# 22.2: Gibbs Energy

The Helmholtz energy \(A\) is developed for isochoric changes and as we have often said before it is much easier to deal with isobaric ones where P= constant. We can therefore repeat the above treatment for the enthalpy and introduce another state function the Gibbs (free) energy

\[G≡ H -TS = U + PV - TS = A + PV\]

If we take both T and P constant (pleasantly achievable conditions!) we get

\[dU-TdS + PdV \le 0\]

\[dG \le 0\]

G either decreases (spontaneously) or is constant (at equilibrium). Calculating the state function between two end points we get:

\[ΔG = ΔH - TΔS ≤ 0 (T,P constant)\]

This quantity is key to the question of spontaneity under the conditions we usually work under. If for a process ΔG is positive it does not occur spontaneous and can only be made to occur if it is 'pumped', i.e. coupled with a process that has a negative ΔG. The latter is spontaneous.

Note

If ΔG=0 we are at * equilibrium*.

### Direction of the spontaneous change

Because the \(ΔS\) term contains the temperature \(T\) as coefficient the spontaneous direction of a process, e.g. a chemical reaction can **change** with temperature depending on the values of the enthalpy and the entropy change ΔH and ΔS. This is true for the melting process, e.g. for water below 0^{o}C we get water=>ice, above this temperature ice melts to water, but it also goes for chemical reactions.

Example

Consider

\[NH_{3(g)} + HCl_{(g)} \rightleftharpoons NH_4Cl_{(s)}\]

Δ_{r}H at 298K / 1 bar is -176.2 kJ. The change in entropy is -0.285 kJ/K so that at 298K ΔG is -91.21 kJ. Clearly this is a reaction that will proceed to the depletion of whatever is the limiting reagent on the left.

However at 618 K this is a different story. Above this temperature ΔG is positive! (assuming enthalpy and entropy have remained the same, which is almost but not completely true) The reaction will not proceed. Instead the *reverse* reaction would proceed spontaneously. The salt on the right would decompose in the two gases -base and acid- on the left.

### Meaning of the ΔG term

As we have seen, ΔA can be related to the maximal amount of work that a system can perform at constant V and T. We can hold an analogous argument for ΔG except that V is not constant so that we have to consider volume work (zero at constant volume) .

\[dG = d(U+PV-TS) = dU -TdS - SdT - PdV +VdP\]

As dU = TdS + δw_{rev}

\[dG = δw_{rev} -SdT + VdP + PdV\]

As the later term is -δw_{volume}

\[dG = δw_{rev} -SdT + VdP - δw_{volume}\]

At constant T and P the two middle terms drop out

\[dG = δw_{rev} - δw_{volume} = δw_{other useful work}\]

Note

ΔG stands for the (maximal) reversible, isobaric isothermal non-PV work that a certain spontaneous change can perform. The volume work may not be zero, but is corrected for.

### Natural variables of G

Because \(G ≡ H-TS\), we can write

\[dG = dH -TdS -SdT\]

\[dG = TdS +VdP -TdS -SdT = VdP - SdT\]

The natural variables of \(G\) are pressure \(P\) and temperature \(T\). This is what makes this function the most useful of the four U, H, A, and G: these are the natural variables of most of your laboratory experiments!

### Summary

We now have developed the basic set of concepts and functions that together form the framework of thermodynamics. Let's summarize four very basic state functions:

state function | natural variables |
---|---|

dU = -PdV + TdS | U(V,S) |

dH = +VdP + TdS | H(P,S) |

dA = -PdV - SdT | A(V,T) |

dG = +VdP - SdT | G(P,T) |

Note:

- The replacement of δq by TdS was based on
*reversible*heat. This means that in the irreversible case the expressions for dU and dH become inequalities - We only unclude
*volume*work in the above expressions. If other work (elastic, electrical e.g.) is involved extra terms need to be added: dU = TdS - PdV + xdX etc.

We are now ready to begin applying thermodynamics to a number of very diverse situations, but we will first develop some useful partial differential machinery.