deletions | additions       diff --git a/subsubsection_Finding_the_activation_energy__.tex b/subsubsection_Finding_the_activation_energy__.tex  index 38a6fcf..09451bb 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 must be constant for a given reaction, so a straight line was plotted regardless of the apparent curve.  \\\\  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}.