Gibbs free energy and the equilibrium constant

Result

Gibbs free energy and the equilibrium constant

T-035Home CU-201Threads thermo

Statement

The standard Gibbs free energy change of a reaction relates to its equilibrium constant by \(\Delta G^\circ=-RT\ln K\); more generally, at any (not necessarily standard) composition, \(\Delta G=\Delta G^\circ+RT\ln Q\), where \(Q\) is the reaction quotient. This relation is not an independent postulate but a direct, derivable consequence of how Gibbs free energy depends on pressure (or concentration) for each species, combined with gibbs-free-energy's equilibrium condition \(\Delta G=0\) — it explains, rather than merely asserts, why the equilibrium constant takes its familiar "products over reactants" multiplicative form.

Why it matters

Introductory treatments typically introduce the equilibrium constant \(K\) as a memorised ratio written directly from a balanced equation, with no justification for why it takes that specific mathematical form. This result derives it instead, and in doing so provides the single most important quantitative bridge between thermodynamics (how favourable a reaction is, \(\Delta G^\circ\)) and equilibrium chemistry (how far a reaction proceeds toward products, \(K\)): a large negative \(\Delta G^\circ\) corresponds to a large \(K\) (strongly product-favoured), and vice versa.

Hypotheses

Every species behaves as an ideal gas or ideal solution.This licenses the simple logarithmic dependence of chemical potential on pressure or concentration used in the derivation; real, non-ideal gases and concentrated solutions require a more general treatment using activity coefficients in place of bare pressures or concentrations, an important refinement beyond this introductory scope. Temperature is held constant when comparing \(\Delta G^\circ\), \(K\), and any measured composition (\(Q\)).Both \(K\) and \(\Delta G^\circ\) are temperature-dependent quantities (gibbs-free-energy establishes \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\)); combining a \(K\) value measured at one temperature with a \(\Delta G^\circ\) computed at a different temperature via this result's equation would not be valid.

Proof

dG = VdP \quad (\text{constant } T)

Combining the definitions \(H=U+PV\) (calorimetry) and \(G=H-TS\) (gibbs-free-energy) with the reversible-process relation \(dU=T\,dS-P\,dV\) (from the first and second laws, first-law-chemistry and entropy-second-law) gives, after full expansion, \(dG=-S\,dT+V\,dP\); at constant temperature, the \(-S\,dT\) term vanishes, leaving \(dG=V\,dP\). A

\mu(P) = \mu^\circ + RT\ln\frac{P}{P^\circ}

For one mole of ideal gas, \(V=RT/P\) (ideal-gas-law); substituting into Step 1 and integrating from the standard pressure \(P^\circ\) to \(P\) gives the molar Gibbs free energy (chemical potential) \(\mu\) as an explicit logarithmic function of pressure, relative to its standard-state value \(\mu^\circ\). A

\Delta G_{\text{rxn}} = \sum(\text{coeff.}\times\mu_{\text{products}}) - \sum(\text{coeff.}\times\mu_{\text{reactants}}) = \Delta G^\circ_{\text{rxn}} + RT\ln Q

Summing Step 2 over every species in a balanced reaction \(aA+bB\to cC+dD\), weighted by stoichiometric coefficient, and collecting all the logarithmic terms together defines the reaction quotient \(Q\equiv\dfrac{(P_C/P^\circ)^c(P_D/P^\circ)^d}{(P_A/P^\circ)^a(P_B/P^\circ)^b}\) — the exact "products over reactants, raised to stoichiometric powers" form is derived here directly, not assumed in advance. A

\text{At equilibrium: } \Delta G=0,\ Q=K \ \Longrightarrow\ 0 = \Delta G^\circ + RT\ln K \ \Longrightarrow\ \Delta G^\circ = -RT\ln K

Applying gibbs-free-energy's equilibrium condition (\(\Delta G=0\)) to Step 3, and using the definition of \(K\) as the specific value \(Q\) takes at equilibrium, gives the central relation directly. A

Result

\Delta G^\circ = -RT\ln K, \qquad \Delta G = \Delta G^\circ + RT\ln Q

Reading. The equilibrium constant is not an independently defined ratio but a direct mathematical consequence of the pressure- (or concentration-) dependence of Gibbs free energy combined with the equilibrium condition already established in gibbs-free-energy; the more general relation additionally predicts \(\Delta G\) — and hence the direction a reaction proceeds — from any arbitrary starting composition, not only the standard state.

