# Question 2e1b0

Apr 23, 2016

You can say that the rate of the reaction approximately doubles for each 10 °C rise in the temperature.

#### Explanation:

This rule of thumb comes from the Arrhenius equation:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \ln \left({k}_{2} / {k}_{1}\right) = {E}_{a} / R \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right) \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where ${E}_{a}$ is the activation energy and ${T}_{2}$ is the higher temperature.

For example, if ${E}_{a} = \text{53 kJ//mol}$ and we increase the temperature from 25 °C to 35 °C, we get

ln(k_2/k_1) = ("53 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))))(1/(298 color(red)(cancel(color(black)("K")))) -1/(308 color(red)(cancel(color(black)("K"))))) = 6400 × 1.09 × 10^"-4" = 0.70#

${k}_{2} / {k}_{1} = {e}^{0.70} = 2$

That is, an increase of 10 °C in temperature doubles the rate.

The rule holds approximately at ambient temperatures for activation energies between 35 kJ/mol and 65 kJ/mol.