# 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?

Jan 28, 2017

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

$k = A {e}^{- \frac{{E}_{a}}{R T}}$

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

$A {e}^{- \frac{{E}_{a}}{R {T}_{2}}}$ from $A {e}^{- \frac{{E}_{a}}{R {T}_{1}}}$

all of which can be written in ratio forms as

${k}_{2} / {k}_{1} = {e}^{\left(- {E}_{a} / R\right) \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)}$

Taking the natural log of each side:

$\ln 2 = - \left({E}_{a} / R\right) \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)$

or

${E}_{a} = \frac{\ln 2 \left(R\right)}{\frac{1}{T} _ 2 - \frac{1}{T} _ 1}$

Using 308K for ${T}_{2}$ and 298K for ${T}_{1}$

${E}_{a} = \frac{0.693 \left(8.314\right)}{\frac{1}{308} - \frac{1}{298}}$

${E}_{a} = 52.4 k \frac{J}{m o l}$