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# 8.6: Temperature Dependence of A and G

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

$dA = -pdV - SdT$

and

$dG = -Vdp - SdT$

So by inspection, it is easy to see that

$\left( \dfrac{\partial A}{\partial T} \right)_V = -S$

and

$\left( \dfrac{\partial G}{\partial T} \right)_p = -S$

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

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

and

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

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.

$A= U -TS$

and

$G = H -TS$

Dividing by $$T$$ yields

$\dfrac{A}{T} = \dfrac{U}{T} -S$

and

$\dfrac{G}{T} = \dfrac{H}{T} -S$

Now differentiating each expression with respect to $$T$$ at constant $$V$$ or $$p$$ respectively yields

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

and

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

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

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

and

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

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:

$d \left( \dfrac{G}{T} \right) = H d \left( \dfrac{1}{T} \right)$

Or for finite changes $$\Delta G$$ and $$\Delta H$$:

$d \left( \dfrac{\Delta G}{T} \right) = \Delta H d \left( \dfrac{1}{T} \right)$

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

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

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

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.

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

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

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

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

$\Delta G_{500\,K} = -787.9\,kJ$

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!

## Contributors and Attributions

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

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