I think sometimes that equilibrium would be better taught without le Chatelier! But his Principle is with us for the foreseeable future, so there it is. You can find an extract from his original paper here.

A common question is the effect of a change in conditions on an equilibrium. This can be wrapped up in a variety of ways, each with its own pitfall.

Consider the following equilibrium;

2SO₂(g) + O₂(g) ó 2SO₃(g) DH = -197 kJ mol^-1

(a) Write the expression for K_c for this reaction.

(b) What is the effect of an increase in temperature on

        (i) the position of equilibrium

        (ii) the rate of the reaction?

(a) The common error here is one of notation; [X] means 'the concentration of X', and you must use the convention. You aren't there to explain your non-standard notation!

(b)(i). Le Chatelier predicts that an increase in temperature will favour the endothermic direction in the equilibrium, in this case the reverse reaction; the concentration of sulphur dioxide and of oxygen will rise - the reaction will 'move to the left'. The common problems are:

• it is not always made clear by a candidate which is the exothermic process - 'the reaction is exothermic' is open to some argument;
• the answer is given that 'the reaction moves in a way so as to reduce the temperature'.

The reason why the 'direction' of the reaction talked about has to be made clear is that the answers offered by many people are quite long and often quite difficult to unravel. Sometimes they are self-contradictory. The act of defining the reaction in question in your mind will help to clarify and abbreviate your answer.

The second point made above is of course quite true; a candidate who had written quite a lot of the script in this style would think it quite good and go home satisfied. The script would not score highly. The candidate is asking the examiner to second-guess the answer - which way does the writer mean? There is nothing to show whether there is any understanding of the principles at all.

There is also a fundamental misunderstanding of the whole question. When the temperature is changed, the temperature meant is the equilibrium temperature; and the change is assumed to come from immersion of the reacting system in a heat reservoir of such a large capacity that its temperature does not change. A thermostat, in fact.

The shift in the equilibrium position does not come from any 'desire' of the system to reduce the temperature again. The system acquires a new composition so that it is once more in equilibrium, but the absorbtion of the necessary energy from the thermostat does not change the thermostat's temperature, by definition. Another criticism that could be made of the suggestion that it does alter this temperature is that it suggests a confusion in the writer's mind between heat energy and temperature.

(b)(ii) The problem here is (and was!) that candidates remain solidly in equilibrium mode, and link the fall in yield of the forward reaction (correct) to a reduction of the rate. NOT correct! The rate goes up if the temperature rises, always. So you get less sulphur trioxide (equilibrium constant) but you get that small amount quickly (rate constant).

The equilibrium position and the rate at which equilibrium is attained are often confused. Get them clear in your mind.

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