation constant for him is 2.5 x 10-5

Answer: 2. Ph of the respiting solution is 9.4

Question 67. At a given temperature, If the solubility products for MX, MA₂ and M., B2 In the water are 10-22, and 10-33 respectively and their solubilities are S₁, S₂ and S₃ mol-1 respectively, then

1. S₁<S₃
2. S₂>S₃
3. S₂>S₁
4. S₂=S₃

Answer: 1. S1<S3

Question 68. At 25-C temperature, the solubility products for BaCrO₄ and SrCrO₄ salts are 2,4×10-10 and 3.6 x 10-6 respectively. If an aqueous solution of K₂CrO₄ is added drop by drop to an aqueous solution containing Ba²⁺ and Sr²₄ ions with concentrations of 10-4 and 10-3(M) respectively, then—

1. BaCrO₄ will be precipitated first
2. SrCrO₄ will be precipitated first
3. The concentration of sr²⁺ ions will be 6.6 x 10-8 mol-l-1 when ba²⁺ ions start precipitating
4. The concentration of ba²⁺ ions will be 6,6 x lct8 mol-l-1 when sr²⁺ ions start precipitating.

Answer: 1. BaCrO₄ will be precipitated Erst

Equilibrium Constant Calculation

Question 69. In a buffer solution composed of NaCN and HCN [pKa = 9.4), [NaCN] = 0.2(M) and [HCN] = 0.4(M). An aqueous solution contains Zn²⁺, Ca²⁺, Mn²⁺, and Cr³⁺ ions, each of which has a concentration of 0.1(M). If 500mL of the buffer solution is added to 500mL of this aqueous solution, then the ions that will precipitate in the resulting solution are—

1. Zn⁺
2. Ca²⁺
3. Mn²⁺
4. Cr³⁺

Answer: 1. Zn⁺

Equilibrium Constant Calculation

Question 70. At 25°C, pKb(NH₃) = 4.74, pCa(HF) = 3.14 and pATa(HCN) = 9.4. Hence—

1. An aqueous solution of nh₄f is acidic
2. Aqueous solution of nh₄cn is acidic
3. The pH of an aqueous solution of nh4cn is greater than that of an aqueous solution of nh4f
4. Ph values of the solutions of both nh4cn and nh4f are independent of the concentrations ofthe solutions

Answer: 1. Aqueous solution of nh4f is acidic

Question 71. Which of the following comments is true—

1. If pKa values for the acids HA and HB are 4 and 5 respectively, then the concentration of OH- ions in 0.1(M) aqueous solution of HB will be greater than that in 0.1(M) aqueous solution of HA
2. pH of pure water at 0°C is smaller than that at 25°C
3. The degree of hydrolysis of NH₄F in its 0.1 (M) and 0.2(M) aqueous solutions is the same at a particular temperature
4. pH of an acid is 5, implying that the acid is weak.

Answer: 1. If pKa values for the acids HA and HB are 4 and 5 respectively, then the concentration of OH- ions in 0.1(M) aqueous solution ofHB will be greater than that in 0.1(M) aqueous solution of HA

Equilibrium Constant Calculation

Question 72. A certain buffer solution contains equal concentrations of A- and HA. Kb for A- is 10-10. Hence—

1. Ka for HA is 10^-3
2. ka for HA is 10^-4
3. pH of the buffer is 4
4. PH of the better is 9

Answer: 2. ka for HA is 10-4

Question 73. A buffer solution containing NH3 and NHÿCl has a pH value of 9. pKb for NH3 is 4.7. If in the buffer solution, the total concentration of buffering reagents is 0.6 mol I.-1, then the amount of—

1. NH₃ in the solution is 3.4g.L^-1
2. NH₄C1 in the solution is 8.9g.L^-1
3. NH₄C1 in the solution is 21.4g.L^-1
4. NH₃ in the solution is 17.5g.L^-1

Answer: 1. NH₃ in the solution is 3.4g.L^-1

Equilibrium Fill In The Blanks

Question 1. At a given temperature for a reversible reaction, K< 1. The rate constant of the forward reaction is___________________ than that ofthe reverse reaction.
Answer: Lower

Question 2. If K<1 for a reaction at a particular temperature, then the value AG° is___________________
Answer: Positive

Question 3. For a gaseous reaction, K > Kc. An increase in pressure at constant temperature will ___________________ the product.
Answer: Decrease

Equilibrium Constant Calculation

Question 4. If C(s) is added the concentration of to the reaction system C(s) + CO₂(g) 2CO(g) at constant temperature and volume, then the concentration of CO₂(g) will___________________.
Answer: Remain Unchanged

Question 5. For the reaction, COCl₂(g) ?=± CO(g) + Cl₂(g); AH > 0, an increase in temperature at equilibrium will increase the concentration of___________________
Answer: CO and Cl₂

Question 6. The addition of inert gas to the reaction system, PCl5(g)⇌ PCl₃(g) + Cl₂(g) at constant___________________does do not affect the state of equilibrium.
Answer: Temperature, Volume.

Equilibrium Constant Calculation

Question 7. At a particular temperature, pKw for pure water =12. Its pH will be___________________
Answer: 6

Question 8. The first and second ionization constants of H2S in its aqueous solution are x and y respectively. So, x is _ than y.
Answer: Greater

Equilibrium Constant Calculation

Question 9. The conjugate base of [Al(H20)g]3+ is___________________.
Answer: [Al(H₂O)₅OH]₂+

Question 10. The conjugate acid and conjugate base of HPO2- are respectively.
Answer: \(\mathrm{H}_2 \mathrm{PO}_4^{-}, \mathrm{PO}_4^{3-}\)

Question 11. At 25°C, in an aqueous solution of HA, Ka for HA = 1(T6. Kb for A+ ion is___________________
Answer: 10-8

Question 12. If the solubility of Ag3P04 in its saturated aqueous solution is S(M), then, its solubility product will be ___________________
Answer: 27S₄

Question 13. The solubility of Ag₂CrO₄ ___________________in an aqueous solution of than its solubility in pure water.
Answer: Lower

Question 14. Addition of CH₃COONa to an aqueous solution of CH₆COOH ___________________AgNO₃ is the pH value.
Answer: Increases