Chemguide: Support for CIE A level Chemistry

Learning outcomes 5.4(c) and 5.4(d)

These statements are about using ΔG° to decide whether or not a change will be spontaneous, and how changing the temperature might affect the spontaneity.

Before you go on, you should find and read the statements in your copy of the syllabus.

Spontaneous changes

In everyday life, something is spontaneous if it happens of its own accord, without any input from outside. The same thing is true in chemistry, but there is one major difference which defies everyday common sense.

If you drop marble chips (calcium carbonate) into dilute hydrochloric acid, there is an immediate fizzing. You don't need to do anything else - the reaction happens entirely of its own accord. It is a spontaneous change.

But in chemistry, a spontaneous change doesn't have to be rapid; in fact, it can be very, very, very slow indeed - even infinitely slow!

For example, carbon burns in oxygen to make carbon dioxide, but a piece of carbon will stay totally unchanged however long you keep it unless you first heat it. The energetics are right for a reaction to happen, but there is a huge activation energy.

Chemistry counts the reaction between carbon and oxygen as spontaneous! Personally, I think that is daft, and I prefer the word "feasible", which is often used in this topic. However, CIE uses the technical term "spontaneous", and so that is what we will use too.

Spontaneous changes and ΔG°

Whether or not a reaction (or other physical change) is spontaneous depends on the sign of ΔG°. If ΔG° is positive, then the reaction isn't spontaneous under standard conditions - it can't happen.

For a reaction to be spontaneous under standard conditions, ΔG° has to be negative.

Remember that although it may be feasible (a better word than spontaneous!), the reaction may not actually happen in any sensible time scale if there is a high activation energy barrier.

ΔG changes with temperature

You have met the equation:


Note:  Notice that I have removed the standard symbols. We are no longer going to be talking about standard conditions, and so these don't apply any more. At this level, we always make the approximation that the values of ΔH and ΔS aren't affected by temperature.

Using this equation is easily illustrated using the decomposition of calcium carbonate:

The value of ΔS for this reaction is +160.4 J K-1 mol-1.

Note:  CIE use this example as a part of a specimen paper, but quote a slightly different value for ΔS. The value I quote is consistent with a calculation result from my calculation book. Don't worry about it! Nobody is expecting you to remember these values.

Remember that if you are using this in ΔG calculations, you first have to convert into kJ.

ΔS = +0.1604 kJ K-1 mol-1.

Suppose you had some calcium carbonate in the lab at 293 K. You can calculate a value of ΔG as:

ΔG = +178 - 293(+0.1604) = +131 kJ mol-1

The value is positive and so the reaction isn't spontaneous (feasible). It cannot happen at this temperature.

But suppose you heated it to 1000°C (1273 K). Recalculating gives:

ΔG = +178 - 1273(+0.1604) = -26.2 kJ mol-1

This value is negative, and so the reaction is spontaneous at this temperature. And you know, of course, that if you heat calcium carbonate strongly enough, it decomposes to give calcium oxide and carbon dioxide.

So how strongly do you have to heat it? You can work out an approximate temperature by finding out at what point ΔG becomes negative (i.e. less than 0).

Note:  It will only be an "approximate temperature" because of the approximations we make about ΔH and ΔS not changing with temperature. Don't worry about this for CIE purposes.

For a reaction to be spontaneous, the value of ΔG has to be less than 0. In mathematical terms, it is spontaneous if:

ΔG < 0

Because ΔG = ΔH - TΔS, that means that for a spontaneous reaction:

ΔH - TΔS < 0

If you know values for ΔH and ΔS, then you can work out a value for T which makes this expression less than 0.

In the case we are looking at

ΔH = +178 kJ mol-1, and

ΔS = +0.1604 kJ K-1 mol-1

Putting those values into the expression ΔH - TΔS < 0 gives

178 - T x 0.1604 < 0

You can treat the "less than" sign just like an equals sign, and so rearranging this gives:

178 < T x 0.1604

178 / 0.1604 < T

1110 < T

That's a strange way of looking at it, of course ("1110 is less than T."). But that is just the same as saying that T has to be greater than 1110 K.

Working out the effect of temperature without doing calculations

Look again at the equation:


Remember that for a reaction to be spontaneous, ΔG has to be negative.

ΔH could be negative (an exothermic reaction) or positive (an endothermic reaction). Similarly ΔS could be either positive or negative.

There are four possible combinations of the signs of ΔH and ΔS. I want to look at those in turn.

Where ΔH is negative and ΔS is positive

In the equation ΔG = ΔH - TΔS:

ΔH is negative. TΔS is positive, and so -TΔS is negative.

Both terms are negative irrespective of the temperature, and so ΔG is also bound to be negative. The reaction will be spontaneous at all temperatures.

Where ΔH is positive and ΔS is negative

In the equation ΔG = ΔH - TΔS:

ΔH is positive. TΔS is negative, and so -TΔS is positive.

Both terms are positive irrespective of the temperature, and so ΔG is also bound to be positive. The reaction will not be spontaneous at any temperature.

Where ΔH is positive and ΔS is positive

In the equation ΔG = ΔH - TΔS:

ΔH is positive. TΔS is positive, and so -TΔS is negative.

Now increasing the temperature will change things. At higher temperatures, -TΔS will become more and more negative, and will eventually outweigh the effect of ΔH.

The reaction won't be spontaneous at low temperatures, but if you heat it, there will be a temperature at which it becomes spontaneous, because ΔG becomes negative.

The decomposition of calcium carbonate is a case like this, and we have done three calculations around it.

Where ΔH is negative and ΔS is negative

In the equation ΔG = ΔH - TΔS:

ΔH is negative. TΔS is negative, and so -TΔS is positive.

Again there will be a temperature effect. As temperature increases, -TΔS will become more and more positive, and will eventually outweigh the effect of ΔH.

At low temperatures, ΔG will be negative because of the effect of the negative ΔH, but as you increase the temperature, the effect of the positive -TΔS will eventually outweigh that. The value of ΔG will then become positive, and the reaction will no longer be spontaneous.

In summary

I really wouldn't suggest you tried to learn this - it is too confusing. Make sure that you understand it, so that when a question comes up you can work it out at the time.

If you want to practise on a real case, look at Q7(c) in the specimen paper for the 2016 syllabus. That is the only question available at the time of writing, and there won't be anything else available until after the June 2016 exams.

But it isn't too difficult to make up questions for yourself which would help you through the whole topic. Choose a simple reaction where you can find entropy values for everything involved. Use a data book if you have one, or find the equivalent information online, and avoid anything which involves solutions.

Use enthalpies of formation to work out the enthalpy change for the reaction. (If you use combustion reactions, for example, then you can just look up the enthalpy of combustion to save you the bother!)

Then work out the entropy change during the reaction. Decide what the effect of temperature will be on the feasibility of the reaction, and then work out values of ΔG at some widely different temperatures to see if you are right. (When you are doing those calculations, don't forget to correct the entropy change value from joules to kJ!)

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© Jim Clark 2014