Chemistry 2710 Spring 2000 Quiz 10 Solution

The Arrhenius parameters are obtained by plotting $\ln k$ vs T^-1. Our first task is therefore to calculate these quantities:

T-1 ( K-1)	$3.661\times 10^{-3}$	$3.570\times 10^{-3}$	$3.470\times 10^{-3}$	$3.354\times 10^{-3}$
$\ln k$	-10.117	-9.469	-8.956	-8.294

Here is the graph of the data:

The line has slope $-5845\,\mathrm{K}$ and intercept 11.33. The activation energy and preexponential factor are therefore

$$\begin{eqnarray*}\bar{E}_a & = & -R(\mathrm{slope}) = 48.6\,\mathrm{kJ/mol},\\ ... ...y & = & e^{11.33} = 8.33\times 10^4\,\mathrm{L\,mol^{-1}s^{-1}}. \end{eqnarray*}$$

It should be noted that this reaction has unusually small activation parameters.