JEE Chemistry Equilibrium MCQs Set 35

Answer: (b) \( S = (K_{sp}/6912)^{1/7} \)

Explanation:
1. Zr₃(PO₄)₄ ⇌ 3Zr⁴⁺ + 4PO₄³⁻, so [Zr⁴⁺] = \( 3S \) and [PO₄³⁻] = \( 4S \).
2. \( K_{sp} = (3S)^{3}(4S)^{4} = 27 \times 256 S^{7} = 6912 S^{7} \).
3. \( S = (K_{sp}/6912)^{1/7} \).

Teacher's Note:
1. Raise each ion concentration to its stoichiometric power.
2. \( 27 \times 256 = 6912 \).

Answer: (d) I, II and III

Explanation:
1. I: \( H^{+} = 80 \) mmol, OH⁻ = 40 mmol, leaving 40 mmol in 800 mL = \( 0.05 M \), so pH = 1.3 (true).
2. II: \( K_w \) rises with temperature (true). III: \( [H^{+}] = 10^{-5} = \alphaC \); \( K_a = 10^{-5} = 10^{-5}\alpha/(1 - \alpha) \), so \( \alpha = 0.5 \) (true).
3. IV is false, since the common-ion effect is an application of Le Chatelier's principle. So I, II and III are correct.

Teacher's Note:
1. Work through each statement independently.
2. The common ion shifts the equilibrium and suppresses ionisation.

Answer: (d) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required

Explanation:
1. The reaction is exothermic, so \( K \) decreases with temperature (\( a \) is true).
2. Inert gas at constant volume changes no concentrations (\( b \) is true); fewer gas moles on the product side means higher pressure favours the forward direction (\( c \) is true).
3. A large \( K \) does not mean a fast reaction; the reaction needs a catalyst \( (V_{2}O_{5}) \) to proceed at a useful rate, so \( (d) \) is incorrect.

Teacher's Note:
1. \( K \) tells about the extent, not the rate, of a reaction.
2. Catalysts change the rate but not \( K \).

Answer: (c) 5.35

Explanation:
1. NH₄Cl is a salt of a weak base and a strong acid: pH = \( 7 - \frac{1}{2}(pK_b + \log C) \).
2. \( pK_b = 5 \) and \( \log C = \log(2 \times 10^{-2}) = -1.699 \).
3. pH = \( 7 - \frac{1}{2}(5 - 1.699) = 7 - 1.65 = 5.35 \).

Teacher's Note:
1. Hydrolysis of NH₄⁺ makes the solution acidic.
2. Use the formula for the salt of a weak base and strong acid.

Answer: (d) 2.49 × 10⁻¹⁰ M

Explanation:
1. \( K_{sp} = 4S^{3} = 4 \times (1.84 \times 10^{-5})^{3} = 2.49 \times 10^{-14} \).
2. At pH 12, pOH = 2 and [OH⁻] = \( 10^{-2} M \).
3. [Cd²⁺] = \( K_{sp}/[OH^{-}]^{2} = 2.49 \times 10^{-14}/10^{-4} = 2.49 \times 10^{-10} M \).

Teacher's Note:
1. Find \( K_{sp} \) first, then use the fixed [OH⁻] from the buffer.
2. High pH lowers the solubility of a hydroxide.

Answer: (a) 2C(s) + O₂(g) ⇌ 2CO(g)

Explanation:
1. \( K_p = K_c(RT)^\Deltan_g \), so \( K_p \ne K_c \) when \( \Deltan_g \ne 0 \).
2. \( (a) \Deltan_g = 2 - 1 = 1 \) (carbon is a solid). In \( (b), (c) \) and \( (d), \Deltan_g = 0 \).

Teacher's Note:
1. Count only gaseous species.
2. Equal moles of gas on both sides give \( K_p = K_c \).

Answer: (c) K₂ = K₁⁻³

Explanation:
1. The second reaction is the reverse of the first multiplied by 3.
2. Reversing gives \( 1/K_{1} \); multiplying the equation by 3 cubes the constant, so \( K_{2} = (1/K_{1})^{3} = K_{1}^{-3} \).

Teacher's Note:
1. Reversal inverts \( K \); multiplying by \( n \) raises \( K \) to the nth power.
2. \( K_{1}^{3}K_{2} = 1 \).

Answer: (b) 1, 24.62 dm³ atm mol⁻¹, 1.65 × 10⁻³ dm⁻⁶ atm⁻² mol²

Explanation:
1. \( K_p/K_c = (RT)^\Deltan_g \).
2. \( (i) \Deltan_g = 0 \), so the ratio is 1. (ii) \( \Deltan_g = 1 \), so the ratio is RT = 24.62 dm³ atm mol⁻¹.
3. (iii) \( \Deltan_g = -2 \), so the ratio is (RT)⁻² = \( 1/(24.62)^{2} = 1.65 \times 10^{-3} \) dm⁻⁶ atm⁻² mol².

Teacher's Note:
1. \( \Deltan_g \) is the change in gaseous moles.
2. Negative \( \Deltan_g \) gives a ratio less than one.

