Question

Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?

Answer 1

The Arrhenius equation will tell us that the activation energy in this case must be 52.4 kJ/mol.

Explanation:

The Arrhenius equation says that the dependence of the rate constant `k` on temperature reads

`k=Ae^(-(E_a)/(RT))`

For the sake of this problem, let us say that `k_1` is the initial value of the constant at a temperature `T_1`, and `k_2` is the value at temperature `T_2`

To double the reaction rate, we must double the value of k, i.e. the ratio `k_2/k_1` must equal 2.

This will require the right hand side of the Arrhenius equation to change to

`Ae^(-(E_a)/(RT_2))` from `Ae^(-(E_a)/(RT_1))`

all of which can be written in ratio forms as

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

Taking the natural log of each side:

`ln2 = -(E_a/R)(1/T_2-1/T_1)`

or

`E_a= (ln2 (R))/(1/T_2-1/T_1)`

Using 308K for `T_2` and 298K for `T_1`

`E_a= (0.693 (8.314))/(1/308-1/298)`

`E_a=52.4 kJ/(mol)`