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Multiple Choice Questions

1. $\text{Ag}^+ + \text{NH}_3 \rightleftharpoons [\text{Ag}(\text{NH}_3)]^+ ; k_1 = 3.5 \times 10^{-3}$

$[\text{Ag}(\text{NH}_3)]^+ + \text{NH}_3 \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+ ; k_2 = 1.7 \times 10^{-3}$

Then the formation constant of $[\text{Ag}(\text{NH}_3)_2]^+$ is [JEE 2006]

(a) $6.08 \times 10^{-6}$

(b) $6.08 \times 10^6$

(d) None

Answer: (a)

Explanation: $K_f = k_1 \times k_2 = (3.5 \times 10^{-3}) \times (1.7 \times 10^{-3}) = 5.95 \times 10^{-6} \approx 6.08 \times 10^{-6}$.

2. Which is correct statement if catalyst is added at equilibrium condition? [JEE 2006]

(a) The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction

(b) The condition for equilibrium is $\sum \nu_i G_i = 0$ where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent

(d) Catalyst will not alter the rate of either of the reaction

Answer: (b)

Explanation: The condition for equilibrium is $\sum \nu_i G_i = 0$ where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.

3. The equilibrium constant at 298 K for a reaction $A + B \rightleftharpoons C + D$ is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in $\text{mol L}^{-1}$) will be: [JEE Main 2016]

(a) 0.182

(b) 0.818

(d) 1.182

Answer: (c)

Explanation:

$$A + B \rightleftharpoons C + D$$

At $t = 0$: $1$, $1$, $1$, $1$

At $t = t_{eq}$: $1-x$, $1-x$, $1+x$, $1+x$

$$\frac{(1+x)^2}{(1-x)^2} = 100 \implies \frac{1+x}{1-x} = 10 \implies 1+x = 10 - 10x \implies 11x = 9 \implies x = \frac{9}{11}$$

$$[\text{D}] = 1 + \frac{9}{11} = \frac{20}{11} = 1.818\text{ M}$$

4. The thermal dissociation equilibrium of $\text{CaCO}_3(s)$ is studied under different conditions. $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ For this equilibrium, the correct statement(s) is (are): [JEE Adv. 2013 P-II]

(a) $\Delta H$ is dependent on T

(b) K is independent of the initial amount of $\text{CaCO}_3$

(d) $\Delta H$ is independent of the catalyst, if any

Answer: (a, b, d)

Explanation: For the equilibrium $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, the equilibrium constant ($K$) is independent of initial amount of $\text{CaCO}_3$ whereas at a given temperature it is independent of pressure of $\text{CO}_2$. $\Delta H$ is independent of catalyst and it depends on temperature. Hence (a), (b) and (d) are correct.

5. The % yield of ammonia as a function of time in the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), \Delta H < 0$ at $(P, T_1)$ is given below: If this reaction is conducted at $(P, T_2)$ with $T_2 > T_1$, the % yield of ammonia as a function of time is represented by: [JEE Adv. 2015 P-I]

(a)

(b)

(c)

(d)

Answer: (b)

Explanation: Increasing the temperature lowers equilibrium yield of ammonia ($\Delta H < 0$). However, at higher temperature the initial rate of forward reaction would be greater than at lower temperature that is why the percentage yield of $\text{NH}_3$ would be more initially.

6. For the following reactions, equilibrium constants are given [JEE Main 2019]

$$\text{S}(s) + \text{O}_2(g) \rightleftharpoons \text{SO}_2(g) ; K_1 = 10^{52}$$

$$2\text{S}(s) + 3\text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) ; K_2 = 10^{129}$$

The equilibrium constant for the reaction, $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ is:

(a) $10^{181}$

(b) $10^{154}$

(d) $10^{77}$

Answer: (c)

Explanation:

$$\text{S}(s) + \text{O}_2(g) \rightleftharpoons \text{SO}_2(g) ; K_1 = 10^{52} \quad \text{... (i)}$$

$$2\text{S}(s) + 3\text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) ; K_2 = 10^{129} \quad \text{... (ii)}$$

$$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) ; K = x$$

Multiplying equation (i) by 2:

$$2\text{S}(s) + 2\text{O}_2(g) \longrightarrow 2\text{SO}_2(g) ; K_1' = (10^{52})^2 = 10^{104} \quad \text{... (iii)}$$

