Chemistry · Ch 4 — Chemical Thermodynamics

Gibbs energy and spontaneity

4.11.6

Gibbs energy and spontaneity

The total entropy change that accompanies a process is given by

\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = \Delta S + \Delta S_{surr} \qquad \text{...(4.36)}

The subscript "sys", which refers to the system, is dropped hereafter.

Relation between $\Delta G$ and $\Delta S_{total}$

According to the second law of thermodynamics, for a process to be spontaneous, $\Delta S_{total} > 0$ . If $\Delta H$ is the enthalpy change accompanying a reaction (the system), the enthalpy change of the surroundings is $-\Delta H$ . With this,

\Delta S_{surr} = -\frac{\Delta H}{T} \qquad \text{...(4.37)}

Substituting the above into Eq. (4.36),

\Delta S_{total} = \Delta S - \frac{\Delta H}{T} \qquad \text{...(4.38)}

Thus, $\Delta S_{total}$ is expressed in terms of the properties of the system only. Rearranging,

-T\,\Delta S_{total} = \Delta H - T\,\Delta S \qquad \text{...(4.39)}

Substituting in Eq. (4.35),

\Delta G = -T\,\Delta S_{total} \qquad \text{...(4.40)}

Note

The book's printed Eq. (4.39) reads " $T\,\Delta S_{total} = \Delta H - T\Delta S$ " — with the leading minus sign missing (zoom-verified). Multiplying Eq. (4.38) by $T$ actually gives $T\,\Delta S_{total} = T\Delta S - \Delta H$ , i.e. $-T\,\Delta S_{total} = \Delta H - T\Delta S$ ; only this corrected form leads to the book's own next step, Eq. (4.40). We show the correct sign; the printed slip is disclosed here so the derivation and the book can be reconciled.

For a spontaneous reaction $\Delta S_{total} > 0$ , and hence $\Delta G < 0$ . At constant temperature and pressure, the Gibbs energy of the system decreases in a spontaneous process. …

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