Equilibrium & Spontaneity — $\Delta G^\circ$, $K$, Temperature

Theoretical background (quick recall)

  • Standard Gibbs criterion:
    \(\Delta G_r^\circ = \Delta H_r^\circ - T\Delta S_r^\circ\)

  • Link to equilibrium:
    \(\Delta G_r^\circ = -RT\ln K \;\;\Rightarrow\;\; K=\exp\!\left[-\frac{\Delta G_r^\circ}{RT}\right]\)

  • For gas-phase reactions, the pressure-based constant (with $p_0=1\,\text{bar}$) is \(K_p=\frac{\left(\dfrac{p_{\text{NO}_2}}{p_0}\right)^2}{\left(\dfrac{p_{\text{NO}}}{p_0}\right)^2\left(\dfrac{p_{\text{O}_2}}{p_0}\right)}.\) Equivalently, $K_p=K_c\,(RT)^{\Delta n_\text{gas}}$ with $\Delta n_\text{gas}=\sum\nu_\text{products}-\sum\nu_\text{reactants}$.

  • Direction from standard conditions: if $\Delta G_r^\circ<0$ at $T$, the reaction is spontaneous toward products; if $K\gg1$, products strongly prevail at equilibrium.

  • Temperature effect (van ’t Hoff):
    \(\frac{d\ln K}{dT}=\frac{\Delta H_r^\circ}{RT^2}\) For exothermic reactions $(\Delta H_r^\circ<0)$, $K$ decreases as $T$ increases.

Exercise — Oxidation of NO: $K_p$ and direction of spontaneity

For \(2\,\mathrm{NO}(g)+\mathrm{O}_2(g)\;\rightleftharpoons\;2\,\mathrm{NO}_2(g)\) at $T=298.15\,\text{K}$, use
$\Delta H_f^\circ(\mathrm{NO})=+90.2\,\text{kJ mol}^{-1}$,
$\Delta H_f^\circ(\mathrm{NO}_2)=+33.2\,\text{kJ mol}^{-1}$, and
$\Delta S_r^\circ=-145.0\,\text{J mol}^{-1}\,\text{K}^{-1}$

to evaluate $\Delta G_r^\circ$, then $K_p$. State the spontaneous direction from standard conditions and discuss the effect of increasing temperature.

Step-by-step solution (with explanations)

1) Write $K_p$ (definition).
Using partial pressures normalized by $p_0=1\,\text{bar}$: \(K_p=\frac{\left(\dfrac{p_{\text{NO}_2}}{p_0}\right)^2}{\left(\dfrac{p_{\text{NO}}}{p_0}\right)^2\!\left(\dfrac{p_{\text{O}_2}}{p_0}\right)}.\) Here $\Delta n_\text{gas}=2-(2+1)=-1$, so $K_p=K_c\,(RT)^{-1}$.

This means that when we pass from concentrations ($K_c$) to pressures ($K_p$), the units are corrected by a factor $(RT)^{\Delta n}$.

2) Compute $\Delta H_r^\circ$ from formation enthalpies.
Remember $\Delta H_f^\circ(\mathrm{O}_2,g)=0$: \(\Delta H_r^\circ=2\,\Delta H_f^\circ(\mathrm{NO}_2)-\big[2\,\Delta H_f^\circ(\mathrm{NO})+1\cdot\Delta H_f^\circ(\mathrm{O}_2)\big] =2(33.2)-2(90.2)= -114.0\,\text{kJ mol}^{-1}.\)

3) Compute $\Delta G_r^\circ$ at $298.15\,\text{K}$.
Use $\Delta S_r^\circ=-145.0\,\text{J mol}^{-1}\,\text{K}^{-1}=-0.145\,\text{kJ mol}^{-1}\,\text{K}^{-1}$: \(\Delta G_r^\circ=\Delta H_r^\circ-T\Delta S_r^\circ = -114.0 - (298.15)(-0.145) \approx -70.8\,\text{kJ mol}^{-1}.\)

Negative $\Delta G_r^\circ$ already indicates spontaneity toward products under standard conditions.

4) Convert $\Delta G_r^\circ$ into $K_p$.
With $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$: \(K_p=\exp\!\left[-\frac{\Delta G_r^\circ}{RT}\right] =\exp\!\left(\frac{70.8\times10^3}{(8.314)(298.15)}\right) =\exp(28.56)\approx 2.5\times10^{12}.\)

Such a huge $K_p$ means that, at equilibrium, the concentration of $\mathrm{NO}_2$ is overwhelmingly larger than that of $\mathrm{NO}$ and $\mathrm{O}_2$.

5) Direction from standard conditions.
Because $\Delta G_r^\circ<0$ and $K_p\gg1$, the reaction is spontaneous toward products from standard-state reactants; equilibrium lies far to the side of $\mathrm{NO}_2$.

6) Temperature effect (sign analysis).
Here $\Delta H_r^\circ<0$ (exothermic) and $\Delta S_r^\circ<0$ (gas moles decrease: $3\to2$).
Increasing $T$ makes $-T\Delta S_r^\circ$ more positive, so $\Delta G_r^\circ$ becomes less negative (eventually positive at high $T$): the reaction is less favorable as temperature increases, and $K_p$ decreases with $T$ (consistent with van ’t Hoff and Le Châtelier).

Conceptual notes

  • Why $\Delta S_r^\circ<0$ here? Gas-phase stoichiometry gives $\Delta n_\text{gas}=-1$: fewer gas moles at products generally implies lower entropy of the system.
  • Standard formation data reminder: $\Delta H_f^\circ(\mathrm{O}_2,g)=0$ by convention; only $\mathrm{NO}$ and $\mathrm{NO_2}$ contribute to $\Delta H_r^\circ$.
  • Large $K_p$ ($\sim10^{12}$ at $298\,\text{K}$) means that, at equilibrium, $p_{\text{NO}2}$ dominates compared to $p{\text{NO}}$ and $p_{\text{O}_2}$ under standard-state scaling.
  • For different $T$, you may estimate $K_p(T)$ using the van ’t Hoff equation with (piecewise) constant $\Delta H_r^\circ$ in the temperature range of interest.
  • Pedagogical point: Students should train to interpret both numbers ($\Delta G^\circ$, $K$) and signs ($\Delta H^\circ$, $\Delta S^\circ$) to predict spontaneity and temperature effects qualitatively, not only numerically.