# The rate constant for a reaction at 25.0 degrees C is 0.010 s^-1 and its activation energy is 35.8 KJ. How do you find the rate constant at 50.0 degrees C?

Oct 24, 2016

You compare them at two temperatures and determine the Arrhenius plot.

## $\textcolor{w h i t e}{\implies} {k}_{2} = {k}_{1} {e}^{\frac{{E}_{a}}{R} \left[\frac{1}{{T}_{1}} - \frac{1}{{T}_{2}}\right]}$

Be sure to use $8.314472 \times {10}^{- 3} \text{kJ/mol"cdot"K}$ if ${E}_{a}$ is in $\text{kJ}$, and to convert ${T}_{1}$ and ${T}_{2}$ into $\text{K}$.

The Arrhenius equations are:

${k}_{1} = A {e}^{- {E}_{a} \text{/} R {T}_{1}}$
${k}_{2} = A {e}^{- {E}_{a} \text{/} R {T}_{2}}$

The same reaction has the same activation energy ${E}_{a}$ and the same universal gas constant $R$. But at different temperatures, the rate constant $k$ differs.

Now, divide these equations.

${k}_{1} / {k}_{2} = {e}^{- {E}_{a} \text{/"RT_1)/e^(-E_a"/} R {T}_{2}}$

taking the natural logarithm allows you to separate the right side.

$\ln \left({k}_{1} / {k}_{2}\right) = \ln \left({e}^{- {E}_{a} \text{/"RT_1)/e^(-E_a"/} R {T}_{2}}\right)$

$= \ln \left({e}^{- {E}_{a} \text{/"RT_1)) - ln(e^(-E_a"/} R {T}_{2}}\right)$

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

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

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

To find the rate constant at a new temperature, try switching the sign on the left and right:

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

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

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

There, now it's easier to solve for ${k}_{2}$:

${k}_{2} / {k}_{1} = {e}^{\frac{{E}_{a}}{R} \left[\frac{1}{{T}_{1}} - \frac{1}{{T}_{2}}\right]}$