deletions | additions       diff --git a/subsubsection_Finding_the_activation_energy__.tex b/subsubsection_Finding_the_activation_energy__.tex  index b8c81cb..38a6fcf 100644  --- a/subsubsection_Finding_the_activation_energy__.tex  +++ b/subsubsection_Finding_the_activation_energy__.tex 
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The Arrhenius equation states that  \begin{equation}  k=Ae^{\frac{-E_a}{RT}  \end{equation} By taking natural logarithms of both sides of this equation, it can be shown that:  \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 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}.