Subtracting (iii) from (ii), we get:

$$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) ; K_{eq} = \frac{10^{129}}{10^{104}} = 10^{129 - 104} = 10^{25}$$

7. 5.1g $\text{NH}_4\text{SH}$ is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid $\text{NH}_4\text{SH}$ decomposed to $\text{NH}_3$ and $\text{H}_2\text{S}$ as gases. The $K_p$ of the reaction at 327°C is ($R = 0.082\text{ L atm mol}^{-1}\text{K}^{-1}$, Molar mass of $\text{S} = 32\text{ g mol}^{-1}$, molar mass of $\text{N} = 14\text{ g mol}^{-1}$) [JEE Main 2019]

(a) $1 \times 10^{-4}\text{ atm}^2$

(b) $4.9 \times 10^{-3}\text{ atm}^2$

(d) $0.242 \times 10^{-4}\text{ atm}^2$

Answer: (c)

Explanation:

$$\text{NH}_4\text{SH}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)$$

$$n = \frac{5.1}{51} = 0.1\text{ mole}$$

$\alpha = 30\% = 0.3$

So number of moles at equilibrium = $0.1 \times 0.3 = 0.03\text{ mol}$ of $\text{NH}_3$ and $0.03\text{ mol}$ of $\text{H}_2\text{S}$.

Now use $PV = nRT$ at equilibrium:

$$P_{\text{total}} \times 3 = (0.03 + 0.03) \times 0.082 \times 600$$

$$P_{\text{total}} = 0.984\text{ atm}$$

At equilibrium $P_{\text{NH}_3} = P_{\text{H}_2\text{S}} = \frac{P_{\text{total}}}{2} = 0.492\text{ atm}$

$$K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = (0.492)(0.492) = 0.242\text{ atm}^2$$

8. Consider the reaction, $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. The equilibrium constant of the above reaction is $K_P$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $P_{\text{NH}_3} \ll P_{\text{total}}$ at equilibrium) [JEE Main 2019]

(a) $\frac{3^{3/2} K_P^{1/2} P^2}{4}$

(b) $\frac{3^{3/2} K_P^{1/2} P^2}{16}$

(d) $\frac{K_P^{1/2} P^2}{4}$

Answer: (b)

Explanation: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$$K_P = \frac{P_{\text{NH}_3}^2}{P_{\text{N}_2} P_{\text{H}_2}^3}$$

Since $\text{NH}_3$ dissociates into $\text{N}_2$ and $\text{H}_2$ in a 1:3 ratio, $P_{\text{H}_2} = 3 P_{\text{N}_2} = \frac{3}{4} P$ and $P_{\text{N}_2} = \frac{P}{4}$.

$$K_P = \frac{P_{\text{NH}_3}^2}{\left(\frac{P}{4}\right) \left(\frac{3P}{4}\right)^3} = \frac{256 P_{\text{NH}_3}^2}{27 P^4}$$

$$P_{\text{NH}_3} = \frac{3^{3/2} K_P^{1/2} P^2}{16}$$

9. For the equilibrium, $2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$, the value of $\Delta G^\circ$ at 298 K is approximately: [JEE Main 2019]

(a) $-80\text{ kJ mol}^{-1}$

(b) $-100\text{ kJ mol}^{-1}$

(d) $80\text{ kJ mol}^{-1}$

Answer: (d)

Explanation: $2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-, K = 10^{-14}$

$$\Delta G^\circ = -RT \ln K = -8.314 \times 298 \ln 10^{-14} \approx 80\text{ kJ/mol}$$

10. In a chemical reaction, $A + 2B \rightleftharpoons 2C + D$, the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is: [JEE Main 2019]

(a) 16

(b) 4

(d) 1/4

Answer: (b)

