Question

A hydrogenation reaction is carried out at $500\text{K}$. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is $400\text{K}$. Calculate the activation energy of the reaction, if the catalyst lowers the activation energy by $20\text{kJ/mol}$ (Assume Arrhenius factor to be same during the above temperature change)

• A. $1000\text{kJ/mole}$
• B. $500\text{kJ/mole}$
• C. $250\text{kJ/mole}$
• D. $100\text{kJ/mole}$

Answer 1

The correct option is D $100\text{kJ/mole}$

According to the question,
$\text{Rate of reaction at}500\text{K}=\text{rate of catalysed reaction at}400\text{K}$

$\therefore K_{500}=K_{400}^{\prime }$

$\Rightarrow A.e^{-\frac{E_a}{R\times 500}}=A{e}^{-\frac{E_a^{\prime }}{R\times 400}}$
$\Rightarrow ln{e}^{\frac{-E_a}{R\times 500}}=ln{e}^{-\frac{E_a^{\prime }}{R\times 400}}$
$\Rightarrow \frac{E_a}{500}=\frac{E_a^{\prime }}{400}$
$\Rightarrow \frac{E_a}{500}=\frac{E_a-20}{400}$
$\Rightarrow 400E_a=500E_a-10000$

$\therefore E_a=100\text{kj/mol}$