R2.3.7 Gibbs free energy and equilibrium (Higher Level Only) Notes

The Relationship Between $ \Delta G^\circ $ and $ K $

Standard Gibbs Free Energy Change ($ \Delta G^\circ $) and Its Significance

1. As discussed in Reactivity 1.4, the Gibbs free energy change ($ \Delta G $) is a thermodynamic quantity that determines whether a reaction is spontaneous.
2. A spontaneous reaction proceeds without requiring external energy input.
   1. $ \Delta G< 0 $: The reaction is spontaneous in the forward direction.
   2. $ \Delta G >0 $: The reaction is non-spontaneous in the forward direction (but spontaneous in reverse).
   3. $ \Delta G = 0 $:The system is at equilibrium.

But how does $ \Delta G^\circ $ relate to the equilibrium constant ($ K $)?

The Equation Relating $ \Delta G^\circ $ and $ K $

1. The connection between $ \Delta G^\circ $ and $ K $ is expressed mathematically as: $$
   \Delta G^\circ = -RT \ln K
   $$ where:
   1. $ \Delta G^\circ $: Standard Gibbs free energy change (in $\text{J mol}^{-1}$).
   2. $ R $: Gas constant ($ 8.31 \, \mathrm{J \, mol^{-1} \, K^{-1}} $).
   3. $ T $: Absolute temperature in Kelvin.
   4. $ K $: Equilibrium constant (unitless).
2. This equation links thermodynamics ($ \Delta G^\circ $) to equilibrium ($ K $) and provides critical insights into reaction behavior:
3. When $ \Delta G^\circ < 0 $:
   1. The reaction is product-favored.
   2. $ K > 1 $, indicating that the equilibrium lies toward the products.
4. When $ \Delta G^\circ > 0 $:
   1. The reaction is reactant-favored.
   2. $ K < 1 $, indicating that the equilibrium lies toward the reactants.
5. When $ \Delta G^\circ = 0 $:
   1. The system is at equilibrium.
   2. $ K = 1 $, meaning that reactants and products are present in comparable amounts.

Interpreting $ \Delta G^\circ $ and $ K $

$ \Delta G^\circ < 0 $ (Spontaneous Forward Reaction)

1. Suppose a reaction has $ \Delta G^\circ = -25 \, \text{kJ mol}^{-1} $ at 298 K.
2. The negative value indicates that the forward reaction is spontanous under standard conditions.
3. Using the equation: $$
   K = e^{-\frac{\Delta G^\circ }{ RT}}
   $$
4. Substitute the values:
   1. $ \Delta G^\circ = -25,000 \, \mathrm{J mol^{-1}} $
   2. $ R = 8.31 \, \mathrm{J mol^{-1} K^{-1}} $
   3. $ T = 298 \, \mathrm{K} $ $$
      K = e^{-(-25,000) / (8.31 \times 298)}
      $$ $$
      K = e^{10.08} \approx 23000
      $$
5. Since $ K \gg 1 $, this equilibrium strongly favors the products.

$ \Delta G^\circ > 0 $ (Non-Spontaneous Forward Reaction)

1. Now consider a reaction with $ \Delta G^\circ = +10 \, \mathrm{kJ mol^{-1}} $.
2. Substituting into the equation: $$
   K = e^{-10,000 / (8.31 \times 298)}
   $$ $$
   K = e^{-4.03} \approx 0.018
   $$
3. Here, $ K \ll 1 $, meaning the equilibrium strongly favors the reactants.

$ \Delta G^\circ = 0 $ (Equilibrium)

1. If $ \Delta G^\circ = 0 $, then: $$
   K = e^{0} = 1
   $$
2. This indicates that the concentrations of reactants and products are balanced at equilibrium, depending on their stoichiometric coefficients.