6.6: Temperature Dependence of A and G

In differential form, the free energy functions can be expressed as

\[dA = -pdV - SdT \nonumber \]

and

\[dG = -Vdp - SdT \nonumber \]

So by inspection, it is easy to see that

\[\left( \dfrac{\partial A}{\partial T} \right)_V = -S \nonumber \]

and

\[\left( \dfrac{\partial G}{\partial T} \right)_p = -S \nonumber \]

And so, it should be fairly straightforward to determine how each changes with changing temperature:

\[\Delta A = - \int_{T_1}^{T_2} \left( \dfrac{\partial A}{\partial T} \right)_V dT = - \int_{T_1}^{T_2} S\,dT \nonumber \]

and

\[\Delta G = - \int_{T_1}^{T_2} \left( \dfrac{\partial G}{\partial T} \right)_p dT = - \int_{T_1}^{T_2} S\,dT \nonumber \]

But the temperature dependence of the entropy needed to be known in order to evaluate the integral. A convenient work-around can be obtained starting from the definitions of the free energy functions.

\[ A= U -TS \nonumber \]

and

\[G = H -TS \nonumber \]

Dividing by \(T\) yields

\[\dfrac{A}{T} = \dfrac{U}{T} -S \nonumber \]

and

\[\dfrac{G}{T} = \dfrac{H}{T} -S \nonumber \]

Now differentiating each expression with respect to \(T\) at constant \(V\) or \(p\) respectively yields

\[ \left( \dfrac{\partial \left( \frac{A}{T}\right)}{\partial T} \right)_V = - \dfrac{U}{T^2} \nonumber \]

and

\[ \left( \dfrac{\partial \left( \frac{G}{T}\right)}{\partial T} \right)_p = - \dfrac{H}{T^2} \nonumber \]

Or differentiating with respect to \(1/T\) provides a simpler form that is mathematically equivalent:

\[ \left( \dfrac{\partial \left( \frac{A}{T}\right)}{\partial \left( \frac{1}{T}\right)} \right)_V = U \nonumber \]

and

\[ \left( \dfrac{\partial \left( \frac{G}{T}\right)}{\partial \left( \frac{1}{T}\right)} \right)_p = H \nonumber \]

Focusing on the second expression (since all of the arguments apply to the first as well), we see a system that can be integrated. Multiplying both sides by \(d(1/T)\) yields:

\[ d \left( \dfrac{G}{T} \right) = H d \left( \dfrac{1}{T} \right) \nonumber \]

Or for finite changes \(\Delta G\) and \(\Delta H\):

\[ d \left( \dfrac{\Delta G}{T} \right) = \Delta H d \left( \dfrac{1}{T} \right) \nonumber \]

and integration, assuming the enthalpy change is constant over the temperature interval yields

\[ \int_{T_1}^{T_2} d \left( \dfrac{\Delta G}{T} \right) = \Delta H \int_{T_1}^{T_2} d \left( \dfrac{1}{T} \right) \nonumber \]

\[\dfrac{\Delta G_{T_2}}{T_2} - \dfrac{\Delta G_{T_1}}{T_1} = \Delta H \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{GH1} \]

Equation \ref{HG1} is the Gibbs-Helmholtz equation and can be used to determine how \(\Delta G\) changes with changing temperature. The equivalent equation for the Helmholtz function is

\[ \dfrac{\Delta A_{T_2}}{T_2} -\dfrac{\Delta A_{T_1}}{T_1} = \Delta U \left(\dfrac{1}{T_2} -\dfrac{1}{T_1} \right) \label{GH2} \]

Example \(\PageIndex{1}\):

Given the following data at 298 K, calculate \(\Delta G\) at 500 K for the following reaction:

\[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + H_2O(g) \nonumber \]

Compound	\(\Delta G_f^o\) (kJ/mol)	\(\Delta H_f^o\) (kJ.mol)
CH4(g)	-50.5	-74.6
CO2(g)	-394.4	-393.5
H2O(g)	-228.6	-241.8

Solution

\(\Delta H\) and \(\Delta G_{298\, K}\) and can be calculated fairly easily. It will be assumed that \(\Delta H\) is constant over the temperature range of 298 K – 500 K.

\[\Delta H = (1 \,mol)(-393.5 \,kJ/mol) + (2\, mol)(-241.8\, kJ/mol) – (1\, mol)(-74.5\, kJ/mol) = -820.6\, kJ \nonumber \]

\[\Delta G_{298} = (1\, mol)(-394.4\, kJ/mol) + (2\, mol)(-228,6\, kJ/mol) – (1\, mol)(-50.5\, kJ/mol) = -801.1\, kJ \nonumber \]

So using Equation \ref{GH1} with the data just calculated gives

\[\dfrac{\Delta G_{500\,K}}{500 \,K} - \dfrac{-801.1\,kJ}{298\,K} = (-820.6\, kJ) \left( \dfrac{1}{500\,K} - \dfrac{1}{298\,K} \right) \nonumber \]

\[ \Delta G_{500\,K} = -787.9\,kJ \nonumber \]

Note: \(\Delta G\) became a little bit less negative at the higher temperature, which is to be expected for a reaction which is exothermic. An increase in temperature should tend to make the reaction less favorable to the formation of products, which is exactly what is seen in this case!