Gibbs–Helmholtz equation

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The Gibbs–Helmholtz equation is a thermodynamic equation useful for calculating changes in the Gibbs energy of a system as a function of temperature. It is named after Josiah Willard Gibbs and Hermann von Helmholtz:

$\left( \frac{\partial ( \frac{G} {T} ) } {\partial T} \right)_{p\,} = - \frac {H} {T^2}$

With:

$H\,$ the enthalpy

$T\,$ the absolute temperature

$G\,$ the Gibbs free energy

at constant pressure $P\,$ . The equation states that the change in the G/T ratio at constant pressure as a result of an infinitesimally small change in temperature is a factor (H/T²).

For a chemical reaction the equation reads:

$\left( \frac{\partial ( \frac{\Delta G} {T} ) } {\partial T} \right)_{p\,} = - \frac {\Delta H} {T^2}$

with $\Delta G\,$ as the change in Gibbs energy and $\Delta H\,$ as the enthalpy change (which is considered independent of temperature).

which can rearrange to:

$\frac{\Delta G,T_2}{T_2} - \frac{\Delta G^\circ,T_1}{T_1} = \Delta H^\circ(P)*(\frac{1}{T_2} - \frac{1}{T_1})$

This equation quickly enables the calculation of the Gibbs free energy change for a chemical reaction at any temperature T₂ with knowledge of just the Standard Gibbs free energy change of formation and the Standard enthalpy change of formation for the individual components at 25°C and 1 bar.

Through:

$\ \Delta G^\circ = -RT \ln K$

which relates Gibbs energy to an equilibrium constant, the van 't Hoff equation is derived.

[edit] Proof

The Gibbs free energy for a closed system

$dG = - SdT + VdP \,$

at constant pressure $P\,$ (dP = 0) reduces to

$dG_{p\,} = - SdT \,$

$\left(\frac{\partial G}{\partial T}\right)_{p\,} = - S \,$

The dependence of the G/T ratio on T is found with the aid of the quotient rule:

$\left( \frac{\partial ( \frac{G} {T} ) } {\partial T} \right)_{p\,}= \frac{\left( \frac{\partial G}{\partial T} \right)_{p\,}}{T} - \frac{G}{T^2} = \frac{T\left ( \frac{\partial G}{\partial T} \right)_{p\,}- G}{T^2} = \frac{-ST - G}{T^2} = \frac{-H}{T^2}$

Sometimes it can be found like this:

$\left( \frac{\partial ( \frac{G} {T} ) } {\partial\left(\frac{1}{T}\right)}\right)_{p\,}= H$

Here, the chain rule has been used with a bit of rearrangement. If $y$ and $u$ are functions of $x$ , the chain rule says that $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$ . Dividing both sides through by $\frac{du}{dx}$ gives $\frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}}$ . In the equation above, $y$ is $\frac{G}{T}$ , $x$ is $T$ , and u is $\frac{1}{T}$ . Thus, $\frac{du}{dx}$ is $\frac{-1}{T^2}$ .