JEE Chemistry Equilibrium MCQs Set 07

Answer: (a) \( 1/K^2 \)

Explanation:
1. Reversing the reaction gives \( 1/K \).
2. Doubling the coefficients squares the constant, so \( K' = 1/K^2 \).

Teacher's Note:
1. Reciprocal for reversal and a power for scaling.
2. Apply both in either order.

Answer: (b) mostly products

Explanation:
1. A very large K means the ratio of products to reactants is huge.
2. The mixture contains mostly products.

Teacher's Note:
1. K much greater than 1 favours products.
2. K much less than 1 favours reactants.

Answer: (c) \( K^{1/2} \)

Explanation:
1. Halving all the coefficients takes the square root of K.
2. So the new constant is \( K^{1/2} \).

Teacher's Note:
1. Scaling by n gives \( K^n \).
2. Here n = 1/2.

Answer: (c) 4/5 atm

Explanation:
1. Let \( p_B = x \), so \( p_A = 1 - x \).
2. \( K_p = x/(1-x) = 4 \), giving x = 0.8 = 4/5 atm.

Teacher's Note:
1. Use a pressure table for the reaction.
2. \( K_p \) is dimensionless here because \( \Delta n = 0 \).

Answer: (c) 0.625

Explanation:
1. Reacted = 0.2 mol, so A = B = 0.8 mol and C = 0.2 mol in 2 L.
2. \( [A] = [B] = 0.4 \) M and \( [C] = 0.1 \) M.
3. \( K_c = 0.1/(0.4 \times 0.4) = 0.625 \).

Teacher's Note:
1. Divide by the volume because \( \Delta n \neq 0 \).
2. Then substitute into \( K_c \).

Answer: (a) low temperature and high pressure

Explanation:
1. The forward reaction is exothermic, so low temperature favours it.
2. Gas moles fall from 2 to 1, so high pressure favours the product.

Teacher's Note:
1. Both rules point in the same direction here.
2. Always check enthalpy and moles separately.

Answer: (d) solubility of a gas in a liquid increases

Explanation:
1. By Henry's law, higher pressure of a gas increases its solubility in a liquid.
2. Ammonia formation increases with pressure and \( \ce{CO2 + C <=> 2CO} \) shifts backward (fewer gas moles on the left).
3. Melting under pressure depends on the substance and is not general.

Teacher's Note:
1. Gas solubility rises with pressure.
2. Check gas moles for each equilibrium.

Answer: (a) \( \ce{N2O4(g) <=> 2NO2(g)} \)

Explanation:
1. For \( \ce{A <=> 2B} \) starting with 1 mol: A = 1 − α, B = 2α, total = 1 + α.
2. \( K_p = \dfrac{(2\alpha P/(1+\alpha))^2}{(1-\alpha)P/(1+\alpha)} = \dfrac{4\alpha^2P}{1-\alpha^2} \).
3. Only \( \ce{N2O4 <=> 2NO2} \) fits this form.

Teacher's Note:
1. Match the stoichiometry of dissociation \( A \rightarrow 2B \).
2. The factor 4 comes from the coefficient 2 squared.

Answer: (a) 4 times

Explanation:
1. Increasing the volume 16 times reduces the pressure to 1/16.
2. \( \alpha \propto P^{-1/2} \), so α becomes \( (1/16)^{-1/2} = 4 \) times.

Teacher's Note:
1. Pressure is inversely proportional to volume at constant T and n.
2. Apply the square-root relation carefully.

Answer: (b) the amounts of C and D increase

Explanation:
1. Gas moles are 3 on the left and 4 on the right (A is solid).
2. Lower pressure favours the side with more gas moles, so the forward reaction occurs.
3. The amounts of C and D increase.

Teacher's Note:
1. Ignore solids when counting gas moles.
2. Low pressure favours more gas moles.

Answer: (d) no effect whatever the value of \( \Delta n \)

Explanation:
1. An inert gas added at constant volume leaves the concentrations of reacting species unchanged.
2. Hence Q remains equal to K and there is no shift whatever \( \Delta n \) is.

Teacher's Note:
1. Constant volume means no change in concentration.
2. Constant pressure with inert gas is different.

Answer: (c) c, d, e

Explanation:
1. (a) has no effect. (b) Adding \( \ce{Cl2} \) shifts the reaction backward.
2. (c) An inert gas at constant pressure increases the volume and favours the side with more gas moles, so it favours forward. (d) Larger volume also favours forward.
3. (e) Adding the reactant \( \ce{PCl5} \) favours forward. So (c), (d), (e).

