Answer to Question #127644 in Chemistry for Amalraj

Question

What is the activation energy (in J ) of the reaction A B for which the rate constant is 13 sec -1 at 600K and 1.9 sec -1 at 300K

Answer 1

According to the Arrhenius law, the rate constant and the temperature are related through the activation energy:

$k=A\text{e}^{-\frac{E_a}{k_bT}}$.

Therefore, the logarithm of the ratio of the rate constants at temperatures $T_1$ and $T_2$ is:

$\text{ln}(k_1/k_2) = \frac{E_a}{k_b}(\frac{1}{T_2}-\frac{1}{T_1})$.

From this expression , the activation energy will be:

$E_a= k_b\frac{\text{ln}(k_1/k_2)}{(\frac{1}{T_2}-\frac{1}{T_1})}$

$E_a= 1.38\cdot10^{-23}\cdot\frac{\text{ln}(13/1.9)}{(\frac{1}{300}-\frac{1}{600})}=1.6\cdot10^{-20}$ J.

Answer: the activation energy is 1.6x10^-20 J.

Actually, this energy is in joules per one molecule. If you want the activation energy in J/mol, you must use universal gas constant $R$ (8.314 J/( K mol)) instead of Boltzmann constant $k_b$ .