JEE Chemistry Equilibrium MCQs Set 09

Answer: (a) −6 kJ mol⁻¹

Explanation:
1. K = k_f/k_b = 10³/10² = 10.
2. ΔG° = −RT ln K = −8.3 × 300 × 2.3 = −5727 J ≈ −6 kJ mol⁻¹.

Teacher's Note:
1. Equilibrium constant equals the ratio of forward and backward rate constants.
2. K greater than 1 gives a negative ΔG°.

Answer: (d) 2 %

Explanation:
1. Start with 1 mol H₂O, degree of decomposition α: H₂O = 1 − α, H₂ = α, O₂ = α/2, total = 1 + α/2.
2. \( K_p=\dfrac{\alpha\,(\alpha/2)^{1/2}}{(1-\alpha)\,(1+\alpha/2)^{1/2}} \) at P = 1 bar. For small α, \( K_p\approx\alpha^{3/2}/\sqrt2 \).
3. \( \alpha^{3/2}=2.83\times10^{-3} \), so α ≈ 0.02, i.e. about 2 %.

Teacher's Note:
1. For small degree of dissociation, neglect α compared with 1.
2. Write partial pressures from mole fractions times total pressure.

Answer: (c) 4 × 10⁻³³

Explanation:
1. Required reaction = (reverse of reaction 1) + (reaction 2) + 3 × (reaction 3).
2. \( K=\dfrac{K_2K_3^3}{K_1}=\dfrac{1.6\times10^{12}\times10^{-39}}{4\times10^{5}} \).
3. K = 4 × 10⁻³³.

Teacher's Note:
1. Reversing a reaction inverts K; multiplying the equation by n raises K to the power n; adding equations multiplies K values.
2. Check that the species cancel correctly.

Answer: (c) 3

Explanation:
1. The reaction is exothermic, so raising the temperature shifts it backward (A no).
2. Fewer gas moles on the right (3 → 2), so higher pressure favours SO₃ (B yes).
3. Adding a reactant (SO₂ or O₂) shifts it forward (C, D yes). A catalyst only speeds up equilibrium (E no).
4. Factors that help: B, C, D = 3.

Teacher's Note:
1. Apply Le Chatelier's principle one factor at a time.
2. A catalyst never changes the equilibrium composition.

Answer: (b) 12

Explanation:
1. (i) \( K_{p1}=\dfrac{\alpha^2p_1}{1-\alpha^2}=3 \); (ii) \( K_{p2}=\dfrac{4\alpha^2p_2}{1-\alpha^2}=1 \).
2. Dividing: \( \dfrac{p_1}{4p_2}=3 \), so \( p_1/p_2=12 \).
3. So x = 12.

Teacher's Note:
1. For A ⇌ 2B, the total moles at equilibrium are 1 + α and K_p = 4α²p/(1 − α²).
2. Dividing the two expressions cancels the α-dependent term.

Answer: (d) addition of helium does not affect the equilibrium

Explanation:
1. Helium is inert and added at constant volume, so the partial pressures (concentrations) of the reacting gases do not change.
2. The reaction quotient stays equal to K, so there is no shift.

Teacher's Note:
1. At constant pressure, adding an inert gas increases the volume and favours the side with more moles.
2. At constant volume an inert gas has no effect on the position of equilibrium.

Answer: (a) 10

Explanation:
1. Concentrations after mixing: salt = 5 mmol/50 mL, acid = 0.5 mmol/50 mL; the ratio [salt]/[acid] = 10.
2. Henderson equation: 5 = pKₐ + log 10, so pKₐ = 4.
3. Kₐ = 10⁻⁴ = 10 × 10⁻⁵, so x = 10.

Teacher's Note:
1. Both volumes double, so the ratio of salt to acid equals the ratio of moles.
2. pH = pKₐ + log([salt]/[acid]).