Answer: (b) 1.9 g, 0.28 mol L⁻¹

Explanation:
1. Na₂SO₄ = \( 2/143 = 14 \) mmol, the limiting reagent against 100 mmol Ca(OH)₂. Ca(OH)₂ + Na₂SO₄ → CaSO₄ + 2NaOH.
2. CaSO₄ formed = 14 mmol × \( 136 = 1.9 g \).
3. NaOH formed = 28 mmol in 100 mL, so [OH⁻] = \( 0.28 M \).

Teacher's Note:
1. Identify the limiting reagent first.
2. Each mole of Na₂SO₄ produces 2 mol of OH⁻.

Answer: (b) \( 3^{3/2} K_p^{1/2} P^{2}/16 \)

Explanation:
1. From dissociating NH₃: \( p_N_{2} \) : \( p_H_{2} = 1 \) : 3. Since \( p_{NH}_{3} \) ≪ \( P, p_N_{2} + p_H_{2} \approx P \), so \( p_N_{2} = P/4 \) and \( p_H_{2} = 3P/4 \).
2. \( K_p = p_{NH}_{3}^{2}/(p_N_{2} p_H_{2}^{3}) = p_{NH}_{3}^{2}/[(P/4)(3P/4)^{3}] = 256 p_{NH}_{3}^{2}/(27P^{4}) \).
3. \( p_{NH}_{3} = (27K_p/256)^{1/2} P^{2} = 3^{3/2} K_p^{1/2} P^{2}/16 \).

Teacher's Note:
1. Use the stoichiometric ratio of the dissociation products.
2. Substitute the partial pressures into the \( K_p \) expression.

Answer: (a) 0.4736 V

Explanation:
1. At equilibrium \( E_{cell} = 0 \), so \( E^{\circ}_{cell} = (0.059/n) \log K_c \) with \( n = 2 \).
2. \( E^{\circ}_{cell} = (0.059/2) \log(10 \times 10^{15}) = 0.0295 \times 16 = 0.472 V \), which corresponds to \( 0.4736 V \).

Teacher's Note:
1. Use the Nernst equation at equilibrium.
2. Check that the unit is volts, not millivolts.

Answer: (d) 2√(x + y) atm

Explanation:
1. Let \( p_B = p_{1} \) and \( p_E = p_{2} \); then \( p_C = p_{1} + p_{2} \).
2. \( K_{p1} = p_{1}(p_{1} + p_{2}) = x \) and \( K_{p2} = p_{2}(p_{1} + p_{2}) = y \), so \( x + y = (p_{1} + p_{2})^{2} \).
3. Total pressure = \( p_{1} + p_{2} + (p_{1} + p_{2}) = 2(p_{1} + p_{2}) = 2\sqrt{x + y} \) atm.

Teacher's Note:
1. \( C(g) \) is common to both equilibria.
2. Add the two \( K_p \) expressions.

Answer: (d) 4

Explanation:
1. Let initial \( [A] = a, [B] = 1.5a \), and \( x \) be the extent of reaction: \( [A] = a - x, [B] = 1.5a - 2x \).
2. Equal at equilibrium: \( a - x = 1.5a - 2x \), so \( x = 0.5a \). Then \( [A] = [B] = 0.5a, [C] = 2x = a, [D] = x = 0.5a \).
3. \( K = [C]^{2}[D]/([A][B]^{2}) = (a^{2})(0.5a)/(0.5a \times 0.25a^{2}) = 4 \).

Teacher's Note:
1. Use an ICE table with the stoichiometric coefficients.
2. Equal equilibrium concentrations determine the extent.

Answer: (a) a positive slope

Explanation:
1. \( \ln K \) = −ΔH°/RT \( + \DeltaS^{\circ}/R \), so the slope is \( -\DeltaH^{\circ}/R \).
2. For an exothermic reaction \( \DeltaH^{\circ} \) is negative, so the slope is positive.

Teacher's Note:
1. Compare with \( y \) = mx \( + c \), with \( y = \ln K \) and \( x = 1/T \).
2. For an endothermic reaction the slope is negative.

Answer: (b) 3 × 10⁻²⁰

Explanation:
1. \( H_{2}S \) ⇌ \( 2H^{+} + S^{2-} \) has \( K = K_{1}K_{2} = 1.2 \times 10^{-20} \).
2. \( [S^{2-}] = K[H_{2}S]/[H^{+}]^{2} = 1.2 \times 10^{-20} \times 0.1/(0.2)^{2} = 3 \times 10^{-20} M \).

Teacher's Note:
1. Multiply the two step constants for the overall equilibrium.
2. The \( H^{+} \) from HCl represses the ionisation of \( H_{2}S \).

Answer: (c) 1.1 × 10⁻⁹ M

Explanation:
1. [SO₄²⁻] in final solution = \( 1 \times 50/500 = 0.1 M \).
2. At the onset of precipitation [Ba²⁺] = \( K_{sp}/[SO_{4}^{2-}] = 10^{-10}/0.1 = 10^{-9} M \) in 500 mL.
3. Original [Ba²⁺] (in 450 mL) = \( 10^{-9} \times 500/450 = 1.1 \times 10^{-9} M \).

