5.1.1 (j,k) Effect of temperature on rate constants

Syllabus

(j) a qualitative explanation of the effect of temperature change on the rate of a reaction and hence the rate constant

(k) the Arrhenius equation:

(i) the exponential relationship between the rate constant, k and temperature, T given by the Arrhenius equation,

k = Ae^–Ea/RT

(ii) determination of Ea and A graphically using: ln k = –Ea/RT + ln A derived from the Arrhenius equation.

{Ea = activation energy, A = pre-exponential factor, R = gas constant (provided on the Data Sheet)}

{Explanation of A is not required.}

{Equations provided on the Data Sheet.}

What does this mean?

Qualitative explanation

You should know this from GCSE but it is a perennially poorly answered question in GCSE exams.

At higher temperatures particles move faster.

This leads to a higher collision frequency.

This alone would raise the rate of reaction.

But at higher temperatures collision have more energy and are also more likely to be successful because they are more likely to equal or exceed the Activation Energy.

A typical rate equation might be Rate = k[A]²[B]

We haven't increased any concentrations but the rate has increased anyway.

The only part of the right-hand side of the rate equation that can change to account for the increased rate is the rate constant.

In short, higher temperatures increase the rate constant, lower temperatures decrease in.

Quantitative

Welcome to the Arrhenius Equation, you'll come to love it.

k = Ae^–Ea/RT

You don't have to remember this since it'll be on the data-sheet but you do need to know what it means.

A is simply a constant.

e is the base of natural logarithms (ln)- if you do Maths then you'll become used to it. If not you can ignore its significance and simply use it on your calculator.

E_a - is the Activation Energy

R is the Gas Constant familiar from PV =nRT and also given on the Data Sheet.

T is the Temperature in Kelvin

Note that E_a is generally given in kJ/mol while R is in J/K/mol - so you will need to match these units if using both in a calculation.

As you can see the rate constant, k, rises exponentially with the temperature.

It would be relatively difficult for an examiner to ask you to do much with this graph except use it to find k at any given temperature.

Which is a fairly trivial sort of thing to have to do.

So it's unlikely to feature heavily in future exams.

However, it is quite likely that you'll be given a different graph and use it to find A and E_a.

Why?

Taking the natural log (ln) of both sides of the Arrhenius Equation gives us the following:

ln(k) = ln(Ae^–Ea/RT)

ln k = –Ea/RT + ln A

If we plot ln(k) against ¹/_T we find we get a straight line.

And Year 9 Maths tells us that y=mx + c

So in a graph of ln(k) vs ¹/_T we can calculate the constant A by finding the y-axis intercept (even if we have to extrapolate the line backwards)

The intercept is equal to ln(A).

And the gradient of the graph is equal to -^Ea/_R

There's something scary-looking about the Arrhenius Equation but if you learn to use it you don't really need to understand the Maths very well.

And no one will expect you to explain it.

Videos

1. What is the Arrhenius Equation?