Q.24 What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K?
Best answer
Answer: Ea ≈ 54.66 kJ mol–1
Given \(k_2/k_1 = 2\), \(T_1 = 303\) K, \(T_2 = 313\) K, \(R = 8.314\) J K–1 mol–1.
\(\log_{10}\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2 - T_1}{T_1T_2}\right)\)
\(E_a = \dfrac{2.303 R\, T_1T_2 \log_{10}2}{T_2 - T_1} = \dfrac{2.303\times8.314\times303\times313\times0.3010}{10}\)
\(= \dfrac{19.147\times94839\times0.3010}{10} = 5.466\times10^{4}\ \text{J mol}^{-1} = 54.66\ \text{kJ mol}^{-1}\)
Log in or create a free account to answer this question.
