Rates of reaction - kinetics (A2)
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How does 'y = mx + c' relate to the Arrhenius equation?
We have been asked this question and have decided to use it to take the opportunity to explain how to make sense of any equation for a straight line graph.
What’s the problem?
It's a big jump from GCSE Maths to the Arrhenius equation, and so you can be forgiven for finding the connection difficult to follow. However, the link is there, believe it or not.
Getting on the straight, if not the narrow!
There are many cases in applied science when it helps to try to resolve a numerical solution to a straight line graph.
There are many reasons for this, including the very practical one that, if you have a set of experimental results and you know that they are related by a straight line graph, it is relatively easy to choose the best straight line to draw. Curves are much more difficult!
There are several ways of describing a straight line graph by means of an equation. In A-level Chemistry we only need one of them. This is usually written:
y = mx + c
Explain, please
First of all, let's be clear about what the various terms mean in the equation: y = mx + c.
• 'y' means 'whatever you decide to plot on the y axis (the vertical one).
• 'x' means 'whatever you decide to plot on the x axis.
• 'c' means the reading on the y axis at the point where the graph crosses it. For a given set of values of x and y, c has a constant value.
• 'm' is the crucial one as far as we are concerned. 'm' is the gradient of the graph.
If m is so important, how can you find it?
You can find the value of m by measuring a change in y and dividing it by the corresponding change in the value of x. The steeper the graph is, the bigger is the value of m.
The value of m is positive when the straight line rises from left to right. It is negative if the line falls from left to right. Assuming that the axis and scales are drawn up in the usual way.
So where does the Swedish person come in?
Right now, as it happens! Let's look at the Arrhenius equation. This is normally written
lnk = constant – EA/RT
but I am going to re-arrange it slightly so that it reads:
lnk = -EA/RT + constant
Now, what do all these symbols mean?
• k is the rate constant for the reaction;
• lnk is its logarithm.
• Although it is called the rate 'constant', it does in fact vary with the temperature T (in Kelvins).
• EA is the activation energy of the reaction.
• R is called the gas constant; this one really is a constant with the value 8.31 Joules per Kelvin. You don't need to remember this value of R, it is always given to you when you need it.
A little light relief: The Constant Chemist
Chemists have to put up with a certain amount of mockery from other scientists, particularly physicists, over the word ‘constant’. When a physicist speaks of a constant, the word often means something which has a universal, unchanging value such that you can look it up in a reference book.
Chemists often define a constant and then immediately go into detail about how it varies! The gas constant, R, is a physicist’s constant but the rate constant, k, is only constant for a particular reaction and then only at a particular temperature. Sorry, but we have to live with it!
At last, the connection
OK, we are ready to compare the two equations.
In y = mx + c, y is lnk and x is 1/T, these are the variables.
If you plot a graph of lnk on the y-axis against 1/T on the x axis, you will get a straight line the gradient of which is –EA/R.
If you measure this gradient on your graph you can easily calculate a value for the activation energy.
Note that with the axis drawn up in the usual way the line will have a negative gradient.
Get yourself organised
In the real chemical world you don’t get lnk and 1/T presented to you on a plate, as it were.
The raw results of experiments often give you a list of k values and a list of temperatures, T. T must be in Kelvins, by the way, not Celsius.
It is best to construct a table with headings such as:
| T | 1/T | k | lnk |
It is quite common to measure the rate of the reaction by recording the time t to reach a particular stage in the reaction (such as the disappearance of a piece of magnesium ribbon). The longer this takes, the slower the reaction, so the rate is proportional to 1/t. Given that everything is kept the same except the temperature in a series of experiments, you can use this rate instead of the rate constant - which saves a bit of arithmetic. You will need another column in your table.
| T | 1/T | t | 1/t | ln(1/t) |
You should include the units in your table too. T is in Kelvin (K). The time intervals are usually in seconds (s). So the units of 1/t are s-1
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updated: 12 January 2007
