JEE Main 2013 Paper

Concept:
Arrhenius equation- Temperature Dependence and Rate constant:
- The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation.
- It has been found that for a chemical reaction with the rise in temperature by ${10}^{\circ }\mathit{C}$, the rate constant is nearly doubled.
- Mathematically, Arrhenius equation is given by,
$\mathit{k}=\mathit{A}\mathit{e}-\frac{\mathit{E}\mathit{a}}{\mathit{R}\mathit{T}}$
where $\mathit{A}$ is the Arrhenius factor or the frequency factor. It is also called the preexponential factor. It is a constant specific to a particular reaction, $\mathit{R}$ is gas constant and ${\mathit{E}}_{\mathit{a}}$ is activation energy measured in joules/mole ( $\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{-1}$ ).
Activation energy ( ${\mathit{E}}_{\mathit{a}}$ ):
- It is the minimum energy that must be supplied to the reactants to enable them to cross over the energy barrier between reactants and products.
- For fast reaction; ${\mathit{E}}_{\mathit{a}}$ is low.
- For slow reaction; ${\mathit{E}}_{\mathit{a}}$ is high.
- An alternative form of the Arrhenius equation is given by:
$\log \left(\frac{{\mathit{k}}_2}{{\mathit{k}}_1}\right)=\frac{\mathit{E}\mathit{a}}{2.303\mathit{R}}\left[\frac{{\mathit{T}}_2-{\mathit{T}}_1}{{\mathit{T}}_1{\mathit{T}}_2}\right]$
Where ${\mathit{K}}_1$ and ${\mathit{K}}_2$ are the rate constants for a reaction at two different temperatures ${\mathit{T}}_1$ and ${\mathit{T}}_2$.

Calculation:
Given: ${\mathit{K}}_2∕{\mathit{K}}_1=2$ (The rate of reaction is doubled) ; $\mathit{R}=8.314\mathit{J}{\mathit{K}}^{-1}\mathit{m}\mathit{o}{\mathit{l}}^{-1};{\mathit{T}}_1=$ 300K; ${\mathit{T}}_2=310\mathit{K}$
As per the Arrhenius equation,
$\log \left(\frac{{\mathit{k}}_2}{{\mathit{k}}_1}\right)=\frac{\mathit{E}\mathit{a}}{2.303\mathit{R}}\left[\frac{{\mathit{T}}_2-{\mathit{T}}_1}{{\mathit{T}}_1{\mathit{T}}_2}\right]$
$\log 2=\frac{{\mathit{E}}_{\mathit{a}}}{2.303\mathit{R}}\left[\frac{310-300}{300\times 310}\right]$
${\mathit{E}}_{\mathit{a}}=\frac{\log 2\times 2.303\times 8.314\times 310\times 300}{10}$
Therefore, ${\mathit{E}}_{\mathit{a}}=53603\mathit{J}=53.6\mathit{k}\mathit{J}∕\mathit{m}\mathit{o}\mathit{l}$
Hence, the activation energy ( ${\mathit{E}}_{\mathit{a}}$ ) of the reaction whose rate doubles when its temperature changes from 300 K to 310 K is $53.6\mathit{K}\mathit{J}∕\mathit{m}\mathit{o}\mathit{l}$