# 9.6: The Gibbs-Helmholtz Equation

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### Ideal gas

For a mole of ideal gas we can use the gas law to integrate volume over pressure and we get

$ΔG_{molar} = RT \ln \left(\dfrac{P_2}{P_1}\right) \nonumber$

It is customary to identify one of the pressures (P1} with the standard state of 1 bar and use the plimsoll to indicate the fact that we are referring to a standard state by writing:

$G_{molar}(P) = G^o_{molar} + RT \ln \left(\dfrac{P}{1}\right)=G^o_{molar} + RT \ln[P] \nonumber$

The fact that we are making the function intensive (per mole) is usually indicated by putting a bar over the $$G$$ symbol, although this is often omitted for Gomolar

### Solids

For solids the volume does not change very much with pressure (the isothermal compressibility $$κ$$ is small), so can assume it more or less constant:

$G(P_{final})=G(P_{initial})+ \int VdP \text{(from init to final)} ≈ G(P_{initial})+ V \int dP \text{(from init to final)}=G(P_{initial})+ VΔP \nonumber$

## The Gibbs-Helmholtz Expression

$\frac { G } { T } = \frac { H } { T } - S \nonumber$

Take the derivative under constant pressure of each side to get

$\left( \frac { \partial G / T } { \partial T } \right) _ { P } = - \frac { H } { T ^ { 2 } } + \frac { 1 } { T } \left( \frac { \partial H } { \partial T } \right) _ { P } - \left( \frac { \partial S } { \partial T } \right) _ { P } \nonumber$

We make use of the relationship between $$C_p$$ and $$H$$ and $$C_p$$ and $$S$$

\begin{align} \left( \frac { \partial G / T } { \partial T } \right) _ { P } &= - \frac { H } { T ^ { 2 } } + \cancel{ \frac { C _ { P } } { T }} - \cancel{\frac { C _ { P } } { T }} \\ &= - \frac { H } { T ^ { 2 } } \end{align} \nonumber

We said before that $$S$$ is a first order derivative of $$G$$. As you can see from this derivation the enthalpy $$H$$ is also a first order derivative, albeit not of $$G$$ itself, but of $$G/T$$.

$\left( \frac { \partial \Delta G / T } { \partial T } \right) _ { P } = - \frac { \triangle H } { T ^ { 2 } } \nonumber$

The last step in the derivation simply takes the step before twice -say for the $$G$$ and $$H$$ at the begin and end of a process- and subtracts the two identical equations leading to a $$Δ$$ symbol. In this differential form the Gibbs-Helmholtz equation can be applied to any process.

## Gibbs Energy as a Function of Temperature

If heat capacities are know from 0 K we could determine both enthalpy and entropy by integration:

$S(T) = S(0) + \int_0^T \dfrac{C_p}{T} dT \nonumber$

$H(T) = H(0) + \int_0^T C_p\; dT \nonumber$

As we have seen we must be careful at phase transitions such as melting or vaporization. At these points the curves are discontinuous and the derivative $$C_p$$ is undefined.

$H ( T ) = H ( 0 ) + \int_0^{T_{fus}} C_p(T)_{solid} dT + \Delta H_{fus} + \int_{T_{fus}}^{T_{boil}} C_p(T)_{liquid} dT + \Delta H_{vap} + etc. \label{Hcurve}$

\begin{align} S ( T ) &= S ( 0 ) + \int_0^{T_{fus}} \dfrac{C_p(T)_{solid}}{T} dT + \Delta S_{fus} + \int_{T_{fus}}^{T_{boil}} \dfrac{C_p(T)_{liquid}}{T} dT + \Delta S_{vap} + etc. \\[4pt] &= S(0) + \int_0^{T_{fus}} \dfrac{C_p(T)_{solid}}{T} dT + \dfrac{\Delta H_{fus}}{T_{fus}} + \int_{T_{fus}}^{T_{boil}} \dfrac{C_p(T)_{liquid}}{T} dT + \dfrac{\Delta H_{vap}}{T_{boil}} + etc. \label{Scurve} \end{align}

with $$H(T=0)= \text{undefined}$$ and $$S(T=0)=0$$ from the third law of thermodynamics.

We also discussed the fact that the third law allows us to define $$S(0)$$ as zero in most cases. For the enthalpy we cannot do that so that our curve is with respect to an undefined zero point. We really should plot $$H(T) - H(0)$$ and leave $$H(0)$$ undefined.

Because the Gibbs free energy $$G= H-TS$$ we can also construct a curve for $$G$$ as a function of temperature, simply by combining the $$H$$ and the $$S$$ curves (Equations \ref{Hcurve} and \ref{Scurve}):

$G(T) = H(T) - TS(T) \nonumber$

Interestingly, if we do so, the discontinuties at the phase transition points will drop out for $$G$$ because at these points $$Δ_{trs}H = T_{trs}Δ_{trs}S$$. Therefore, $$G$$ is always continuous.

The $$H(0)$$ problem does not disappear so that once again our curve is subject to an arbitrary offset in the y-direction. The best thing we can do is plot the quantity $$G(T) - H(0)$$ and leave the offset $$H(0)$$ undefined.

We have seen above that the derivative of $$G$$ with temperature is $$-S$$. As entropy is always positive, this means that the G-curve is always descending with temperature. It also means that although the curve is continuous even at the phase transitions, the slope of the G curve is not, because the derivative $$-S$$ makes a jump there. Fig. 22.7 in the book shows an example of such a curve for benzene. Note the kinks in the curve at the mp and the boiling point.

9.6: The Gibbs-Helmholtz Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.