**The Arrhenius Equation**

\(k = A{e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}\)

k = Rate Constant (units vary)

E_{a} = Activation Energy (Jmol^{-1})

T = Temperature (K)

A = Arrhenius Constant (approximate constant)

e = exponential (approximately 2.71 but its a symbol on your calculator a bit like π)

**As E _{a} gets bigger k gets smaller, why!?:**

If you were to increase –E_{a} this would mean the fraction (-E_{a}/RT) would become a larger negative number. If you were to calculate Ae to the power of a more negative number then you will find it will give you a very small number so k has decreased. If you think about it, this makes perfect sense as by having a large activation energy would mean a slow rate, as a smaller fraction of the particles would have the minimum energy to react (you may wish to look back at the Maxwell-Boltzmann distribution).

**As temperature rises k rises, why!?**

If you were to make T bigger, this means that the bottom of the fraction becomes larger, and therefore the entire fraction (-E_{a}/RT) becomes a less negative number. If you were to calculate Ae to the power of this less negative number you will find it will give you a much larger number and will therefore give you a large rate constant. For a similar reason to the above, this makes a lot of sense because if you increase the temperature you will give a larger fraction of the particles the minimum energy needed to react.

**More useful Forms**

In an exam you will usually want to use an altered version of the Arrhenius equation. Below is how these equations are rearranged. If you find this difficult you may wish to just remember the equations highlighted in red.

\(k = A{e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}\)

Is the same as:

\(lnk = ln\left( {Ae - \frac{{{E_a}}}{{RT}}} \right)\)

Cancel ln and e:

\(lnk = lnA - \frac{{{E_a}}}{{RT}}\)

This version is much more user friendly and is usually the form you would use.

**Below is this equation rearranged to make E _{a} the subject:**

\(lnk = lnA - \frac{{{E_a}}}{{RT}}\)

Move lnA across:

\(lnA - lnk = \frac{{{E_a}}}{{RT}}\)

Move RT across to make E_{a} the subject:

\(\left( {lnA - lnk} \right) \times RT = {E_a}\)

**Now here is the same equation rearranged to make T the subject:**

\(lnk = lnA - \frac{{{E_a}}}{{RT}}\)

Move lnA across:

\(lnA - lnk = \frac{{{E_a}}}{{RT}}\)

Move T across to give:

\(\left( {lnA - lnk} \right) \times T = \frac{{{E_a}}}{R}\)

Finally move (lnA – lnk) across to make T the subject:

\(T = \frac{{{E_a}}}{{R \times \left( {lnA - lnk} \right)}}\)

**Now here is the same equation rearranged to make A the subject:**

\(k = A{e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}\)

Move A across:

\(\frac{k}{A} = {e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}\)

Move k across to give 1/A:

\(\frac{1}{A} = \frac{{{e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}}}{k}\)

Finally invert the equation to make A the subject:

\(A = \frac{k}{{{e^{ - \left( {\frac{{{E_a}}}{{RT}}} \right)}}}}\)

To summarise if you did not understand the rearrange process do not panic. You only need to be able to remember the four red forms and use them to find the unknown. It IS a good idea to attempt to understand this though.

The other form of question you may be given is to create an Arrhenius plot. This is when you plot the graph lnk against 1/T. This should give you a straight line where the gradient of this is equal to –E_{a}/R and the y intercept is lnA.

The reason for this relationship is due to the equation:

\(lnk = lnA - \frac{{{E_a}}}{{RT}}\)

Being in the form of a straight line equation:

\(y = c + m \times x\)

Again for the sake of these question you do not necessarily have to understand WHY this equation matches the equation of a straight line, only that the gradient and y intercept help you find the above unknowns.