Energy, entropy and equilibrium

These are some standard entropy values. Notice that the units are J K –1 mol –1. These are different from the units for enthalpy, kJ mol –1.

Let's work out the standard entropy change, Δ S ө, for the decomposition of calcium carbonate.

We need to know the total standard entropy of the products and the total standard entropy of the reactants.

First, add up all the entropy values for the products, in this case calcium oxide and carbon dioxide.

Next, subtract all the entropy values for the reactants. In this case there is only one, calcium carbonate.

The standard entropy change for the decomposition of calcium carbonate is positive.

These are some standard entropy values. Notice that the units are J K –1 mol –1. These are different from the units for enthalpy, kJ mol –1.

Let's work out the standard entropy change, Δ S ө, for the synthesis of ammonia.

We need to know the total standard entropy of the products and the total standard entropy of the reactants.

First, add up all the entropy values for the products, in this case ammonia. Note that we multiply its entropy value by 2.

Next, subtract all the entropy values for the reactants, starting with nitrogen.

Note that we multiply the entropy value for hydrogen by 3.

Remember to multiply where necessary.

The standard entropy change for the synthesis of ammonia is positive.

We can work out the entropy change in the surroundings when a chemical change occurs.

To do this we need to know the enthalpy change and the temperature at which the reaction occurs.

If we know the entropy change for the reaction (the system) and the entropy change for the surroundings, we can work out the total entropy change.

First calculate the enthalpy change, Δ H, for the reaction.

The enthalpy change is 92.4 kJ mol –1. Remember to multiply by 1000 because entropy values are in J K –1 mol –1.

Divide by the temperature, in this example 298 K.

The entropy change in the surroundings in this example is –310 J K –1 mol –1.

Now let's calculate the total entropy change.

First, substitute the entropy change for the surroundings that we have just calculated.

Next, substitute the entropy change for the system. In this example it is +93 J K –1 mol –1.

The total entropy change here is negative. As it would lead to a decrease in entropy, the reaction would not happen spontaneously at 298 K.

The total entropy change must be positive for a reaction to happen. If it is zero, the reaction forms an equilibrium.

We can work out the minimum temperature at which the total entropy change is zero.

This equation shows the relationship between the enthalpy change, the temperature and the entropy change of the system when the total entropy change is zero.

We can rearrange the equation to find the temperature at which the total entropy change is zero.

The enthalpy change when ice melts is 6.01 kJ mol –1, which is 6010 J mol –1. The corresponding change in the entropy of the system is 22 J K –1 mol –1.

The total entropy will be zero at 273 K or 0 °C. At this temperature, ice and liquid water exist together in equilibrium.