# Question 0b94b

May 30, 2015

The activation energy for this reaction is equal to +161 kJ/mol.

You were given the rate constant for a reaction at two different temperatures, 298 K and 425 K. This means tha you can use the Arhennius equation to calculate the activation energy for this reaction

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

$k$ - the rate constant;
$A$ - the Arhennius factor;
${E}_{A}$ - the activation energy for the reaction;
$R$ - the gas constant, equal to $8.314 {\text{J/mol"^(-1)"K}}^{- 1}$
$T$ - the tmperature at which the reaction takes place - expressed in Kelvin.

For the two rate constants given, you can write

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

Take the natural log from both sides of the equations to get

$\ln \left({k}_{1}\right) = \ln A + \left(- {E}_{A} / \left(R {T}_{1}\right)\right)$ $\text{ } \textcolor{b l u e}{\left(1\right)}$

and

$\ln \left({k}_{2}\right) = \ln A + \left(- {E}_{A} / \left(R {T}_{2}\right)\right)$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

Subtract equation $\textcolor{b l u e}{\left(2\right)}$ from equation $\textcolor{b l u e}{\left(1\right)}$ to get

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

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

Now plug your values into the equation and solve for ${E}_{A}$

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

E_A = (ln((3.16 * 10^(-15)cancel("s"^(-1)))/(8.66 * 10^(-7)cancel("s"^(-1)))) * 8.314"J"/("mol" * cancel("K")))/((1/425 - 1/298)cancel("K")) = (-161531"J"/"mol")/((1/425 - 1/298))#

${E}_{A} = \text{161086 J/mol}$

Rounded to three sig figs and expressed in kJ per mol, the answer will be

${E}_{A} = \textcolor{g r e e n}{\text{+161 kJ/mol}}$