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

Solution:

It is given that T1 = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k1 = k and that of k2 = 2k

Also, R = 8.314 J K−1 mol−1

Now, substituting these values in the equation:

$\log \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]$

We get:

$\log \frac{2 k}{k}=\frac{E_{\mathrm{a}}}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right]$

$\Rightarrow \log 2=\frac{E_{\mathrm{a}}}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right]$

$\Rightarrow E_{\mathrm{a}}=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log 2}{10}$

= 52897.78 J mol−1

= 52.9 kJ mol−1

Note: There is a slight variation in this answer and the one given in the NCERT textbook.