# How do you begin from the Gibbs-Helmholtz equation to determine the new temperature at which a reaction ceases to become spontaneous? What fundamental assumptions are involved?

Jan 6, 2018

The Gibbs-Helmholtz equation at a chosen standard pressure ($\circ$) (of $\text{1 atm}$ in this case) is:

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

To check spontaneity, one would need $\Delta {G}^{\circ}$, keeping in mind that ALL of this is NOT at $\text{298.15 K}$, i.e. NOT at standard temperature.

The alternative version is:

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

Keeping in mind that this process is at constant pressure:

$d \left(\Delta {G}^{\circ} / T\right) = - \frac{\Delta {H}^{\circ}}{T} ^ 2 \mathrm{dT}$

Integrating both sides, we get:

${\int}_{\left(1\right)}^{\left(2\right)} d \left(\Delta {G}^{\circ} / T\right) = - {\int}_{{T}_{1}}^{{T}_{2}} \frac{\Delta {H}^{\circ}}{T} ^ 2 \mathrm{dT}$

Assuming that $\Delta {H}^{\circ}$ is constant in the temperature range (and is defined at standard pressure):

$\frac{\Delta {G}^{\circ} \left({T}_{2}\right)}{T} _ 2 - \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{T} _ 1 = \Delta {H}^{\circ} \left({P}^{\circ}\right) \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

A reaction is "no longer spontaneous" when it reaches equilibrium. For that scenario, $\Delta {G}^{\circ} \left({T}_{2}\right) = 0$. So, we solve for the ${T}_{2}$ where that is the case.

$- \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{\Delta {H}^{\circ}} \frac{1}{T} _ 1 = \frac{1}{T} _ 2 - \frac{1}{T} _ 1$

$- \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{\Delta {H}^{\circ}} \frac{1}{T} _ 1 + \frac{1}{T} _ 1 = \frac{1}{T} _ 2$

This becomes:

$\textcolor{b l u e}{{T}_{2}} = {\left[- \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{\Delta {H}^{\circ}} \frac{1}{T} _ 1 + \frac{1}{T} _ 1\right]}^{- 1}$

$= {\left[\left(1 - \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{\Delta {H}^{\circ}}\right) \frac{1}{T} _ 1\right]}^{- 1}$

$= {T}_{1} / \left[1 - \frac{\Delta {G}^{\circ} \left({T}_{1}\right)}{\Delta {H}^{\circ}}\right]$

$= \textcolor{b l u e}{\frac{\Delta {H}^{\circ} {T}_{1}}{\Delta {H}^{\circ} - \Delta {G}^{\circ} \left({T}_{1}\right)}}$

Or, if we use the relation that $\Delta {G}^{\circ} \left({T}_{1}\right) = \Delta {H}^{\circ} - {T}_{1} \Delta {S}^{\circ}$ at the first temperature, also assuming the change in entropy is constant in the temperature range,

$\textcolor{b l u e}{{T}_{2}} = \frac{\Delta {H}^{\circ} {T}_{1}}{\cancel{\Delta {H}^{\circ} - \Delta {H}^{\circ}} + {T}_{1} \Delta {S}^{\circ}}$

$= \textcolor{b l u e}{\Delta {H}^{\circ} / \Delta {S}^{\circ}}$

which should be a familiar result.