Rates of reaction - kinetics
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How would I rearrange the Arrhenius Equation to find activation energy?
Igloo writes … This is a good question.
Although the equation can be rearranged (making EA the subject of the formula) this is not an appropriate procedure in this case. You need to make use of all your experiments to “average out” your results and to eliminate erroneous values, so a graphical method is appropriate for this determination.
You should remember from your GCSE maths that when plotting an equation such as:
In other words, the quantity in front of “x” (known as the “coefficient of x”) is always the gradient of the line.
The Arrhenius Equation is:
ln(rate of reaction) = constant – EA/RT
I assume you know what each of these “letters” stands for.
Next, compare the Arrhenius Equation with either of the simple equations above.
Instead of plotting y on the vertical axis you plot “ln(rate)”, and instead of plotting x on the horizontal axis you plot “1/T”, so, just as before, the coefficient of 1/T must be the gradient of the line. In this case, the gradient is “ – EA/R" (don’t forget the minus sign), so you need to do the following:
[1] Calculate “ln(rate)” values for each temperature (using a calculator)
[2] Calculate “1/T” values for each temperature (converted to Kelvin first)
[3] Plot ln(rate) values on the vertical axis against 1/T values on the horizontal axis
[4] Measure the gradient of your line (which you will find has a negative slope)
[5] Since your gradient is – EA/R, you multiply your (negative) gradient by –R
(-8.31 J K–1 mol–1), and you will have the activation energy, EA (which is positive)
This is all described in the Nuffield Students’ Book on page 258, and there are plenty of practice questions at the end of the topic on pages 268–271.
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updated: 20 August 2003
