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8.6: Temperature Dependence of A and G

  • Page ID
    199223
  • 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)