Explanation:

$$A + 2B \rightleftharpoons 2C + D$$

At $t = 0$: $a_0$, $1.5a_0$, $0$, $0$

At $t = t_{eq}$: $a_0 - x$, $1.5a_0 - 2x$, $2x$, $x$

At equilibrium $[A] = [B] \implies a_0 - x = 1.5a_0 - 2x \implies x = 0.5a_0$

At $t = t_{eq}$:

$[A] = 0.5a_0$, $[B] = 0.5a_0$, $[C] = a_0$, $[D] = 0.5a_0$

$$K = \frac{[C]^2 [D]}{[A][B]^2} = \frac{(a_0)^2 (0.5a_0)}{(0.5a_0)(0.5a_0)^2} = \frac{0.5}{0.25} = 4$$

11. Two solids dissociate as follows [JEE Main 2019]

$$A(s) \rightleftharpoons B(g) + C(g) ; K_{p1} = x\text{ atm}^2$$

$$D(s) \rightleftharpoons C(g) + E(g) ; K_{p2} = y\text{ atm}^2$$

The total pressure when both the solids dissociate simultaneously is:-

(a) $2\sqrt{x + y}\text{ atm}$

(b) $\sqrt{x - y}\text{ atm}$

(d) $\sqrt{x + y}\text{ atm}$

Answer: (c)

Explanation:

$$A(s) \rightleftharpoons B(g) + C(g) ; K_{p1} = P_B \cdot P_C = P_1(P_1 + P_2) = x \quad \text{... (i)}$$

$$D(s) \rightleftharpoons C(g) + E(g) ; K_{p2} = P_C \cdot P_E = (P_1 + P_2)P_2 = y \quad \text{... (ii)}$$

Adding (i) and (ii):

$$x + y = (P_1 + P_2)^2 \implies P_1 + P_2 = \sqrt{x + y}$$

Now total pressure $P_T = P_B + P_C + P_E = P_1 + (P_1 + P_2) + P_2 = 2(P_1 + P_2) = 2\sqrt{x + y}\text{ atm}$

Integer and Subjective

12. $\text{MX}_2$ dissociates into $\text{M}^{2+}$ and $\text{X}^-$ ions in an aqueous solution, with a degree of dissociation ($\alpha$) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is: [JEE Adv. 2014 P-I]

Answer: 2

Explanation: $\text{MX}_2 \rightleftharpoons \text{M}^{2+} + 2\text{X}^-$

Van't Hoff factor $i = 1 + (n - 1)\alpha = 1 + (3 - 1)(0.5) = 1 + 2(0.5) = 2$.

Ratio of observed $\Delta T_f$ to normal $\Delta T_f$ is $i = 2$.

13. For the given compound X, the total number of optically active stereoisomer is _________. [JEE Adv. 2018 P-II]

Answer: 7

Explanation: Only three stereocentres are present. Total isomers $= 2^3 = 8$. But one is optically inactive (meso). So, answer is 7.

14. For the following reaction, equilibrium constant $K_c$ at 298 K is $1.6 \times 10^{17}$

$$\text{Fe}^{2+}(aq) + \text{S}^{2-}(aq) \rightleftharpoons \text{FeS}(s)$$

When equal volume of $0.06\text{ M Fe}^{+2}$ and $0.2\text{ M S}^{-2}$ solution are mixed, then equilibrium concentration of $\text{Fe}^{+2}$ is found be $Y \times 10^{-17}\text{ M}$. Y is: [JEE Adv. 2019 P-II]

Answer: 8.93

Explanation: After mixing:

$[\text{Fe}^{2+}]_0 = 0.03\text{ M}$, $[\text{S}^{2-}]_0 = 0.1\text{ M}$

At equilibrium $[\text{S}^{2-}] \approx 0.1 - 0.03 = 0.07\text{ M}$

$$K_c = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]} \implies 1.6 \times 10^{17} = \frac{1}{[\text{Fe}^{2+}] \times 0.07}$$

$$[\text{Fe}^{2+}] = \frac{1}{1.6 \times 10^{17} \times 0.07} = \frac{100}{11.2} \times 10^{-17} = 8.928 \times 10^{-17}\text{ M}$$

Answer after rounding off is 8.93 (or 8.92 after truncation).