The quantum casino: tutorials

Why is free energy ‘free’?

Why do we use the word ‘free’ energy in the term ‘Gibbs free energy’ and in what sense is it free? Let us look again at the reaction

HCl(g) + NH₃(g) NH₄Cl(s)

We have seen that this is a reaction in which ∆H is negative (energy is given out from system to surroundings) and ∆S_system is negative (the solid product is more ordered than the gaseous reactants). The Second Law of Thermodynamics tells us that this reaction (and all other reactions) is only possible if the entropy of the universe (ie system + surroundings) increases. This happens because the reaction gives out heat energy to the surroundings which increases the entropy of the surroundings to outweigh the entropy decrease of the system.

The figures (which we have calculated earlier) are

∆H = –176 kJ mol^-1

∆S_system = –284 J K^-1 mol^-1

So, ∆S_surroundings must be at least +284 J K^-1 mol^-1 or the reaction cannot take place, according to the Second Law of Thermodynamics.

We have seen above that the heat given out by the reaction is more than enough to do this, which is why the reaction can take place. But just how much of this energy is actually needed to make the reaction feasible?

We know that

∆S_surroundings = –∆H / T

∆H = –T∆S_surroundings

and at room temperature (298 K)

∆H = –298 x 284/1000 (remember to convert entropy in J K^-1 mol^-1 to kJ K^-1 mol^-1)

∆H = –85 kJ mol^-1

Let us be clear what this result means. Of the 176 kJ mol^-1 of energy produced in the reaction by bonds breaking and making by the reaction, 85 kJ mol^-1 must be given out to the surroundings as heat to make the reaction feasible. The rest is ‘free’ energy and could in principle be given out in some form other than heat. Looking at the other side of the coin, of the 176 kJ mol^-1 available from bond breaking and making, 85 kJ mol^-1 must be given out as heat; this part of the energy is not ‘free’ at all, it can only be given out as heat or the reaction cannot happen.

In this example, it is not easy to see how the energy produced by the reaction could be harnessed in any way except heat. However the next example, of an electrochemical cell, illustrates the point in a more practical way.

The reaction

2Ag⁺(aq) + Cu(s) 2Ag(s) + Cu²⁺(aq) ∆H = –147 kJ mol^-1

takes place readily in a test tube.

2Ag⁺(aq) + Cu(s)

2Ag(s) + Cu²⁺(aq)

Simply looking at the reactants and products suggests that ∆S_system will be negative because two mobile ions on the left produce one on the right; the products are more ordered than the reactants. This is in fact the case, ∆S_system = –193 J K^-1 mol^-1. So some heat energy must be transferred from system to the surroundings to make the total entropy change positive. We can calculate this as above

∆S_surroundings = –∆H / T

∆H = –T∆S_surroundings

and at room temperature (298 K)

∆H = –298 x 193/1000 (remember to convert entropy in J K^-1 mol^-1 to kJ K^-1 mol^-1)

∆H = –58 kJ mol^-1

Copper reacting with silver nitrate solution in an electrochemical cell

So of the 147 kJ mol^-1 of energy available from the reaction, 58 KJmol^-1 must be given out as heat to increase the entropy of the surroundings. The rest is ‘free’ to be given out in other forms. In this case we can see how energy might be extracted other than as heat. We can carry out the reaction in an electrochemical cell. This splits the reaction into two halves each of which takes place in a separate container but the overall reaction is exactly the same as are the entropy and enthalpy changes.

The arrangement is shown on the right.

In the left hand beaker the half reaction

Cu(s) Cu²⁺(aq) + 2e^-

The electrons released travel through the wire to the piece of silver in the right hand beaker and drive this half reaction

2Ag⁺(aq) + 2e^- 2Ag(s)

(This is a redox (electron transfer) reaction, something which is made clear by the fact that electrons travel through the wire.) The cell diagram for this arrangement is

Cu(s)|Cu²⁺(aq)¦¦Ag⁺(aq)|Ag(s) E = + 0.46 V

The overall reaction (obtained by adding the two half reactions) and cancelling the electrons is exactly the same as the reaction that takes place in a single beaker by directly mixing the reactants.

Cu(s) + 2Ag⁺(aq) + 2e^- Cu²⁺(aq) + 2Ag(s) + 2e^-

However, we can now see how to extract energy from this reaction in a form other than heat. We can use the electrical energy produced by the electron flow to light a bulb (light energy) or turn a motor (kinetic energy) or lift a weight (mechanical energy).

However, of the 193 kJ mol^-1 of energy produced by the reaction, we have already seen that 58 kJ mol^-1 must be given out as heat to the surroundings but the rest (135 kJ mol^-1) is ‘free’. This is the maximum amount that we can extract as electrical energy and turn into other forms of energy. Although ∆H is –193 kJ mol^-1, no amount of chemical ingenuity or clever engineering can possibly extract more than 135 kJ mol^-1 as electrical energy (or in fact any type of energy except heat).

∆H and ∆G

At the pre–16 level, students may well have developed the idea that exothermic reactions (∆H –ve) tend to occur while endothermic ones (∆H +ve) do not. Indeed, many chemists use this as a rule of thumb. That it works as a rule of thumb is due to the fact that we live and do most of our chemistry in a (relatively) low temperature world.

∆G = ∆H – T∆S_system

If T is relatively small (and room temperature, approximately 298 K is relatively small), the term T∆S_system will also be small and ∆G ≈ H. So the sign of ∆H is generally the same as that of ∆G and it acts a reasonably good predictor of reaction feasibility.

Essentially this means that at relatively low temperatures, the total entropy change of most reactions is dominated by the entropy change of the surroundings brought about by the energy changes due to bond breaking and making. At higher temperatures, the total entropy change of the reaction is dominated by the entropy change of the system.

Why is free energy ‘free’?