For a specific bimolecular gas-phase chemical reaction operating isothermally at , it is statistical — Chemical Kinetics Chemistry Question
Question
For a specific bimolecular gas-phase chemical reaction operating isothermally at $300 \text{ K}$, it is statistically calculated using collision theory that exactly $1$ out of every $10^{10}$ collisions possesses enough internal kinetic energy to successfully overcome the activation barrier. Using the statistical fraction relation $f = e^{-E_a/RT}$, calculate the exact activation energy $E_a$ of this reaction in $\text{kJ mol}^{-1}$. (Use $2.303 \times R \times 300 \approx 5740 \text{ J mol}^{-1}$ and round your final answer to the nearest integer).
💡 Solution & Explanation
The statistical fraction of successful energetic collisions is mathematically given by $f = e^{-E_a/RT}$. We are explicitly given that $f = 10^{-10}$. Therefore, $10^{-10} = e^{-E_a/RT}$. Taking the natural logarithm of both sides gives: $\ln(10^{-10}) = -E_a/RT \Rightarrow -10 \ln(10) = -E_a/RT$. Expanding $\ln(10)$ to $2.303$ gives $-10 \times 2.303 = -E_a / (R \times 300)$. Isolating $E_a$: $E_a = 10 \times 2.303 \times R \times 300 = 10 \times 5740 = 57400 \text{ J mol}^{-1}$, which converts to $57.4 \text{ kJ mol}^{-1}$. Rounded to the nearest integer, it is 57.