# The activation energy of a certain reaction is 87 kJ·mol–1. What is the ratio of the rate constants for this reaction when the temperature is decreased from 37 °C to 15 °C?

### 1 Answer | Add Yours

Let the rate constant for the reaction at `37^o C` be `k_1`

the rate constant for the reaction at `15^o C` be `k_2`

So, `k_1/k_2=e^((E_a/R)(1/T_2-1/T_1))`

Here, `T_1=37^o C=(37+273) K=310 K`

`T_2=15^o C=(15+273) K=288 K`

`E_a=87 kJ*mol^-1=87*10^3 J*mol^-1`

`R=8.3145 J*K^-1*mol^-1`

Plugging in the values in the above equation:

`k_1/k_2=e^2.5784=13.2`

Therefore, the ratio of the rate constants for this reaction when the temperature is decreased from 37 °C to 15 °C is **13.2**.

**Sources:**