# The activation energy of a certain reaction is 46.6 kj/mol. At 20 degrees Celsius the rate constant is 0.0130s^-1. At what temperature would this reaction go twice as fast (T2)?

Mar 19, 2016

At ${31}^{\circ} \text{C}$

#### Explanation:

We can use The Arrhenius Equation:

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

$k$ is the rate constant

$A$ is the frequency factor and is constant

$R$ is the gas constant and $= 8.31 \text{J""/""K""/""mol}$

${E}_{a}$ is the activation energy

$T$ is the absolute temperature.

A more usable form of the equation is obtained by taking natural logs of both sides $\Rightarrow$

$\ln k = \ln A - {E}_{a} / \left(R T\right)$

If the reaction occurs at two temperatures ${T}_{1}$ and ${T}_{2}$ then we can write:

$\ln {k}_{1} = \ln A - {E}_{a} / \left(R {T}_{1}\right) \text{ } \textcolor{red}{\left(1\right)}$

$\ln {k}_{2} = \ln A - {E}_{a} / \left(R {T}_{2}\right) \text{ } \textcolor{red}{\left(2\right)}$

Now we can subtract both sides of $\textcolor{red}{\left(1\right)}$ and $\textcolor{red}{\left(2\right)} \Rightarrow$

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

Since we know that ${k}_{2} = 2 {k}_{1}$ this becomes:

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

:.-0.693=(46.6xx10^3)/ 8.31[1/T_2-1/293]

$\therefore \frac{1}{T} _ 2 - \frac{1}{293} = - 1.235 \times {10}^{-} 4$

$\therefore \frac{1}{T} _ 2 = - 1.245 \times {10}^{-} 4 + 34.13 \times {10}^{-} 4 = 34.13 \times {10}^{-} 4$

$\therefore {T}_{2} = 304 \textcolor{w h i t e}{x} \text{K}$

$\therefore {T}_{2} = 304 - 273 = {34}^{\circ} \text{C}$

This shows how a rise of just 14 degrees can double the rate of reaction.

The graphic below shows how a small increase in temperature results in a large increase in the number of particles which have the minimum energy for a reaction to happen: