The activation energies of two reactions are 18 kJ mol−1 and 4.0 kJ mol−1. Assuming the pre-exponential factor to be the same for both reactions, the ratio k1/k2 of their rate constants at 27 °C (k1 for the reaction with 18 kJ mol−1) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3.65 × 10−3
By the Arrhenius equation, \( k=Ae^{-E_a/RT} \). With the same A and the same temperature, A cancels in the ratio and only the difference in activation energies matters:
\( \displaystyle \frac{k_1}{k_2}=e^{-(E_{a1}-E_{a2})/RT} \)
\( \displaystyle \log\frac{k_1}{k_2}=-\frac{E_{a1}-E_{a2}}{2.303RT} \)
With \( E_{a1}-E_{a2}=18000-4000=14000\,\mathrm{J\,mol^{-1}} \) and T = 27 + 273 = 300 K:
\( \displaystyle \log\frac{k_1}{k_2}=-\frac{14000}{2.303\times8.314\times300}=-\frac{14000}{5744}=-2.437 \)
\( \displaystyle \frac{k_1}{k_2}=10^{-2.437}=10^{0.563}\times10^{-3}\approx3.65\times10^{-3} \)
The reaction with the higher activation energy has the smaller rate constant, so the ratio is well below 1.