The thermodynamic driving force of a reaction depends on the current composition of the reaction mixture, not only on its standard-state free energy.

For a reaction at temperature $T$,

$$\boxed{\Delta G=\Delta G^\circ+RT\ln Q},$$

where

• $\Delta G$ is the Gibbs free energy change for the reaction under the current conditions,
• $\Delta G^\circ$ is the standard Gibbs free energy change at the same temperature,
• $R$ is the universal gas constant,
• $Q$ is the dimensionless thermodynamic reaction quotient.

Thermodynamically, $Q$ is built from activities. For an ideal dilute solution, an activity is approximated by a concentration divided by the standard concentration, $a_i\approx c_i/c^\circ$; for an ideal gas it is approximated by $p_i/p^\circ$. Introductory $Q_c$ or pressure-based expressions therefore represent convenient idealized forms of the same dimensionless quotient.

The sign of $\Delta G$ predicts the net thermodynamic direction:

• $\Delta G<0$: forward reaction is favored;
• $\Delta G>0$: reverse reaction is favored;
• $\Delta G=0$: the mixture is at equilibrium.

Example

Suppose at $298,\mathrm K$ a reaction has

$$\Delta G^\circ=-5.00,\mathrm{kJ/mol}$$

and the current thermodynamic reaction quotient is

$$Q=10.0.$$

Using

$$R=8.314,\mathrm{J,mol^{-1},K^{-1}},$$

$$RT\ln Q=(8.314)(298)\ln(10.0) \approx5.71,\mathrm{kJ/mol}.$$

Therefore

$$\Delta G=-5.00+5.71=+0.71,\mathrm{kJ/mol}.$$

Although the standard-state reaction has $\Delta G^\circ<0$, the current mixture has $\Delta G>0$, so under these actual conditions the net thermodynamic tendency is in the reverse direction.

This distinction is crucial: $\Delta G^\circ$ characterizes a standard reference composition, whereas $\Delta G$ describes the reaction driving force at the composition that actually exists.

The expression also explains why reaction direction changes continuously as a reaction proceeds. As reactants are converted to products, $Q$ changes, which changes $RT\ln Q$ and therefore the free-energy driving force.