# Question a2fab

Mar 5, 2016

${\text{25.4 kJ mol}}^{- 1}$

#### Explanation:

Your starting point here will be the Arrhenius equation, which shows how the rate constant of a given reaction depends on the absolute temperature at which the reaction takes place

$\textcolor{b l u e}{| \overline{\underline{k = A \cdot \exp \left(- {E}_{a} / \left(R T\right)\right)}} |} \text{ }$, where

$k$ - the rate constant
$A$ - the pre-exponential factor, specific to a given reaction
${E}_{a}$ - the activation energy of the reaction
$R$ - the universal gas constant, used as ${\text{8.314 J mol"^(-1)"K}}^{- 1}$
$T$ - the absolute temperature of the gas

To find the activation energy of your reaction, take the natural log of both sides of the Arrhenius equation. This will get you

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

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

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

Now, you know the value of the rate constant at two different temperatures

• ${k}_{1} = 3.2 \cdot {10}^{- 6} \text{M s"^(-1) -> "at T"_1 = "298 K}$
• ${k}_{2} = 4.2 \cdot {10}^{- 5} \text{M s"^(-1) -> "at T"_2 = "398 K}$

This means that you can write

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

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

Divide these two equations to get rid of the $\ln \left(A\right)$ term

$\ln \left({k}_{1}\right) - \ln \left({k}_{2}\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(A\right)}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(A\right)}}} - {E}_{a} / \left(R {T}_{1}\right) + {E}_{a} / \left(R {T}_{2}\right)$

This will be equivalent to

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

This equation will allow you to find the value of ${E}_{a}$ for this reaction. Rearrange and plug in your values to get

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

E_a = ln((3.2 * 10^(-6)color(red)(cancel(color(black)("M s"^(-1)))))/(4.2 * 10^(-5)color(red)(cancel(color(black)("M s"^(-1)))))) * ("8.314 J mol"^(-1)color(red)(cancel(color(black)("K"^(-1)))))/((1/398 - 1/298)color(red)(cancel(color(black)("K"^(-1)))))#

${E}_{a} = {\text{25387 J mol}}^{- 1}$

I'll leave the answer rounded to three sig figs and expressed in kilojoules per mole

${E}_{a} = \textcolor{g r e e n}{| \overline{\underline{{\text{25.4 kJ mol}}^{- 1}}} |}$

Mind you, you should round this off to two sig figs, since that's how many sig figs you have for the two rate constants.