CBSE 2025, Set 1

Question

A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.

$[	ext{R} = 8.314 	ext{ J K}^{-1}	ext{ mol}^{-1};\; \log 4 = 0.602]$

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** For first order reaction, $t_{1/2} = \dfrac{0.693}{k}$ $$k_1 = \frac{0.693}{20} = 3.465 	imes 10^{-2} 	ext{ min}^{-1} \quad (300	ext{ K})$$ $$k_2 = \frac{0.693}{5} = 1.386 	imes 10^{-1} 	ext{ min}^{-1} \quad (350	ext{ K})$$ **Step 2:** Arrhenius equation $$\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[\frac{T_2 - T_1}{T_1 T_2}ight]$$ $$\log 4 = \frac{E_a}{2.303 	imes 8.314} 	imes \frac{50}{350 	imes 300}$$ $$0.602 = \frac{E_a}{19.15} 	imes \frac{50}{105000}$$ $$E_a = \frac{0.602 	imes 19.15 	imes 105000}{50} = \boxed{24209 	ext{ J mol}^{-1} \approx 24.21 	ext{ kJ mol}^{-1}}$$

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