Contact FutureLearn for Support
Skip main navigation
We use cookies to give you a better experience, if that’s ok you can close this message and carry on browsing. For more info read our cookies policy.
We use cookies to give you a better experience. Carry on browsing if you're happy with this, or read our cookies policy for more information.

Skip to 0 minutes and 1 secondAs an example, let's calculate the entropy change of the following reaction. This reaction is the formation reaction of aluminum oxide at standard condition. The enthalpy of formation is -16 hundreds 74 kJ/mole. To calculate entropy of the formation reaction, let's take the reversible path with the same reaction at zero K. This is the formation reaction at zero K. The entropy of formation at zero K is delta S3. The reversible path is involved in cooling of reactants aluminum and oxygen, the entropy of cooling is delta S1 and S2 for Al and O2, respectively. The heating of the product Al2O3 completes the reversible path, and the related entropy change is delta S4.

Skip to 0 minutes and 54 secondsSo the entropy change of the original reaction is the summation of the entropy changes in the reversible path. Delta S1 and S2 is the simple cooling of Al and O2. Delta S4 is the heating of Al2O3. Delta 3 is the entropy change of the reaction at zero K. Then, let's calculate each of them. For materials that do not undergo phase change in the temperature range of interest such as solid aluminum and solid Al2O3, the entropy change can be simply calculated from the heap capacity Cp. So delta S1, the cooling of Al from room temperature to zero K is 2 times integration of Cp Aluminum over T dT from 298 to 0 K.

Skip to 1 minute and 41 secondsAnd delta S4, the heating of Al2O3 from zero K to room temperature is integration of Cp over T dT from zero to 298 K. Delta S3, the reaction entropy at zero K is zero by the 3rd law. Now, delta S2 remains. It is the entropy change involved in cooling of O2 gas from room temperature to zero K. In cooling O2 gas to zero K, we have to consider phase transitions. O2 might be solid at 0 K and gas at 298 K, thus it experiences two phase transitions upon cooling from gas at 298 K to 0 K. So when we plot entropy versus temperature, it looks like this.

Skip to 2 minutes and 45 secondsIn the single phase regime, such as gas O2, liquid O2, and solid O2, the entropy increases gradually with temperature and the slop is Cp over T. So the slop of solid O2 regime is Cp of solid O2 over T. Likewise the slope in liquid O2 regime is Cp of liquid O2 over T, and that of gas O2 is Cp of gas O2 over T. Then between each phases it experience phase change reaction. First, gas to liquid, the condensation reaction at boiling point Tb and then liquid to solid, the freezing reaction at melting point Tm. The entropy change during the first condensation reaction is denoted as delta S condensation, and that of freezing is delta S freezing O2.

Skip to 3 minutes and 43 secondsThe total entropy change is like this. 1.5 is the mole number in the formation reaction. In the parenthesis, the first part is single phase cooling of O2 gas from room temperature to the boiling point. So the entropy change of cooling O2 gas is Cp of gas O2 over T dT. Then gas O2 experience condensation, and the involved entropy change is delta S, condensation. Next liquid O2 cools down from boiling point to melting point. Then it freeze to solid O2. Finally, solid O2 cools down from melting point to zero K. For the entropy change of phase transition reaction, it is simply related to the latent heats.

Skip to 4 minutes and 34 secondsEntropy change of condensation is the heat of condensation, that is the enthalpy of condensation divided by temperature, here the boiling pts. Likewise, the entropy of freezing is enthalpy of freezing over melting point. For most materials, the entropy change of evaporation is 21 cal per mole K, so entropy change of condensation is -21 cal per mole K. Entropy change of fusion is about 2 cal per mole K. Sometimes different from this value but usually it is in the range of 0.2 to 5.

Exercise of calculating entropy

We can calculate the entropy changes between two states of different temperature if we know the heat capacities.

When the pressure is constant, integration of Cp over temperature with respect to temperature gives the entropy changes upon temperature change within a single phase. If there is phase transition within the temperature range of interest, latent heat and the entropy change of phase transition should be considered also.

Share this video:

This video is from the free online course:

Thermodynamics in Energy Engineering

Hanyang University