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Noah Phipps edited subsubsection_Finding_the_activation_energy__.tex
almost 8 years ago
Commit id: 4ae42721e655abedc6a944e77327ce6977910836
deletions | additions
diff --git a/subsubsection_Finding_the_activation_energy__.tex b/subsubsection_Finding_the_activation_energy__.tex
index 09451bb..ff4671d 100644
--- a/subsubsection_Finding_the_activation_energy__.tex
+++ b/subsubsection_Finding_the_activation_energy__.tex
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\begin{equation}
ln(k)=ln(A)-\frac{E_a}{RT}
\end{equation}
By plotting a graph of ln(k) against 1/T, the gradient is equivalent to $\frac{-E_a}{R}$. This gives the reaction an Activation Energy of 85.04 kJmol\textsuperscript{-1}. This is high for a reaction, and suggests an experimental error. The points plotted on the graph suggest that a curve would be more appropriate, as there is a weak linear relationship. However,
E\textsubscripta E\textsubscript{a} must be constant for a given reaction, so a straight line was plotted regardless of the apparent curve.
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The intercept of the graph can be used to calculate the value of the frequency factor, which in this case is 1.13$\times$10\textsuperscript{15}s\textsuperscript{-1}.