JEE Chemistry Equilibrium MCQs Set 06

Answer: (b) 4

Explanation:
1. HI = 2 mol means 1 mol each of \( \ce{H2} \) and \( \ce{I2} \) reacted, leaving 1 mol each.
2. \( K_c = \dfrac{2^2}{1 \times 1} = 4 \).
3. Since \( \Delta n = 0 \), \( K_p = K_c = 4 \).

Teacher's Note:
1. HI formed is twice the amount of \( \ce{H2} \) used.
2. \( K_p = K_c \) when \( \Delta n = 0 \).

Answer: (a) 1

Explanation:
1. HI = 3 mol means 1.5 mol each of \( \ce{H2} \) and \( \ce{I2} \) reacted. Each remaining = 3 mol.
2. \( K = \dfrac{3^2}{3 \times 3} = 1 \).

Teacher's Note:
1. Volume cancels when \( \Delta n = 0 \).
2. Check the mole table before substituting.

Answer: (b) 0.16

Explanation:
1. \( \ce{H2} \) reacted = 0.25 mol, so HI formed = 0.50 mol.
2. \( K_c = \dfrac{0.5^2}{1.25 \times 1.25} = \dfrac{0.25}{1.5625} = 0.16 \).

Teacher's Note:
1. HI formed is twice the \( \ce{H2} \) consumed.
2. The volume cancels.

Answer: (b) 125

Explanation:
1. Let AB lose 2x. Then \( p_A = 2x \) and \( p_{B_2} = x \), so total = 500 − 2x + 2x + x = 500 + x = 625, giving x = 125.
2. \( p_{AB} = 250 \), \( p_A = 250 \), \( p_{B_2} = 125 \).
3. \( K_p = \dfrac{250^2 \times 125}{250^2} = 125 \).

Teacher's Note:
1. Use a pressure table at constant volume.
2. The total pressure rise gives the extent of reaction.

Answer: (b) an adsorbent is used to remove \( \ce{SO3} \) as soon as it is formed

Explanation:
1. Removing a product lowers its concentration, so the equilibrium shifts forward to replace it.
2. A catalyst does not shift equilibrium and using smaller amounts of reactants does not change K or the direction.

Teacher's Note:
1. Removing a product pulls the equilibrium forward.
2. A catalyst only speeds up attainment of equilibrium.

Answer: (b) temperature is reduced and pressure is increased

Explanation:
1. \( \ce{2SO2 + O2 <=> 2SO3} \) is exothermic with fewer moles on the right.
2. Low temperature favours the exothermic forward reaction and high pressure favours the side with fewer moles.

Teacher's Note:
1. Combine the enthalpy rule with the mole-count rule.
2. Industrially a moderate temperature and a catalyst are used for speed.

Answer: (b) low temperature, high pressure and catalyst

Explanation:
1. The reaction is exothermic, so low temperature favours the yield.
2. It reduces moles of gas from 4 to 2, so high pressure favours the yield.
3. A catalyst is used to reach equilibrium faster.

Teacher's Note:
1. Yield is governed by temperature and pressure; speed by the catalyst.
2. In practice a compromise temperature is used.

Answer: (a) a > b + c

Explanation:
1. Low temperature favours the exothermic forward reaction, as stated.
2. High pressure favours products only if the products have fewer gas moles, so b + c < a.

Teacher's Note:
1. High pressure always favours the side with fewer moles.
2. Use this to infer the stoichiometric relationship.

Answer: (b) \( \ce{N2(g) + O2(g) <=> 2NO(g)} \)

Explanation:
1. Pressure affects equilibrium only when \( \Delta n_g \neq 0 \).
2. For \( \ce{N2 + O2 <=> 2NO} \), \( \Delta n_g = 0 \), so pressure has no effect on the yield.
3. The other three have fewer moles of gas on the right and benefit from high pressure.

Teacher's Note:
1. Check \( \Delta n_g \) first.
2. Equal moles on both sides means no pressure effect.

Answer: (a) yes, the backward reaction is favoured

Explanation:
1. Compression increases pressure, which favours the side with fewer gas moles.
2. \( \ce{N2O4} \) (1 mole) is on the left, so the backward reaction is favoured.

Teacher's Note:
1. Always count gas moles on each side.
2. Compression favours the smaller side.

Answer: (b) the equilibrium constant is given by \( K_p = \dfrac{[CO_2]}{[CH_4][O_2]} \)

Explanation:
1. Correct expression is \( K_p = \dfrac{p_{CO_2}}{p_{CH_4}\,p_{O_2}^2} \) because the coefficient of \( \ce{O2} \) is 2 and liquid water is excluded.
2. The expression in (b) misses the square, so it is the false statement.
3. The others are true.

Teacher's Note:
1. Raise each concentration to its stoichiometric coefficient.
2. Pure liquids are excluded.

Answer: (a) \( (2K_p/P)^{1/3} \)

Explanation:
1. Start with 2 mol of \( \ce{AB2} \): AB2 = 2 − 2x, AB = 2x, B2 = x, total = 2 + x.
2. \( K_p = \dfrac{(2x)^2 \cdot x}{(2-2x)^2} \cdot \dfrac{P}{2+x} \).
3. For x ≪ 1 this becomes \( K_p \approx x^3P/2 \), giving \( x = (2K_p/P)^{1/3} \).

