Q.A gas phase reaction has energy of activation $200\ \text{kJ mol}^{-1}$. If the frequency factor of the reaction is $1.6\times10^{13}\ \text{s}^{-1}$. Calculate the rate constant at 600 K. ($e^{-40.09}=3.8\times10^{-18}$)

Step 1. $k=Ae^{-E_a/RT}$, with $A=1.6\times10^{13}\ \text{s}^{-1}$, $E_a=200\ \text{kJ mol}^{-1}=200{,}000\ \text{J mol}^{-1}$, $R=8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $T=600$ K.

Step 2. $\dfrac{E_a}{RT}=\dfrac{200{,}000}{8.314\times600}=\dfrac{200{,}000}{4988.4}=40.09$ -- matching exactly the exponent given in the hint, confirming the setup. …

Compute $E_a/RT$ to confirm it matches the given exponent, then substitute A and the given $e^{-40.09} …

• Forgetting to convert $E_a$ from kJ mol$^{-1}$ to J mol$^{-1}$ before dividing by $R$ (which is in J, not …