The activation energy of a reaction is 9 kcal mol−1. By what percentage does the rate constant increase when the temperature is raised from 295 K to 300 K? (R = 2 cal K−1 mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 28.9%
The Arrhenius equation written for two temperatures gives the ratio of the rate constants:
\( \displaystyle \log\frac{k_2}{k_1}=\frac{E_a}{2.303\,R}\left[\frac{T_2-T_1}{T_1T_2}\right] \)
With Ea = 9 kcal mol−1 = 9000 cal mol−1 and R = 2 cal K−1 mol−1:
\( \displaystyle \log\frac{k_2}{k_1}=\frac{9000\,\mathrm{cal\,mol^{-1}}}{2.303\times2\,\mathrm{cal\,K^{-1}\,mol^{-1}}}\left[\frac{300\,\mathrm{K}-295\,\mathrm{K}}{295\,\mathrm{K}\times300\,\mathrm{K}}\right] \)
\( \displaystyle \log\frac{k_2}{k_1}=1954\,\mathrm{K}\times\frac{5\,\mathrm{K}}{88500\,\mathrm{K^2}}=0.1104 \)
\( \displaystyle \frac{k_2}{k_1}=\text{antilog}\,(0.1104)=1.289 \)
\( \displaystyle \text{Increase}=(1.289-1)\times100=28.9\% \)
The rate constant increases by about 28.9%.