This page looks at the way that rate constants vary with temperature and activation energy as shown by the Arrhenius equation. | ||

Note: If you aren't sure what a rate constant is, you should read the page about orders of reaction before you go on. This present page is at the hard end of the rates of reaction work on this site. If you aren't reasonably confident about the basic rates of reaction work, explore the rates of reaction menu first. | ||

You will remember that the rate equation for a reaction between two substances | ||

Note: If you don't remember this, you must read the page about orders of reaction before you go on. Use the BACK button on your browser to return to this page. | ||

The rate equation shows the effect of changing the concentrations of the reactants on the rate of the reaction. What about all the other things (like temperature and catalysts, for example) which also change rates of reaction? Where do these fit into this equation? These are all included in the so-called This is shown mathematically in the Arrhenius equation.
Starting with the easy ones . . . *Temperature, T*To fit into the equation, this has to be meaured in kelvin. *The gas constant, R*This is a constant which comes from an equation, pV=nRT, which relates the pressure, volume and temperature of a particular number of moles of gas. It turns up in all sorts of unlikely places! *Activation energy, E*_{A}This is the minimum energy needed for the reaction to occur. To fit this into the equation, it has to be expressed in joules per mole - not in kJ mol ^{-1}.
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Note: If you aren't sure about activation energy, you should read the
introductory page on rates of reaction before you go on. Use the BACK button on your browser to return to this page. | ||

And then the rather trickier ones . . . *e*This has a value of 2.71828 . . . and is a mathematical number, a bit like pi. You don't need to worry exactly what it means, although if you have to do calculations with the Arrhenius equation, you may have to find it on your calculator. You should find an *e*button - probably on the same key as "ln".^{x}*The expression, e*^{-(EA / RT)}For reasons that are beyond the scope of any course at this level, this expression counts the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature. You will find a simple calculation associated with this further down the page. *The frequency factor, A*You may also find this called the pre-exponential factor. A is a term which includes factors like the frequency of collisions and their orientation. It varies slightly with temperature, although not much. It is often taken as constant across small temperature ranges.
By this time you've probably forgotten what the original Arrhenius equation looked like! Here it is again: You may also come across it in a different form created by a mathematical operation on the standard one: "ln" is a form of logarithm. Don't worry about what it means. If you need to use this equation, just find the "ln" button on your calculator.
You can use the Arrhenius equation to show the effect of a change of temperature on the rate constant - and therefore on the rate of the reaction. If the rate constant doubles, for example, so also will the rate of the reaction. Look back at the rate equation at the top of this page if you aren't sure why that is. What happens if you increase the temperature by 10°C from, say, 20°C to 30°C (293 K to 303 K)? The frequency factor, A, in the equation is approximately constant for such a small temperature change. We need to look at how e Let's assume an activation energy of 50 kJ mol At 20°C (293 K) the value of the fraction is: By raising the temperature just a little bit (to 303 K), this increases: You can see that the fraction of the molecules able to react has almost doubled by increasing the temperature by 10°C. That causes the rate of reaction to almost double. This is the value in the rule-of-thumb often used in simple rate of reaction work. | ||

Note: This approximation (about the rate of a reaction doubling for a 10 degree rise in temperature) only works for reactions with activation energies of about 50 kJ mol^{-1} fairly close to room temperature. If you can be bothered, use the equation to find out what happens if you increase the temperature from, say 1000 K to 1010 K. Work out the expression -(E_{A} / RT) and then use the e^{x} button on your calculator to finish the job.
The rate constant goes on increasing as the temperature goes up, but the | ||

A catalyst will provide a route for the reaction with a lower activation energy. Suppose in the presence of a catalyst that the activation energy falls to 25 kJ mol If you compare that with the corresponding value where the activation energy was 50 kJ mol It's no wonder catalysts speed up reactions!
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Note: If you read this carefully, you should notice that I am not saying that the reaction will be 30000 times faster. There may well be 30000 times more molecules which can react, but it is highly likely that the frequency factor will have changed in the presence of the catalyst. And the rate constant k is just one factor in the rate equation. You won't just have the original reactants present as before. The catalyst is bound to be involved in the slow step of the reaction, and a new rate equation will have to include a term relating to the catalyst.
Nevertheless, the catalysed reaction is still going to be a lot faster than the uncatalysed one because of the huge increase in sufficiently energetic molecules. | ||

If you have values for the rate of reaction or for the rate constant at different temperatures, you can use these to work out the activation energy of the reaction. Only one UK A' level Exam Board expects you to be able to do these calculations. They are included in my chemistry calculations book, and I can't repeat the material on this site. | ||

Note: There is no way of making this sufficiently different from what is in the book to avoid being in breach of contract with my publishers if I included it on this site.
If you are interested in my chemistry calculations book you might like to follow this link. | ||

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© Jim Clark 2002 (modified October 2013) |