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Entropy

Double six in dice - a random result?

This is a measure of disorder. It is given the symbol S, and values for different chemicals are tabulated in the data book. The units for entropy are J mol-1 K-1. If you check through some values in the data book, you will find that solids have the lowest entropy. This is because they consist of very ordered, regular arrays of particles. Liquids have a higher entropy as they are more disordered - the particles not being in fixed positions, but free to move around. Gases have the highest entropy as they are most disordered. Their particles move randomly, and generally with great speed. Don't make the mistake of assuming that the entropy of the elements is zero. Even the solid elements at room temperature have a degree of disorder due to the vibrations of their constituent atoms.

When a reaction takes place there are two sources of entropy change. First, the chemicals in the reaction change, and may become more or less disordered. This is called the entropy change of the system, DSsystem. We can work out a numerical value from the data book by adding up the entropies of the products, and taking away the entropies of the reactants. This is written as an equation:

DSsystem = SS(products) - SS(reactants)

Where S is the Greek letter sigma, meaning sum of. More often than not, it will be sufficient to decide whether DSsystem is positive (that is there is an increase in disorder) or negative (a decrease in disorder). This is also a popular request on the exam! Look for changes in state. Particularly the change of solid or liquid into a gas - this indicates a clear increase in entropy. Also, look for changes in the total number of particles. If a substance breaks up into a greater number of particles, then there will also be an increase in disorder.

For example:

MgCO3(s) ⇒ MgO(s) + CO2(g)

has a positive DSsystem because a gas is formed from a solid, and one particle becomes two. If we want to find a numerical value for this entropy change we shall need to look up the standard entropy values of the substances involved:

Substance

Entropy (J mol-1 K-1)

MgCO3(s)

65.7

MgO(s)

26.9

CO2(g)

213.6

So DSsystem = 26.9 + 213.6 - 65.7 = + 174 .8 J mol-1 K-1

This is a positive value as we expected. Don't forget to include the sign and correct units with your answer.

As chemical reactions are accompanied by energy exchange - that is they are exothermic or endothermic - there is also a change in the entropy of the surroundings, given the symbol DSsurroundings. If the reaction is exothermic, then the surroundings get hotter, and the particles move faster, and become more disordered. So exothermic reactions have a positive DSsurroundings, and similarly endothermic reactions have a negative DSsurroundings. The relationship is expressed mathematically as:

DSsurroundings = - DH/T

where DH is the enthalpy change for the reaction in joules per mole, and T is the absolute temperature in Kelvin (remember 0°C = 273K). Note the minus sign, which means that an exothermic reaction (negative DH) has a positive DSsurroundings. This calculation is also very popular on the exam. In the case of the above reaction, DH = + 100.6 kJ mol-1, so at 298K:

DSsurroundings = - + 100.6 × 1000/298 = - 338 J mol-1 K-1

Note that the enthalpy value was multiplied by 1000. This is because if we divide a value in kJ mol-1 by one in K, the answer will be in kJ mol-1 K-1. Technically, if you include these units with your answer you will not be wrong, just a little odd! The normal units for entropy are J mol-1 K-1, achieved by dividing a value in J mol-1 by K. Expressing your answer in kJ mol-1 K-1 is a bit like giving your height in kilometres (not necessarily incorrect, but a little strange!). Don't forget again - you need the correct sign and units, and a sensible number of sig figs is needed here too.

We get the total entropy change for a reaction by adding these two entropy changes together. Expressed mathematically:

DStotal = DSsystem+ DSsurroundings

We come to a very important conclusion after the study of many reactions:

For a spontaneous reaction (one that takes place without external help) DStotalis always positive.

In the case above, DStotal = +174.8 - 338 = - 163 J mol-1 K-1

This means that this reaction is not feasible at this temperature. As one of our entropy values was positive and one negative, it was not possible to decide on the sign of the total entropy change unless we did the calculations. We can still get some useful information from assessing the signs of the two entropy changes. If both DSsystem and DSsurroundings are negative then the reaction is not feasible, if they are both positive then the reaction is feasible. When one is positive and one is negative the outcome is uncertain without doing a calculation.

Try to work out DSsystem , DSsurroundings and DStotal for the following reaction:

CH4(g) + 2O2(g) ⇒ CO2(g) + 2H2O(l) DH = - 890.3 kJ mol-1

Substance

Entropy (J mol-1 K-1)

CH4(s)

186.2

O2(g)

205.0

CO2(g)

213.6

H2O(l)

69.9

Is this reaction feasible at 298K? Check your answer here.

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