The rate constant of a reaction is given by
\(\log_{10} k = 14.34 - \frac{1.25 \times 10^{4}\ \text{K}}{T}\)
What is the energy of activation of the reaction? (R = 8.314 J K−1 mol−1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 239.3 kJ mol−1
The Arrhenius equation in its logarithmic form is:
\( \displaystyle \log k=\log A-\frac{E_a}{2.303\,RT} \)
Comparing it term by term with the given equation, the coefficient of \( \frac{1}{T} \) gives:
\( \displaystyle \frac{E_a}{2.303\,R}=1.25\times10^{4}\,\mathrm{K} \)
\( \displaystyle E_a=2.303\times8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}\times1.25\times10^{4}\,\mathrm{K} \)
\( \displaystyle E_a=239340\,\mathrm{J\,mol^{-1}}\approx239.3\,\mathrm{kJ\,mol^{-1}} \)