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Answer the following.
The rate constant of a first order reaction increases five times when the temperature is raised…
The rate constant of a first order reaction increases five times when the temperature is raised from $350\text{ K}$ to $500\text{ K}$. Calculate the activation energy of the reaction. ($R = 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}$)
Answer
Answer
AIFormula:
$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2\cdot303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$
Substitution:
$\log(5) = \frac{E_a}{2\cdot303 \times 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}}\left(\frac{500 - 350}{350 \times 500}\right)$
$0\cdot6990 = \frac{E_a}{19\cdot147} \times \left(\frac{150}{175000}\right)$
$E_a = \frac{0\cdot6990 \times 19\cdot147 \times 175000}{150} = 15614\text{ J mol}^{-1} = 15\cdot61\text{ kJ mol}^{-1}$
Final answer: 15.61 $\text{kJ mol}^{-1}$
From ISC 2026 Chemistry Paper 1, question 16(i).
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