Answer

The rate constant of a reaction is nearly doubled with a $10^{\circ}$ rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

$k=\mathrm{Ae}^{-E \mathrm{a} / R T}$

Where,

$A$ is the Arrhenius factor or the frequency factor

Tis the temperature

Ris the gas constant

$E_{a}$ is the activation energy

Answer

It is given that $T_{1}=298 \mathrm{~K}$

$\therefore T_{2}=(298+10) \mathrm{K}$

$=308 \mathrm{~K}$

We also know that the rate of the reaction doubles when temperature is increased by $10^{\circ}$.

Therefore, let us take the value of $k_{1}=k$ and that of $k_{2}=2 k$

Also, $R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

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:

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

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

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

$=52897.78 \mathrm{~J} \mathrm{~mol}^{-1}$

$=52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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

Answer

In the given case:

$E_{\mathrm{a}}=209.5 \mathrm{~kJ} \mathrm{~mol}^{-1}=209500 \mathrm{~J} \mathrm{~mol}^{-1}$

$T=581 \mathrm{~K}$

$R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as: $x=e-E a / R T \Rightarrow \operatorname{In} x=-E$