Question

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?

Answer 1

A spontaneous process in thermodynamics is one that happens without any external input to the system. The time-evolution of a system in which it releases free energy and goes to a lower, more thermodynamically stable energy state is a more technical definition (closer to thermodynamic equilibrium).

Complete answer:
The Gibbs free energy (or Gibbs energy) is a thermodynamic potential that may be used to calculate the maximum reversible work that a thermodynamic system can accomplish at a fixed temperature and pressure in thermodynamics. The Gibbs free energy $\Delta G=\Delta H-T\Delta S$ is the greatest amount of non-expansion work that may be taken from a thermodynamically closed system, measured in joules in SI (one that can exchange heat and work with its surroundings, but not matter).
The Gibbs-Helmholtz equation is
$\left(\dfrac{\partial\left(G^{\circ} / T\right)}{\partial T}\right)_{P}=-\dfrac{H^{\circ}}{T^{2}}$
To check spontaneity, one would need $\Delta G^{\circ}$. keeping in mind that $\mathrm{ALL}$ of this is NOT at $298.15 \mathrm{~K}$, i.e. NOT at standard temperature.
It can be written as
$\left(\dfrac{\partial\left(\Delta G^{\circ} / T\right)}{\partial T}\right)_{P}=-\dfrac{\Delta H^{\circ}}{T^{2}}$
this process is at constant pressure:
$d\left(\Delta G^{\circ} / T\right)=-\dfrac{\Delta H^{\circ}}{T^{2}} d T$
Integrating, we get:
$\int_{(1)}^{(2)} d\left(\Delta G^{\circ} / T\right)=-\int_{T_{1}}^{T_{2}} \dfrac{\Delta H^{\circ}}{T^{2}} d T$
Assume$\Delta H^{\circ}$ is constant at given temperature
$\dfrac{\Delta G^{\circ}\left(T_{2}\right)}{T_{2}}-\dfrac{\Delta G^{\circ}\left(T_{1}\right)}{T_{1}}=\Delta H^{\circ}\left(P^{\circ}\right)\left[\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}}\right]$
When a response achieves equilibrium, it is no longer "spontaneous."
For that scenario, $\Delta G^{\circ}\left(T_{2}\right)=0 .$ So, we solve for the $T_{2}$ where
$-\dfrac{\Delta {{G}^{{}^\circ }}\left( {{T}_{1}} \right)}{\Delta {{H}^{{}^\circ }}}\dfrac{1}{{{T}_{1}}}=\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}$
$-\dfrac{\Delta {{G}^{{}^\circ }}\left( {{T}_{1}} \right)}{\Delta {{H}^{{}^\circ }}}\dfrac{1}{{{T}_{1}}}+\dfrac{1}{{{T}_{1}}}=\dfrac{1}{{{T}_{2}}}$
This changes to:
$\begin{aligned} &T_{2}=\left[-\dfrac{\Delta G^{\circ}\left(T_{1}\right)}{\Delta H^{\circ}} \dfrac{1}{T_{1}}+\dfrac{1}{T_{1}}\right]^{-1} \\\ &=\left[\left(1-\dfrac{\Delta G^{\circ}\left(T_{1}\right)}{\Delta H^{\circ}}\right) \dfrac{1}{T_{1}}\right]^{-1} \\\ &=\dfrac{T_{1}}{1-\dfrac{\Delta G^{0}\left(T_{1}\right)}{\Delta H^{\circ}}} \\\ &=\dfrac{\Delta H^{\circ} T_{1}}{\Delta H^{\circ}-\Delta G^{\circ}\left(T_{1}\right)} \end{aligned}$
$={{\left[ \left( 1-\dfrac{\Delta {{G}^{{}^\circ }}\left( {{T}_{1}} \right)}{\Delta {{H}^{{}^\circ }}} \right)\dfrac{1}{{{T}_{1}}} \right]}^{-1}}$
$=\dfrac{{{T}_{1}}}{1-\dfrac{\Delta {{G}^{0}}\left( {{T}_{1}} \right)}{\Delta {{H}^{{}^\circ }}}}$
$=\dfrac{\Delta {{H}^{{}^\circ }}{{T}_{1}}}{\Delta {{H}^{{}^\circ }}-\Delta {{G}^{{}^\circ }}\left( {{T}_{1}} \right)}$
Or, if suppose we use relation that $\Delta G^{\circ}\left(T_{1}\right)=\Delta H^{\circ}-T_{1} \Delta S^{\circ}$ at first T, Also, assuming that the entropy change is constant across the temperature range,
$\begin{aligned} &T_{2}=\dfrac{\Delta H^{\circ} T_{1}}{\Delta H^{\circ}-\Delta H^{\circ}+T_{1} \Delta S^{\circ}} \\\ &=\Delta H^{\circ} / \Delta S^{\circ} \end{aligned}$ $=\dfrac{\Delta {{H}^{{}^\circ }}}{\Delta {{S}^{{}^\circ }}}$
which should be a familiar result.

Note:
The free energy is calculated differently depending on the nature of the operation. When examining processes that occur under constant pressure and temperature circumstances, for example, the Gibbs free energy change is employed, but the Helmholtz free energy change is utilised when analysing processes that occur under constant volume and temperature conditions. Temperature, pressure, and volume can affect the value and even the sign of both free energy changes. Because spontaneous processes are defined by a reduction in the system's free energy, they do not require an external source of energy to operate.