Answer: (d) 8.5

Explanation:
1. Calcium lactate gives 2 × 0.005 = 0.01 M lactate ions, the conjugate base of a weak acid.
2. \( \mathrm{pH}=7+\tfrac12(\mathrm{p}K_a+\log c)=7+\tfrac12(5-2)=8.5 \).

Teacher's Note:
1. Salt of a weak acid and a strong base gives a basic solution.
2. Do not forget that one formula unit of Ca(lactate)₂ supplies two anions.

Answer: (c) 9.079

Explanation:
1. HCl converts NH₃ to NH₄⁺: NH₃ = 0.08 mol, NH₄⁺ = 0.12 mol.
2. pOH = 4.745 + log(0.12/0.08) = 4.745 + log 1.5 = 4.745 + 0.176 = 4.921.
3. pH = 14 − 4.921 = 9.079.

Teacher's Note:
1. A buffer neutralises added acid by converting the weak base to its conjugate acid.
2. For a basic buffer: pOH = pK_b + log([salt]/[base]).

Answer: (b) decreases by 3 units

Explanation:
1. pH = −log[H⁺].
2. If [H⁺] becomes 1000 times larger, ΔpH = −log 1000 = −3.
3. The pH decreases by 3 units.

Teacher's Note:
1. pH changes by 1 unit for every tenfold change in [H⁺].
2. More H⁺ means lower pH.

Answer: (b) 0.5

Explanation:
1. pH 12 means pOH = 2, so [OH⁻] = 0.01 M.
2. Millimoles of OH⁻ in 100 mL = 0.01 × 100 = 1 mmol.
3. Ca(OH)₂ gives 2 OH⁻, so millimoles of Ca(OH)₂ = 0.5.

Teacher's Note:
1. Each formula unit of Ca(OH)₂ gives two hydroxide ions.
2. Millimoles = molarity × volume in mL.

Answer: (a) 15 mL

Explanation:
1. Moles of Na = 0.69/23 = 0.03 mol, which gives 0.03 mol NaOH.
2. 73 g L⁻¹ HCl = 73/36.5 = 2 M.
3. Volume of HCl = 0.03/2 = 0.015 L = 15 mL.

Teacher's Note:
1. 2Na + 2H₂O → 2NaOH + H₂: one mole of Na gives one mole of NaOH.
2. Convert g/L to molarity by dividing by the molar mass.

Answer: (d) 1.86

Explanation:
1. Millimoles of H⁺: from HCl = 600 × 0.01 = 6; from H₂SO₄ = 400 × 0.01 × 2 = 8; total = 14.
2. [H⁺] = 14/1000 = 1.4 × 10⁻² M.
3. pH = 2 − log 1.4 = 2 − (log 2 + log 7 − 1) = 2 − 0.14 = 1.86.

Teacher's Note:
1. Sulphuric acid is treated as supplying two H⁺ ions per molecule here.
2. Add the millimoles of H⁺ first, then divide by the total volume.

Answer: (a) Methyl orange may be used for a weak acid vs weak base titration

Explanation:
1. A weak acid vs weak base titration has no sharp pH jump at the equivalence point, so no common indicator works well.
2. The other three statements are correct: methyl orange (pH 3–4.4) suits a strong acid vs weak base titration; phenolphthalein (pH 8–10) suits weak acid vs strong base and also strong acid vs strong base.
3. Hence (a) is incorrect.

Teacher's Note:
1. Choose the indicator whose colour-change range lies in the steep part of the titration curve.
2. Weak acid vs weak base titrations are best followed with a pH meter.

Answer: (c) 10

Explanation:
1. Molar mass of AgCl = 143.4 g mol⁻¹, so s = 1.434 × 10⁻³/143.4 = 1 × 10⁻⁵ mol L⁻¹.
2. K_sp = s² = 10⁻¹⁰.
3. −log K_sp = 10.

Teacher's Note:
1. For a 1:1 salt K_sp = s².
2. Convert solubility from g/L to mol/L before squaring.