Question

The activation energy for a reaction is 84.0kJ/mol. Addition of a catalyst lowers the activation energy by 23.0 kJ/mol, while leaving the A-factor unchanged. By what factor ( uncatalyzed catalyzed k k ) does the rate increase at 25 °C?

Answer 1

Answer : The rate constant of the reaction is increased by factor, $4.93\times 10^{10}$

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  = 23.0 kJ/mol = 23000 J/mol

$Ea_1$ = activation energy without catalyst  = 84.0 kJ/mol = 84000 J/mol

R = gas constant = 8.314 J/mol.K

T = temperature = $25^oC=273+25=298K$

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

$\frac{K_2}{K_1}=e^{\frac{(84000-23000)J/mol}{8.314J/mol.K\times 298K}}$

$\frac{K_2}{K_1}=4.93\times 10^{10}$

Therefore, the rate constant of the reaction is increased by factor, $4.93\times 10^{10}$