# If the temperature coefficient alpha for a certain reaction is 2.5, then if the reaction was run at 283^@ "C" and 293^@ "C", what is the activation energy in "kJ/mol"?

Dec 17, 2016

${E}_{a} = 6.5 \times {10}^{3}$ $\text{kJ/mol}$

From Wikipedia, the definition of the temperature coefficient $\alpha$, in regards to the rate constant $k$ and temperature $T$, is:

$\frac{\mathrm{dk}}{k} = \alpha \mathrm{dT}$

If we integrate this from state 1 to state 2 on the left side, and ${T}_{1}$ to ${T}_{2}$ on the right side:

${\int}_{\left(1\right)}^{\left(2\right)} \frac{1}{k} \mathrm{dk} = {\int}_{{T}_{1}}^{{T}_{2}} \alpha \mathrm{dT}$

When we assume that the temperature coefficient stays constant across this small temperature range, we can pull $\alpha$ out of the integral and obtain (noting that the integral of $\frac{1}{x}$ is $\ln x$):

$\ln \left({k}_{2}\right) - \ln \left({k}_{1}\right)$

$= \textcolor{g r e e n}{\ln \left({k}_{2} / {k}_{1}\right)} = \alpha \left({T}_{2} - {T}_{1}\right)$

$= \left(2.5\right) \left({293}^{\circ} \text{C" - 283^@ "C}\right)$

$= \textcolor{g r e e n}{25}$

where we used the fact that intervals in the celsius temperature scale are equal to intervals on the Kelvin temperature scale.

Thus, we have indirectly calculated $\ln \left({k}_{2} / {k}_{1}\right)$. Recall that this shows up in the form of the Arrhenius equation that demonstrates the temperature dependence of the rate constant:

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

Therefore, we can now algebraically solve for and calculate the activation energy ${E}_{a}$, in $\text{kJ/mol}$, as:

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

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

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

= -("0.008314472 kJ/mol"cdot"K")(25)[(("556.15 K")("566.15 K"))/(556.15 - 5"66.15 K")]

$= \textcolor{b l u e}{{6.5}_{45} \times {10}^{3}}$ $\textcolor{b l u e}{\text{kJ/mol}}$

where the subscripts indicate digits past the last significant figure.

This means that the reactants have about a $\text{6500-kJ}$ energy barrier in order for the reaction to proceed successfully.