Question

The rate at ${27}^{o}C$ of a chemical reaction increases $1000$ times when a suitable catalyst is introduced. Calculate the change in the energy of activation that the catalyst has brought in.

Answer 1

By Arrhenius equation we have

$K=A{e}^{{-E}_A/RT}$

Taking log on both sides, we get

$\log K=\log A=\frac{E_A}{RT}$

where, $E_A$= Activation Energy

$A$= Arrehenius Constant

$T$= Temperature

$R$= Gas Control

$K$= Rate Constant

It is given that after introducting catalyst,rate of reaction increases $1000times$, which means rate constant has increased to $1000times$.

Writing the Arrhenius equation in absence of catalyst we have,

$\log K_1=\log A-\frac{E_{A1}}{RT}$ $\left(1\right)$

Similarly, writing the Arrhenius equation in presence of catalyst,

$\log K_2=\log A-\frac{E_{A2}}{RT}$ $\left(2\right)$ {Ignoring the change in A}

Using (2)-(1) we have

$\log \left(\frac{K_2}{K_1}\right)=\frac{(E_{A2}-E_{A1})}{RT}$

According to the question $K_2$ =$1000K_1$, $T=300K$, putting these values in the above equation we get

${\log }_e\left(1000\right)=\frac{{△E}_A}{\left(8.314\right)\times \left(600\right)}$

$\Rightarrow {△E}_A=17.229KJ$