Scope. Assumes ideal behaviour throughout (Hypotheses); real, non-ideal systems require the activity-based generalisation of this same relation.

Corollaries & converses

\(K>1\) (product-favoured at equilibrium) corresponds to \(\Delta G^\circ<0\); \(K<1\) (reactant-favoured) corresponds to \(\Delta G^\circ>0\); \(K=1\) corresponds to \(\Delta G^\circ=0\) exactly, following directly from the sign of \(\ln K\) in the Result.
Combining this result with \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\) (gibbs-free-energy) gives \(\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R}\), showing explicitly how and why the equilibrium constant depends on temperature — a relation this network's later development of chemical equilibrium builds on directly.
Converse (reaction-direction prediction): comparing \(Q\) (computed from any given, non-equilibrium composition) against \(K\) predicts the sign of \(\Delta G\), and hence which direction the reaction proceeds: \(QK\) gives \(\Delta G>0\) (reverse, toward reactants); \(Q=K\) means the system is already at equilibrium.

Fails without

Apply the simple \(\mu=\mu^\circ+RT\ln(P/P^\circ)\) relation to a genuinely non-ideal gas or concentrated solution (drop Hypotheses): real intermolecular interactions (already flagged for real gases in van-der-waals-equation) mean the true chemical potential deviates from this ideal logarithmic form, particularly at high pressure or concentration; the rigorous generalisation replaces bare pressures or concentrations with activities, a correction beyond this introductory scope.
Combine a \(K\) value and a \(\Delta G^\circ\) value measured or computed at two different temperatures: since both quantities genuinely depend on \(T\) (Hypotheses, second postulate), the Result's equation is only self-consistent when both refer to the identical temperature.

Common errors

Confusing \(\Delta G\) (the actual free energy change at whatever composition currently exists, generally changing as a reaction proceeds) with \(\Delta G^\circ\) (a single fixed value, defined specifically at standard-state composition) — a very frequent source of error, since only \(\Delta G^\circ\), not \(\Delta G\) itself, is the tabulated, table-lookup quantity.
Mixing a \(K\) and \(\Delta G^\circ\) from inconsistent temperatures (Fails without, second bullet).
Getting the direction of the \(K\)-vs-\(\Delta G^\circ\) sign relationship backward — recalling \(\Delta G^\circ=-RT\ln K\) with the wrong overall sign, which flips whether a large \(K\) is associated with a negative or positive \(\Delta G^\circ\).

Discussion

This relation is a direct application and extension of Gibbs's own foundational 1870s work (already discussed in gibbs-free-energy) to chemical equilibrium specifically, developed further by Jacobus van 't Hoff, whose 1884 studies of reaction rates and chemical equilibrium (including the temperature-dependence relation for \(K\) that follows directly from the Corollaries' \(\ln K\) formula) built directly on this thermodynamic foundation. Van 't Hoff is the same chemist credited, a decade earlier, with proposing the tetrahedral carbon model to explain chirality (chirality-optical-activity) — a genuine illustration of how a single scientist's contributions can span what might otherwise seem like distinct sub-fields of chemistry.

The derivation given here (Steps 1–4) ties together nearly every major result of this unit and the preceding one: enthalpy's definition (calorimetry), the first and second laws (first-law-chemistry, entropy-second-law), Gibbs free energy and its equilibrium condition (gibbs-free-energy), and the ideal gas law (ideal-gas-law) all combine into a single chain of reasoning explaining, from first principles, why chemical equilibrium behaves the way it does.

Common misconception: that the equilibrium constant's specific algebraic form (products over reactants, raised to stoichiometric coefficients) is simply a convention or definition to be memorised. As Step 3 shows explicitly, this form is a direct mathematical consequence of chemical potential's logarithmic dependence on pressure or concentration — it could not have taken a different algebraic form (say, a simple difference rather than a ratio) without contradicting the underlying thermodynamics.

