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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:
\includegraphics{quiz10.eps}
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.



Marc Roussel
2000-04-07