Teacher's Note:
1. The Ba²⁺ solution was diluted from 450 mL to 500 mL.
2. Use the ionic product equal to \( K_{sp} \) at the onset.

Answer: (c) Weak base, strong acid, yellow to pinkish red

Explanation:
1. Methyl orange is yellow in basic or weakly acidic media and pinkish red in acidic media (pH range 3.1–4.4).
2. When an acid is added to the alkali in the flask the colour changes from yellow to pinkish red, so the end point is yellow to pinkish red.
3. Methyl orange is also suitable for a weak base titrated with a strong acid; \( (a) \) and \( (d) \) show colours methyl orange never has.

Teacher's Note:
1. Methyl orange works in the acidic pH range, so it suits a weak base and a strong acid.
2. The colour change is from the basic to the acidic colour.

Answer: (b) 6.9

Explanation:
1. For a salt of a weak acid and a weak base, pH = \( 7 + \frac{1}{2}(pK_a - pK_b) \).
2. pH = \( 7 + \frac{1}{2}(3.2 - 3.4) = 7 - 0.1 = 6.9 \).

Teacher's Note:
1. This pH is independent of the concentration of the salt.
2. The acid is slightly stronger than the base, so the pH is below 7.

Answer: (b) reverse direction because Q > K_c

Explanation:
1. \( K_c = e^{-\DeltaG^{\circ}/RT} = e^{-2494.2/2494.2} = e^{-1} = 0.37 \).
2. \( Q = [B][C]/[A]^{2} = (2 \times 0.5)/(0.25) = 4 \).
3. \( Q (4) \) > \( K_c (0.37) \), so the reaction proceeds in the reverse direction.

Teacher's Note:
1. Compare \( Q \) with \( K \): if \( Q \) > \( K \), the reaction goes backward.
2. RT = \( 8.314 \times 300 = 2494.2 J \), so ΔG°/RT = 1.

Answer: (b) −½

Explanation:
1. \( K_p = K_c(RT)^\Deltan_g \), with \( \Deltan_g = 1 - (1 + \frac{1}{2}) = -\frac{1}{2} \).
2. So \( x = -\frac{1}{2} \).

Teacher's Note:
1. \( \Deltan_g \) is moles of gaseous products minus gaseous reactants.
2. Fractional coefficients are allowed.

Answer: (d) 9.0 L

Explanation:
1. pH 1 means \( [H^{+}] = 0.1 M \) and pH 2 means \( 0.01 M, a \) tenfold dilution.
2. \( M_{1}V_{1} = M_{2}V_{2} \): \( 0.1 \times 1 = 0.01 \times V_{2} \), so \( V_{2} = 10 L \).
3. Water to be added = \( 10 - 1 = 9 L \).

Teacher's Note:
1. A tenfold dilution raises pH by one unit for a strong acid.
2. Water added = final volume − initial volume.

Answer: (c) 1 × 10⁻⁵

Explanation:
1. \( [H^{+}] = [Q^{-}] = 10^{-3} M \) and [HQ] ≈ \( 0.1 M \).
2. \( K_a = (10^{-3})^{2}/0.1 = 10^{-5} \).

Teacher's Note:
1. The acid is weak, so [HQ] ≈ initial concentration.
2. Check α = 1% to confirm.

Answer: (a) 1.8 atm

Explanation:
1. CO₂(g) \( + C(s) \) ⇌ 2CO(g): let \( p \) = pressure of CO₂ converted; \( p_{CO}_{2} = 0.5 - p \) and \( p_{CO} = 2p \).
2. Total = \( 0.5 + p = 0.8 \), so \( p = 0.3 \). Then \( p_{CO}_{2} = 0.2 \) and \( p_{CO} = 0.6 \) atm.
3. \( K_p = (0.6)^{2}/0.2 = 1.8 \) atm.

Teacher's Note:
1. Solid carbon does not appear in \( K_p \).
2. The total pressure rises by \( p \) because gas moles increase.

Answer: (a) 1 × 10⁻¹⁰

Explanation:
1. \( [H^{+}] = [A^{-}] = 10^{-5} M \) and [HA] ≈ \( 1.0 M \).
2. \( K_a = (10^{-5})^{2}/1 = 1 \times 10^{-10} \).

Teacher's Note:
1. For a weak acid \( K_a = [H^{+}]^{2}/C \).
2. The ionisation is negligible compared with \( 1 M \).

Answer: (a) II only

Explanation:
1. An acid donates a proton. In reaction II, H₂PO₄⁻ gives \( H^{+} \) to \( H_{2}O \), so it acts as an acid.
2. In I it is the conjugate base formed; in III it is shown as a proton acceptor.

Teacher's Note:
1. Look for the species that loses \( H^{+} \).
2. H₂PO₄⁻ is amphiprotic.