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What is Le Chatelier's principle and how do you apply it to reversible reactions?

Le Chatelier's principle is commonly known as the quilibrium law and it is useful in order to be able to analyse reversible reactions and how changing the conditions will affect which way the reaction will go.

First, a little on reversible reactions. A reversible reaction is a chemical reaction in which the reaction can go both ways, i,e, the reactants can react to make the products and the products can also react to make the reactants again. In equation form, a reversible reaction looks like this:

A+B ↔C+D where A and B are the original reactants and C and D are the products. 

A reversible reaction taking place in a closed system will mean that eventually a position of equilibrium will be reached. All this means is that the quantity of the reactants and products will reach a constant level and they will not change. Does this mean that the reaction has stopped? The answer is NO! Even though it may seem the reaction has stopped, the reaction is still taking place in both directions: A and B are still reacting to make C and D, and C and D are still reacting to make A and B, but the rates of both the forward and reverse reactions are equal meaning there is no net change in the quantities. Think of it as if you are running on a treadmill in the opposite direction. The treadmill is pushing you one way, but you are running in the opposite direction and so your position does not change if your speed matches the treadmill's speed, but you are still moving your legs. The equilibirum is said to be a dyamic equilibrium whereby both reactions are still happening, but they 'cancel' each other out.

Now lets get back to Le Chatelier's principle. This principle simiplified states that in a dynamic equilibirum under closed conditions, if there is any change in the surrounding conditions, the position of the equilibirum will shift to oppose that change. Let's take a look at the Haber process to better explain Le Chatelier's principle. The Haber process reaction is 

N2 + 3 H2 → 2 NH3.

The forward reaction here is exothermic and the reverese reaction is endothermic. So if we were to increase the temperature at which this reactions occurs, the system will try to oppose the change by removing the extra heat. Which process removes the extra heat? It is the endothermic reaction as that is the reaction which takes in the heat energy. So the reverse reaction is favoured which will push the position of equilibirum to the left meaning there will be a higher quantity of nitrogen and hydrogen produced at equilibirum. 

If we were to decrease the temperature, the system would try to increase the temperature by throwing out heat. The exothermic forward reaction throws out heat so this would be favoured and the position of equilibirum would shift to the right.

Now let's talk about changes to the pressure in the system. The side of the reaction with more gaseous moles is the side that is at a higher pressure. In this case, there is one mole of nitrogen and 3 moles of hydrogen on the left which is 4 moles of gas overall. On the right, there is only 2 moles of ammonia. So the left side is at a higher pressure. If we were to increase the pressure, the system would try to oppose the change by decreasing the pressure and it would do this by favouring the reaction that produces the lesser number of gaseous moles which is the forward reaction as 4 moles of gas are reacting to produce 2. So the position of equilibrium would shift to the right so more ammonia would be produced once equilibrium has been reached.

If the pressure is decreased, the vice versa is true and so the position of equilibirum would shift to the left.

To summarise, an decrease in temperature favours exothermic reactions and an increase would favour an endothermic reaction. An increase in the pressure would favour the reaction producing a lesser number of gaseous moles and a decrease in pressure would favour the reaction producing a greater number of gaseous moles.

A catalyst has no impact whatsoever on the quilibirum position. It simply speeds up the rate of both the forward and reverse reactions equally meaning that the equilibrium position is attained quicker but there is no change to the extent of the reaction, i.e. the equilibirum position.

Mohammad A. 13 Plus  Maths tutor, GCSE Maths tutor, A Level Maths tut...

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