Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, $\mathrm{T(K)}$.

$$
\ln k = 14.34 - \frac{1.25 imes10^{4}}{T}
$$

The energy of activation in $ ext{kcal mol}^{-1}$ for the reaction is :

(Given: $k$ in $ ext{s}^{-1}$, $R = 1.987\ ext{cal mol}^{-1} ext{K}^{-1}$)

Question

Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, $\mathrm{T(K)}$.

$$
\ln k = 14.34 - \frac{1.25	imes10^{4}}{T}
$$

The energy of activation in $	ext{kcal mol}^{-1}$ for the reaction is :

(Given: $k$ in $	ext{s}^{-1}$, $R = 1.987\ 	ext{cal mol}^{-1}	ext{K}^{-1}$)

Accepted Answer

A. 1. **Compare the given equation with the Arrhenius equation** in its logarithmic form, since matching the two expressions lets us identify $E_a$.

$$
\ln k = \ln A - \frac{E_a}{RT}
$$

2. Matching the coefficient of $\frac{1}{T}$ in both expressions:

$$
\frac{E_a}{R} = 1.25	imes10^{4}
$$

Therefore:

$$
E_a = 1.25	imes10^{4}	imes R
$$

3. **Substitute** $R = 1.987\ 	ext{cal mol}^{-1}	ext{K}^{-1}$ to get $E_a$ in cal/mol, then convert to kcal/mol.

$$
E_a = 1.25	imes10^{4}	imes 1.987 = 24.84	imes10^{3}\ 	ext{cal mol}^{-1} = 24.84\ 	ext{kcal mol}^{-1}
$$

Hence, the answer is **A, 24.84**.