Worked examples

\text{Water autoionisation: } K_w=1.0\times10^{-14}\text{ at }298\,\text{K}: \quad \Delta G^\circ=-RT\ln K_w=-(8.314)(298)\ln(1.0\times10^{-14})\approx+79.9\,\text{kJ/mol}

A direct application of the Result; the large positive \(\Delta G^\circ\) correctly reflects that water's autoionisation is strongly reactant-favoured at equilibrium (\(K_w\) is extremely small), matching the well-established standard value for this quantity closely. A

\text{N}_2\text{O}_4(g)\rightleftharpoons2\text{NO}_2(g),\ \Delta G^\circ=+5.4\,\text{kJ/mol at }298\,\text{K}: \quad K=e^{-\Delta G^\circ/RT}=e^{-5400/(8.314\times298)}\approx0.113

A small \(K<1\), correctly reflecting a modestly reactant-favoured equilibrium at standard conditions, consistent with the positive (though comparatively small) \(\Delta G^\circ\). A

\Delta G^\circ(\text{H}_2\text{O autoionisation})\approx+79.9\,\text{kJ/mol}; \qquad K(\text{N}_2\text{O}_4\rightleftharpoons2\text{NO}_2)\approx0.113

Reading. The same equation runs equally well in either direction — from a known \(K\) to \(\Delta G^\circ\), or from a known \(\Delta G^\circ\) to \(K\) — using nothing but temperature and the universal gas constant.

Scope. Both directions of calculation apply to any reaction with either quantity independently known, at a single, consistent temperature.

Problems

A reaction has \(\Delta G^\circ=-32.9\,\text{kJ/mol}\) at \(298\,\text{K}\). Compute \(K\), and state whether the reaction is product- or reactant-favoured at equilibrium.
Solution
\(K=e^{-\Delta G^\circ/RT}=e^{32{,}900/(8.314\times298)}\approx5.84\times10^5\), a very large value, indicating the reaction is strongly product-favoured at equilibrium, consistent with the substantially negative \(\Delta G^\circ\) (Corollaries, first bullet).
A reaction has \(\Delta G^\circ=-10.0\,\text{kJ/mol}\) at \(298\,\text{K}\) (giving \(K\approx56.6\)). If the reaction is currently at a composition with reaction quotient \(Q=50.0\), compute \(\Delta G\) and state which direction the reaction will proceed.
Solution
\(\Delta G=\Delta G^\circ+RT\ln Q=-10{,}000+(8.314)(298)\ln(50.0)\approx-307\,\text{J/mol}\), a small negative value. Since \(\Delta G<0\) (equivalently, \(Q=50.0
Using the Corollaries' relation \(\ln K=-\Delta H^\circ/(RT)+\Delta S^\circ/R\), explain qualitatively why an exothermic reaction (\(\Delta H^\circ<0\)) generally has a smaller equilibrium constant at higher temperature than at lower temperature, all else being equal.
Solution
For \(\Delta H^\circ<0\), the term \(-\Delta H^\circ/(RT)\) is positive and inversely proportional to \(T\); as \(T\) increases, this positive contribution to \(\ln K\) shrinks, tending to decrease \(\ln K\) (and hence \(K\)) as temperature rises, all else (specifically \(\Delta S^\circ\), assumed roughly constant) being equal — consistent with the general, well-known chemical principle that raising temperature tends to disfavour exothermic reactions at equilibrium, a preview of Le Châtelier-type reasoning developed more fully in this network's later equilibrium content.
For a reaction with \(K=1.0\times10^3\) at \(298\,\text{K}\), a chemist prepares a mixture with reaction quotient \(Q=1.0\times10^5\). Predict, without calculating \(\Delta G\) explicitly, which direction the reaction will proceed.
Solution
Since \(Q=1.0\times10^5>K=1.0\times10^3\), the Converse predicts \(\Delta G>0\) for the forward reaction, meaning the reaction proceeds in reverse, toward reactants, until \(Q\) decreases back down to equal \(K\) at the new equilibrium.