# Reversible reactions

A simple model of a reversible reaction is a bit like the ‘two box’ simulation above, with the difference that the probability of a particle moving from left to right is different from the probability of a particle moving from right to left. This results in more counters in one box than the other once the system has settled down to a steady state. Notice again that even when a steady state (‘an equilibrium’) has been reached, there is still movement of counters between the two boxes. However, the number of particles moving L R in a given time is exactly the same as the number moving R L. The simulation allows you to investigate the effect of changing the two rates on the position of equilibrium.

 Note. It is important here to distinguish between the overall rate of particles moving from box to box and the intrinsic rate, which is related to the probability of an individual particle moving from one box to the other. The latter is analogous to the rate constant of a chemical reaction while the former is analogous to the observed rate. The overall rate is the intrinsic rate multiplied by the number of particles in the relevant box.

## Real reversible reactions

Real reactions are more complex.

As in the simulation above, the equilibrium position is governed by the intrinsic rates (rate constants) rates of the forward and back reactions. As there is usually more than one reactant and product, their proportions can vary. Temperature also affects the equilibrium position. The Equilibrium mixtures simulation allows you to investigate the effect of changing the initial concentrations of reactant and products and also the temperature on some important reversible reactions.

## ∆G and equilibrium

We measure the position of equilibrium in a reversible reaction by the equilibrium constant. This is a ratio of the general form [products] / [reactants]. So the bigger the value, the further the equilibrium is over to the right (products) side.

More formally, we find that for a reaction

aA + bB + cC xX + yY + zZ

the expression

 [X]eqmx [Y]eqmy [Z]eqmz _______________________ [A]eqma [B]eqmb [C]eqmc

is a constant, at a particular temperature, and is called Kc, the subscript ‘c’ meaning that it is expressed in terms of concentration.

 ie Kc = [X]eqmx [Y]eqmy [Z]eqmz _______________________ [A]eqma [B]eqmb [C]eqmc

 For gas reactions we often use the partial pressure as a measure of concentration but that is a detail.

Perhaps not surprisingly, we find that the equilibrium constant is linked to ∆G. The more negative is ∆G, the more products predominate and the bigger the equilibrium constant.

The mathematical expression is

G = –RT lnK or ∆G = –2.3 RT log10K (The factor of 2.3 converts between the use of natural logs, ln, and logs to the base 10, log10, sometimes just written ‘log’.) R is the gas constant. It has a value of 8.3 J K-1 mol-1.

The negative sign means that as ∆G becomes more negative, Kc gets bigger and the reaction gets nearer completion, ie with a greater proportion of products in the reaction mixture.

Using the above expression, a value of ∆G of –57 kJ mol-1 corresponds to an equilibrium constant of 1 x 1010 (the units depend on the reaction), ie the reaction is effectively complete. A ∆G of +57 kJ mol-1 corresponds to an equilibrium constant of 1 x 10-10, ie the reaction does not take place at all in effect. This is the basis of the ±60 kJ mol-1 rule of thumb mentioned above.

## ∆G and temperature

G depends on temperature and so, therefore does the equilibrium constant.

We have seen for the reaction

CaO(s) + CO2(g) CaCO3(s)

that we can work out the temperature when ∆G = 0 – the tipping point between the reaction going to the right or the left. It can be useful to plot this on a graph of ∆G against temperature.

Note that it slopes up from left to right, ie as the temperature increase, the reaction becomes less feasible. This is because the graph is really a graph of the expression.

G = ∆H TS

So the slope or gradient of the graph is –∆S. As we have seen, ∆S for this reaction is negative so the slope is positive. The slope remains the same because there is no change in ∆S with temperature.

The temperature when ∆G is zero is approximately 1100 K which is what we worked out earlier. We can plot the same graph for other reactions such as the equilibrium reactions looked at earlier.

## Ellingham diagrams

An important industrial problem is extracting metals from their ores. Ores are usually metal oxides or compounds like sulfides and carbonates which can easily be converted to oxides by heating them in air (roasting). The problem is to find a suitable reducing agent which will convert the oxide to the metal. The reducing agent should ideally be cheap and do the reduction at as low a temperature as possible to reduce fuel energy costs. Coke (an impure form of carbon) is a cheap and common reducing agent as it removes oxygen from metal oxides as carbon monoxide (or carbon dioxide).

We can predict which reducing agents can reduce which oxides and at which temperatures by using the graphing facility to plot ∆G against T for a variety of relevant reactions. All the reactions are written so that they involve 1 mole of oxygen molecules. Such a diagram is called an Ellingham diagram but the graph is exactly the same as before, ∆G against temperature.

For example – can carbon be used to reduce zinc oxide to zinc at a) 1000 K b) 1500K?

The reaction we require is:

2C + 2ZnO 2CO + 2Zn

The graph shows that at 1000 K we have the following approximate ∆G values:

1. 2C + O2 2CO ∆G = –400 kJ mol-1

2. 2Zn + O2 2ZnO ∆G = –550 kJ mol-1

So reversing reaction 2:

3. 2ZnO 2Zn + O2G = +550 kJ mol-1

Hint Reversing the reaction reverses the sign of ∆G

Adding 1. and 3. we get:

2C + O2+ 2ZnO 2CO + 2ZnO + O2

G = –400 + 550 = +150 kJ mol-1

so at 1000 K the reaction is not feasible.

At 1500 K we have the following ∆G values:

4. 2C + O2 2CO ∆G = –500 kJ mol-1

5. 2Zn + O2 2ZnO ∆G = –250 kJ mol-1

so:

6. 2ZnO 2Zn + O2 ∆G = +250 kJ mol-1

Adding 4.and 6. and cancelling the oxygens as above we get

2ZnO + 2C 2CO + 2Zn ∆G = –250 kJ mol-1

At 1500 K the reaction is feasible.

At the point on the graph that where the two lines cross, ∆G = 0. Thus the reaction above is feasible at temperatures above 1200 K.

It is possible to short circuit the above procedure. Plot just the lines for the two processes concerned. At any temperature, the reaction represented by the lower line will have a more negative value of ∆G than the reaction represented by the upper line. So the reaction represented by the lower line will go forwards and drive the reaction represented by the upper line in reverse giving a negative value of ∆G for the overall reaction. So carbon going to carbon monoxide will only reduce magnesium oxide at temperatures above 2000 K, for example.

Carbon, in the form of coke, is used as a reducing agent in a number of industrial processes, most notably in the production of iron, where iron oxides (largely iron(III) oxide, Fe2O3) are reduced to iron in the blast furnace.

Note. Because

G = ∆HTSsystem

a graph of ∆G against T has a slope of –∆Ssystem. Changes of slope indicate that the value of ∆Ssystem. has changed. This happens at temperatures where one of the reactants or products changes state (melts or boils) and hence the value of its entropy changes.

 Home Videos Tutorials The direction of chemical reactions Entropy The Second Law of Thermodynamics Is the Second Law wrong? The role of energy The system and the surroundings Total entropy change The Gibbs free energy ∆G and temperature Reversible reactions Why is free energy 'free'? Simulations Entropy and disorder Arranging objects The distribution of energy The approach to equilibrium Equilibrium mixtures ∆G and temperature 