Question

In presence of catalyst , the activation energy is lowered by 3 kcal at 27° C . Hence , the rate of reaction will increase by

Answer 1

in the absence of catalyst, suppose rate constant =k.then logk=logA-(Ea/2.303RT)………(1)
In the presence of catalyst,suppose rate constant =k'.
Now activation energy=Ea-2(if Ea is in kcal mole-¹)

Answer 2

Answer : The rate of reaction will increases by, 148.4

Solution :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst

$Ea_1$ = activation energy without catalyst

R = gas constant = 2 cal

T = temperature = $27^oC=273+27=300K$

Now put all the given values in this formula, we get

$\frac{K_2}{K_1}=e^{\frac{3000cal}{2cal\times 300}}=148.4$

Therefore, the rate of reaction will increases by, 148.4