# The first order rate constant for hydrolysis of CH3Cl in H2O has a value of 3.32x10-10 s-1 at 25oC and 3.13x10-9 s-1 at 40oC. What is the activation energy?

Feb 2, 2015

The activation energy for the hydrolysis of $C {H}_{3} C l$ is $\text{+115.9 kJ/mol}$.

Like the other answer suggested, you need to use the Arrhenius equation to determine the activation energy for this reaction.

k =A * e^((-E_a/(RT)), where

$k$ - the rate constant of the reaction;
$A$ - the pre-exponential factor;
${E}_{a}$ - the activation energy;
$R$ - the universal gas constant - you'll need to use the one given in $\text{J/mol K}$ $\to$ $\text{8.314 J/mol K}$ to be exact;
$T$ - the temperature in Kelvin.

This equation allows you easily calculate the activation energy when you have two values for the rate constant measured at two different temperatures. Let's take ${k}_{1}$ measured at ${T}_{1}$

${k}_{1} = A \cdot {e}^{\frac{- {E}_{a}}{R {T}_{1}}}$

and ${k}_{2}$ measured at ${T}_{2}$

${k}_{2} = A \cdot {e}^{\frac{- {E}_{a}}{R {T}_{2}}}$

Take the natural log on both sides for both of these equations, you'll get

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

We should find a way to remove $\ln \left(A\right)$ from the equations since no value for $A$ is given to you; the easiest way to do so is to substrat equation (2) from equation (1)

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

You'll end up with

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

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

Now just plug in your values and solve for ${E}_{a}$ - do not forget to use temperature in Kelvin

$\ln \left(\frac{3.32 \cdot {10}^{- 10}}{3.13 \cdot {10}^{- 9}}\right) = {E}_{a} / R \cdot \left(\frac{1}{313.15} - \frac{1}{298.15}\right)$

E_a = (2.244 * "8.314 J/mol")/(0.000161) = "115.9 kJ/mol"