Question

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

Answer 1

Given:

$T_1=298K$

After the increase in temperature by $10K$

$T_2=(T_1+10)K$

$T_2=298+10=308K$

Let us take the value of $K_1=K$

Now, $K_2=2K$​

Also, $R=8.314J{K}^{-1}mo{l}^{-1}$

Now, substituting these values in the Arrhenius equation:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

We get:

$\log \frac{2k}{k}=\frac{E_a}{2.303\times 8.314}\left[\frac{308-298}{308\times 298}\right]$

$\log 2=\frac{E_a}{2.303\times 8.314}\left[\frac{10}{308\times 298}\right]$

$\therefore E_a=\frac{\log 2\times 2.303\times 8.314\times 308\times 298}{10}$

$=52897.78Jmo{l}^{-1}$

$=52.9kJmo{l}^{-1}$