Question

The activation energy of a reaction is 75.2 kJ mol⁻¹ in the absence of a catalyst and 50.14 kJ mol⁻¹ with a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at 25°C? (R = 8.314 JK⁻¹ mol⁻¹).

Answer 1

We can solve this using the Arrhenius equation in the ratio form:

`k_"catalyst"/k_"no catalyst" = "exp" ((E_a^circ  - E_a^"cat")/(RT))`

Where:

`E_a^circ` = 75.2 kJ/mol = 75200 J/mol

`E_a^"cat"` = 50.14 kJ/mol = 50140 J/mol

R = 8.314 J K−1 mol−1

T = 25°C = 298 K

`k_"catalyst"/k_"no catalyst" = "exp" ((75200  - 50140)/(8.314 xx 298))`

= `"exp" ((25060)/(2478.572))`

= exp(10.11)

= 24633

∴ The rate of reaction increases by approximately 24663 times in the presence of the catalyst.