Answer :
Answer : The rate constant of the reaction is increased by factor, [tex]4.93\times 10^{10}[/tex]
Solution :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
The expression used with catalyst and without catalyst is,
[tex]\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}[/tex]
[tex]\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}[/tex]
where,
[tex]K_2[/tex] = rate of reaction with catalyst
[tex]K_1[/tex] = rate of reaction without catalyst
[tex]Ea_2[/tex] = activation energy with catalyst = 23.0 kJ/mol = 23000 J/mol
[tex]Ea_1[/tex] = activation energy without catalyst = 84.0 kJ/mol = 84000 J/mol
R = gas constant = 8.314 J/mol.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in this formula, we get
[tex]\frac{K_2}{K_1}=e^{\frac{(84000-23000)J/mol}{8.314J/mol.K\times 298K}}[/tex]
[tex]\frac{K_2}{K_1}=4.93\times 10^{10}[/tex]
Therefore, the rate constant of the reaction is increased by factor, [tex]4.93\times 10^{10}[/tex]