Question

A hydrogenation reaction is carried out at 5000 k. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is $400K$ $.Whatwouldbethevalueoftheactivationenergyofthereactionifthecatalystlowerstheactivationbarrierby$20 kJ { mol\quad L }^{ -1}$?

Answer 1

We have if $E_a$ and ${E_a}^{\prime }$ be the energy of activation in presence and absence of catalyst for hydrogenation reaction then,

$K=A{e}^{-E_a/RT}$

In presence of catalyst-

Required temperature $\left(T\right)=400K$

$K_1=A{e}^{-E_a/(R\times 400)}$

In absence of catalyst-

Required temperature $\left(T\right)=5000K$

$K_2=A{e}^{-{E_a}^{\prime }/(R\times 5000)}$

The two rates are given to be same, i.e.,

$r_1=r_2$

$\Rightarrow K_1=K_2$

$\therefore A{e}^{-E_a/(R\times 400)}=A{e}^{-{E_a}^{\prime }/(R\times 5000)}$

$\Rightarrow \frac{E_a}{R\times 400}=\frac{{E_a}^{\prime }}{R\times 5000}$

Given that catalyst lowers the activation energy by $20KJ/mol$.

$\therefore \frac{{E_a}^{\prime }-20}{400}=\frac{{E_a}^{\prime }}{5000}$

$\Rightarrow {E_a}^{\prime }=21.73KJ/mol$

Hence the required answer is $21.73KJ/mol$.