# The rate constant, k, for a particular reaction is 1.3xx10^(-4)"M"^(-1)"s"^(-1) at 100^@"C" and 1.5xx10^(-3)"M"^(-1)"s"^(-1) at 150^@"C". What is the E_a (in kJ) for this reaction?

Mar 17, 2016

${\text{64 kJ mol}}^{- 1}$

#### Explanation:

Your tool of choice here will be the Arrhenius equation, which looks like this

color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" ", where

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

As you can see, the Arrhenius equation establishes a relationship between the rate constant and the absolute temperature at which the reaction takes place.

In other words, this equation allows you to determine how a change in temperature affects the rate of the reaction.

Now, convert the two temperatures from degrees Celsius to Kelvin by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will thus have

${T}_{1} = {100}^{\circ} \text{C" + 273.15 = "373.15 K}$

${T}_{2} = {150}^{\circ} \text{C" + 273.15 = "423.15 K}$

So, you know that at ${T}_{1}$, the rate constant for your reaction takes the form

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

Likewise, at ${T}_{2}$ the rate constant takes the form

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

Divide these two equations to get rid of the pre-exponential factor, $A$

${k}_{1} / {k}_{2} = \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{A}}} \cdot \text{exp"(-E_a/(R * T_1)))/(color(red)(cancel(color(black)(A))) * "exp} \left(- {E}_{a} / \left(R \cdot {T}_{2}\right)\right)\right)$

This will be equivalent to - using the properties of exponents

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

Take the natural log of both sides of the equation to get

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

Rearrange to solve for ${E}_{a}$, the activation energy of the reaction

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{E}_{a} = R \cdot \ln \frac{{k}_{1} / {k}_{2}}{\frac{1}{T} _ 2 - \frac{1}{T} _ 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finally, plug in your values to get

E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))

${E}_{a} = {\text{64,212.3 J mol}}^{- 1}$

Expressed in kilojoules per mole, the answer will be

${E}_{a} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{64 kJ mol}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.