Question

The rate of a reaction $A$ doubles on increasing the temperature from $300$ to $310K$. By how much, the temperature of reaction $B$ should be increased from $300K$ so that rate doubles if activation energy of the reaction $B$ is twice to that of reaction $A$?

• A. $4.92K$
• B. $19.67K$
• C. $2.45K$
• D. $9.84K$

Answer 1

The correct option is A $4.92K$

For the reaction A
$log\frac{k^{\prime }}{k}=\frac{E}{2.303R}\left[\frac{T^{\prime }-T}{T{T}^{\prime }}\right]$
$k^{\prime }=2k$
$T=300K{T}^{\prime }=310K$
Substitute values in the above equation.
$log\frac{2k}{k}=\frac{E}{2.303R}\left[\frac{310-300}{300\times 310}\right]$
$log2=\frac{E}{2.303R}\left[\frac{10}{300\times 310}\right]$ ......(1)
For the reaction B
$log\frac{k^{\prime }}{k}=\frac{E}{2.303R}\left[\frac{T^{\prime }-T}{T{T}^{\prime }}\right]$
$k^{\prime }=2k$
$E_a\left(B\right)=2E_a\left(A\right)$
$T=300K{T}^{\prime }=?$
Substitute values in the above equation.
$log\frac{2k}{k}=\frac{2E}{2.303R}\left[\frac{T^{\prime }-300}{300\times T^{\prime }}\right]$
$log2=\frac{2E}{2.303R}\left[\frac{T^{\prime }-300}{300\times T^{\prime }}\right]$ ......(2)
Divide equation (2) with (1)
$\frac{log2}{log2}=\frac{\frac{2E}{2.303R}}{\frac{E}{2.303R}}\times \frac{\left[\frac{T^{\prime }-300}{300\times T^{\prime }}\right]}{\left[\frac{10}{300\times 310}\right]}$
$1=2\times \frac{\left[\frac{T^{\prime }-300}{T^{\prime }}\right]}{\left[\frac{1}{31}\right]}$
$\frac{1}{62}=\left[\frac{T^{\prime }-300}{T^{\prime }}\right]$
$T^{\prime }-300=0.01613T^{\prime }$
$0.9838T^{\prime }=300$
$T^{\prime }=304.92K$
Hence, the temperature of reaction B should be increased from 300K by $304.92-300=4.92K$.