More practice questions

4. Q4. The rate of a reaction quadruples when the temperature rises from 293 K to 313 K. Find the activation energy, assuming it does not change with temperature.

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   log 4 = (E_a/(2.303 × 8.314)) × (20/(293 × 313)), so E_a = 0.602 × 19.15 × 4585 = 52 860 J mol⁻¹ = 52.9 kJ mol⁻¹ (NCERT Exercise 3.30, page 88).

5. Q5. For the first order decomposition of H₂O₂, log(k/s⁻¹) = 14.34 − 1.25 × 10⁴ K/T. Find E_a, and the temperature at which the half-life is 256 minutes.

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   Compare with log k = log A − E_a/(2.303RT): E_a = 2.303 × 8.314 × 1.25 × 10⁴ = 239 kJ mol⁻¹. For t_1/2 = 256 min, k = 0.693/15 360 s = 4.51 × 10⁻⁵ s⁻¹, log k = −4.35, so T = 1.25 × 10⁴/(14.34 + 4.35) ≈ 669 K (NCERT Exercise 3.27, page 88).

6. Q6. A reaction has k = 4.5 × 10³ s⁻¹ at 10 °C and E_a = 60 kJ mol⁻¹. At what temperature is k = 1.5 × 10⁴ s⁻¹?

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   log(10/3) = 0.523 = (60 000/19.15)(1/283 − 1/T), so 1/T = 3.534 × 10⁻³ − 1.67 × 10⁻⁴ and T ≈ 297 K, about 24 °C (NCERT Exercise 3.28, page 88).

7. Q7. Assertion: a catalyst does not change the enthalpy change of a reaction. Reason: a catalyst lowers the activation energy of the reaction. Are both true, and does the reason explain the assertion?

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   Both are true, but the reason does not explain the assertion: ΔH is unchanged because the reactant and product energies are the same on either path (NCERT pages 79 and 82; reasoned).

8. Q8. Why can a catalyst alone not make a thermodynamically unfavourable net direction favourable?

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   A catalyst does not alter ΔG of a reaction, so, as NCERT puts it, it catalyses spontaneous reactions but not non-spontaneous ones (page 82). It can accelerate microscopic events in both directions but cannot supply the external work or coupling needed to drive an unfavourable net change (reasoned).

9. Q9. Hydrocarbon decomposition has k = 2.418 × 10⁻⁵ s⁻¹ at 546 K and E_a = 179.9 kJ mol⁻¹. Find the pre-exponential factor.

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   A = ke^E_a/RT = 2.418 × 10⁻⁵ × e^{179 900/(8.314 × 546)} = 2.418 × 10⁻⁵ × e^39.63 ≈ 3.93 × 10¹² s⁻¹ (NCERT Exercise 3.23, page 87).

10. Q10. A first order reaction goes 10% to completion at 298 K in the same time as 25% at 308 K. With A = 4 × 10¹⁰ s⁻¹, find E_a and k at 318 K.

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    Equal times give k₃₀₈/k₂₉₈ = log(100/75)/log(100/90) = 2.73, so E_a = 2.303R log 2.73 × (298 × 308)/10 ≈ 76.7 kJ mol⁻¹. Then k₃₁₈ = Ae^−E_a/RT ≈ 1.02 × 10⁻² s⁻¹ (NCERT Exercise 3.29, page 88).

11. Q11. For the decomposition of N₂O₅, 10⁵ × k/s⁻¹ is 0.0787, 1.70, 25.7, 178 and 2140 at 0, 20, 40, 60 and 80 °C. Plot ln k against 1/T, find A and E_a, and predict k at 30 °C and 50 °C.

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    With T = t + 273, 10³/T = 3.663, 3.413, 3.195, 3.003 and 2.833 K⁻¹, and ln k = −14.06, −10.98, −8.27, −6.33 and −3.84. The points lie close to a straight line. The best straight line (least squares) has slope −1.21 × 10⁴ K, so E_a = 1.21 × 10⁴ × 8.314 ≈ 101 kJ mol⁻¹, and its intercept, ln A = 30.3, gives A ≈ 1.4 × 10¹³ s⁻¹. Then the line gives ln k ≈ −9.65 at 303 K (10³/T = 3.300) and −7.18 at 323 K (3.096), so k ≈ 6.4 × 10⁻⁵ s⁻¹ at 303 K and 7.6 × 10⁻⁴ s⁻¹ at 323 K; read k from the line, since rounding E_a to 101 kJ mol⁻¹ first lowers it by about 15%. A line drawn by eye gives values close to these (NCERT Exercise 3.22, page 87).

12. Q12. A hydrocarbon decomposes with k = (4.5 × 10¹¹ s⁻¹) e^{−28 000 K/T}. Find E_a and A.

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    Compare with k = Ae^−E_a/RT: E_a/R = 28 000 K, so E_a = 28 000 × 8.314 = 232 792 J mol⁻¹ ≈ 232.8 kJ mol⁻¹, and A = 4.5 × 10¹¹ s⁻¹ (NCERT Exercise 3.26, page 87).