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Calcualting Activation Energy using Arrhenius equation and plot
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[QUOTE="FaraDazed, post: 4664209, member: 438851"] This is part of a lab report so not sure if the template agrees with what I am asking but basically using the Arrhenius equation in the form of [itex]ln(k)=ln(A)+(\frac{-E_a}{R})(\frac{1}{T})[/itex] I plotted a straight line graph of ln(k) vs 1/T and found the gradient to be -5525. Its the units that are confusing me so not sure if it is as straight forward as what I have done below. [tex] gradient=\frac{-E_a}{R} \\ (gradient)(R)=-E_a \\ (-5525)(8.3145)=-E_a \\ -45937.6=-Ea \\ ∴ E_a=45937.6 Joules [/tex] any help is really appreciated, thanks. [/QUOTE]
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Calcualting Activation Energy using Arrhenius equation and plot
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