Teacher's Note:
1. Inert gas at constant pressure acts like a volume increase.
2. Adding reactant or removing product favours the forward reaction.

Answer: (b) backward

Explanation:
1. \( Q_p = \dfrac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3} = \dfrac{3^2}{1 \times 2^3} = 1.125 \) atm⁻².
2. \( Q_p \gg K_p \), so the system has too much product and the reaction proceeds backward.

Teacher's Note:
1. Compare Q with K to decide the direction.
2. Q greater than K means a backward shift.

Answer: (d) \( 1.75 \times 10^{-4} \)

Explanation:
1. \( \Delta n_g = 3 - 2 = 1 \) and T = 700 K.
2. \( K_p = K_c RT = 3 \times 10^{-6} \times 0.0821 \times 700 \).
3. \( K_p \approx 1.7 \times 10^{-4} \), which is closest to \( 1.75 \times 10^{-4} \).

Teacher's Note:
1. Convert °C to K.
2. Use R = 0.0821 L atm mol⁻¹ K⁻¹ with \( K_c \) in mol L⁻¹.

Answer: (d) \( 1.44 \times 10^{-5}/(0.082 \times 773)^{-2} \)

Explanation:
1. \( \Delta n_g = -2 \) and T = 500 + 273 = 773 K.
2. \( K_p = K_c(RT)^{-2} \), so \( K_c = K_p/(RT)^{-2} \).
3. With R = 0.082 L atm mol⁻¹ K⁻¹, \( K_c = 1.44 \times 10^{-5}/(0.082 \times 773)^{-2} \).

Teacher's Note:
1. Use R in L atm units with pressures in atm.
2. Convert °C to K.

Answer: (c) \( \ce{2H2O(g) <=> 2H2(g) + O2(g)} \)

Explanation:
1. Increasing pressure moves the equilibrium toward the side with fewer gas moles.
2. In \( \ce{2H2O <=> 2H2 + O2} \) the left side has 2 moles and the right 3, so it shifts left.
3. (a) has no shift, (b) and (d) shift right.

Teacher's Note:
1. Count gas moles on both sides.
2. Solids do not count.

Answer: (a) 0.01

Explanation:
1. The reverse reaction has the reciprocal constant.
2. \( K_c' = 1/100 = 0.01 \).

Teacher's Note:
1. Reverse means reciprocal.
2. Do not change the coefficients here.

Answer: (d) 150

Explanation:
1. At equilibrium \( [Fe^{3+}] = 3.1 - 3.0 = 0.1 \) M and \( [SCN^-] = 3.2 - 3.0 = 0.2 \) M.
2. \( K_c = \dfrac{3.0}{0.1 \times 0.2} = 150 \).

Teacher's Note:
1. Subtract the amount of complex formed from each reactant.
2. The volume is 1 L, so moles equal molarity.

Answer: (c) 2/3

Explanation:
1. Let x mol of \( \ce{N2O4} \) dissociate. \( \ce{NO2} \) = 2x, total = 1 + x.
2. \( 2x/(1+x) = 1/2 \) gives x = 1/3.
3. \( [NO_2] = 2/3 \) and \( [N_2O_4] = 2/3 \), so \( K_c = (2/3)^2/(2/3) = 2/3 \).

Teacher's Note:
1. Use the mole fraction to find the extent of reaction.
2. The volume is 1 L so moles equal molarity.

Answer: (d) 0.02

Explanation:
1. Dissociation is the reverse of synthesis for the same stoichiometric equation, so \( K' = 1/50 = 0.02 \).

Teacher's Note:
1. Reverse gives reciprocal.
2. Keep the same coefficients.

Answer: (c) Q < \( K_c \)

Explanation:
1. When Q is less than K the product concentrations are lower than at equilibrium.
2. The reaction therefore moves forward to reach equilibrium.

Teacher's Note:
1. Q less than K means forward.
2. Q greater than K means backward.

Answer: (a) 1

Explanation:
1. Before the disturbance \( K_c = (1.5 \times 1.5)/(1.5 \times 1.5) = 1 \).
2. Removing C and D lowers Q to 4/9, so the reaction shifts forward until Q = K again.
3. \( K_c \) depends only on temperature, so it remains 1.

Teacher's Note:
1. Do not confuse Q with K.
2. A change in concentrations does not change K.

Answer: (d) \( \sqrt{1/41} \)

Explanation:
1. The new reaction is the reverse of the original with the coefficients halved.
2. \( K_c' = (1/41)^{1/2} \).

Teacher's Note:
1. Reverse gives the reciprocal and halving gives the square root.
2. Apply both.