# Is it really appropriate to measure entropy change using enthalpy?

One statement by Peter Atkins in his book Elements of Physical Chemistry confused me quite a bit:

If a chemical reaction or phase transition takes place at constant pressure , we can identify $Q$ with the change in enthalpy of the system, and obtain

For a process at constant pressure: $$\Delta S = \frac{\Delta H}{T}.$$

Is it really correct to measure the entropy change using change in enthalpy?

I'm telling so because enthalpy change only measures how much heat energy enters or gets expelled from the system. It doesn't measure how much energy gets expelled out of the system when the system does $PV$ work.

As Peter Atkins points out:

If $10~\text{kJ}$of energy is supplied as the heat to the system that is free to change its volume at constant pressure, then the enthalpy of the system increases by $10~\text{kJ}\; ,$ regardless of how much energy enters or leaves by doing work and we write $\Delta H = +10~\text{kJ}\; .$

I know it is the heat energy that is concerned in the definition of entropy change. But, suppose the system is heated which means it is in more disorder; however if it does work, then the energy the system got as heat would decrease which would therefore decrease the chaos & disorder of the system. If we use only the enthalpy change, wouldn't we exclude that work which decreases the heat energy gained by the system?

I'm rather confused on the use of enthalpy for defining the entropy change. Could anyone please help me explain this how enthalpy change actually measures the entropy change?

• Please articulate your understanding of how the change in entropy of a closed system is determined, starting from thermodynamic equilibrium state A and ending at thermodynamic equilibrium state B. – Chet Miller Nov 9 '15 at 16:50
• @Chester Miller: It should be $$\int_{A}^{B} dS = \int_{A}^{B} \frac{Q_\text{rev}}{T} dT$$ where $Q$ is the net heat energy received by the system reversibly. – user5764 Nov 9 '15 at 16:54

Yes. So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B. If it is a phase change, you start out with liquid at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B). This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor. During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor. Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$. From the first law, the change in internal energy is equal to the heat added minus the work done: $(U_v-U_l)=Q-P(V_v-V_l)$. Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T). Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$. This completes what I wanted to say about the phase change situation. I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.
• Firstly, thanks for the answer, sir. I conceived the maths. However, I've a query: Basically what is entropy? Entropy is the chaos or the disorderliness in the system caused due to the heat energy supplied. First the liquid got thermal energy $Q$ which increased the chaos in the liquid (soon to be vapour). But in the course of doing so, it does work which decreases the heat energy available in the system to $Q- P\Delta V.$ And it is heat which determines the chaos & now there is only $Q- P\Delta V$ left in the system which determines the chaos & not $Q.$ ... – user5764 Nov 9 '15 at 18:13
• ... So, why should the definition of entropy takes $Q\;?$ As the heat energy available for the chaos is $Q- P\Delta V$ & not $Q\; ?$ – user5764 Nov 9 '15 at 18:17