CBSE Class 12 Chemistry 2025: question 24

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. For a first order reaction, $t_{1/2}=\frac{0.693}{k}$, so $k=\frac{0.693}{t_{1/2}}$.

   ½ mark

2. At 300 K: $k_1=\frac{0.693}{20}=3.465\times 10^{-2}$ min${}^{-1}$

   ½ mark

3. At 350 K: $k_2=\frac{0.693}{5}=1.386\times 10^{-1}$ min${}^{-1}$

   ½ mark

4. Arrhenius equation: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
   $\log \frac{0.1386}{0.03465}=\log 4=\frac{E_a}{2.303\times 8.314}\left[\frac{350-300}{350\times 300}\right]$

   ½ mark

5. $0.602=\frac{E_a}{19.15}\times \frac{50}{105000}$
   $E_a=\frac{0.602\times 19.15\times 105000}{50}\approx 24209$ J mol${}^{-1}$ $=24.21$ kJ mol${}^{-1}$

   1 mark

Answer: Eₐ ≈ 24.21 kJ mol⁻¹ (24209 J mol⁻¹)

Revise the chapter

• Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)
• Topic: Arrhenius Equation and Activation Energy