Teacher's Note:
1. Neglect x compared with 1 only after writing the full expression.
2. Learn the \( x^3P/2 \) result.

Answer: (a) 1/64

Explanation:
1. Reversing the reaction gives \( 1/8 \), and doubling the coefficients squares the constant.
2. \( K = (1/8)^2 = 1/64 \).

Teacher's Note:
1. Reverse first, then scale the coefficients.
2. Powers of K follow the multiplier.

Answer: (b) 36 : 1

Explanation:
1. For \( \ce{X <=> Y + Z} \): \( K_{p_1} = \alpha^2P_1/(1-\alpha^2) \).
2. For \( \ce{A <=> 2B} \): \( K_{p_2} = 4\alpha^2P_2/(1-\alpha^2) \).
3. Ratio \( K_{p_1}/K_{p_2} = P_1/(4P_2) = 9 \), so \( P_1 : P_2 = 36 : 1 \).

Teacher's Note:
1. Write \( K_p \) in terms of \( \alpha \) and P for each stoichiometry.
2. The α terms cancel in the ratio.

Answer: (b) 64 times

Explanation:
1. \( K = [Fe^{3+}][OH^-]^3 \) is constant.
2. If \( [OH^-] \) becomes 1/4, \( [OH^-]^3 \) becomes 1/64.
3. So \( [Fe^{3+}] \) must rise by a factor of 64.

Teacher's Note:
1. Cube the factor for the hydroxide.
2. Take the reciprocal to find the change in the other ion.

Answer: (a) \( \dfrac{(0.75)^3(0.25)}{(0.50)^2(0.75)} \)

Explanation:
1. D = 0.25 M, so B reacted = 0.25 (B left = 0.75), A reacted = 0.50 (A left = 0.50), C formed = 0.75.
2. \( K = \dfrac{[C]^3[D]}{[A]^2[B]} = \dfrac{(0.75)^3(0.25)}{(0.50)^2(0.75)} \).

Teacher's Note:
1. Use stoichiometric ratios 2 : 1 : 3 : 1 in the table.
2. Substitute equilibrium, not initial, values.

Answer: (a) \( \ce{2C(s) + O2(g) <=> 2CO(g)} \)

Explanation:
1. \( K_p = K_c \) only when \( \Delta n_g = 0 \).
2. In (a) gas moles go from 1 to 2, so \( \Delta n_g = 1 \).
3. The other three have equal gas moles on both sides.

Teacher's Note:
1. Do not count solids when finding \( \Delta n_g \).
2. Only (a) has a net change in gas moles.

Answer: (d) 36

Explanation:
1. \( \ce{A2} \) reacted = 1.5 mol, so \( \ce{B2} \) left = 0.5 and AB = 3 mol.
2. \( K_c = \dfrac{3^2}{0.5 \times 0.5} = 36 \).
3. \( \Delta n = 0 \), so \( K_p = 36 \).

Teacher's Note:
1. AB formed is twice the \( \ce{A2} \) used.
2. \( K_p = K_c \) here.

Answer: (c) \( [1/(K_1K_2)]^{1/2} \)

Explanation:
1. Adding the two given reactions gives \( \ce{N2 + 2O2 <=> 2NO2} \) with \( K = K_1K_2 \).
2. Reversing and halving it gives \( \ce{NO2 <=> 1/2 N2 + O2} \).
3. Hence \( K = (1/(K_1K_2))^{1/2} \).

Teacher's Note:
1. Add first, then reverse, then halve.
2. Halving the coefficients takes the square root of K.

Answer: (a) 0.62

Explanation:
1. \( K_c = \dfrac{[AB]^2}{[A_2][B_2]} = \dfrac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} \).
2. \( = 7.84/12.6 = 0.62 \).

Teacher's Note:
1. Powers of ten cancel when \( \Delta n = 0 \).
2. Compute the ratio of the mantissas.

Answer: (c) \( 1/\sqrt{RT} \)

Explanation:
1. \( \Delta n_g = 1 - (1 + 1/2) = -1/2 \).
2. \( K_p/K_c = (RT)^{-1/2} = 1/\sqrt{RT} \).

Teacher's Note:
1. Fractional coefficients are allowed in \( \Delta n \).
2. Use \( K_p = K_c(RT)^{\Delta n} \).

Answer: (a) 1.33 atm

Explanation:
1. Start with 1 mol: \( \ce{N2O4} \) = 0.5, \( \ce{NO2} \) = 1.0, total = 1.5 mol.
2. \( p_{NO_2} = (1.0/1.5) \times 1 = 2/3 \) atm and \( p_{N_2O_4} = 1/3 \) atm.
3. \( K_p = (2/3)^2/(1/3) = 4/3 = 1.33 \) atm.

Teacher's Note:
1. Use \( K_p = 4\alpha^2P/(1-\alpha^2) \) as a check.
2. With α = 0.5 and P = 1 it gives 1.33.

Answer: (d) by increasing pressure and decreasing temperature

Explanation:
1. The forward reaction reduces gas moles from 4 to 2 and releases heat.
2. High pressure and low temperature both favour it.
3. Adding product, lowering pressure, or removing reactants shifts the equilibrium backward.

Teacher's Note:
1. Apply the concentration, pressure and temperature rules one at a time.
2. Exothermic reactions are favoured by cooling.