Standard Reaction Gibbs Energy and ΔG°–K Relation

Drop a flask of colorless N₂O₄ into a warm room and it turns brown as some converts to NO₂. Stop it in an ice bath and the brown fades. Somewhere between "all reactant" and "all product" the system settles into a fixed ratio, and that ratio is what chemists call the equilibrium constant $K$ . This article derives the equation that connects $K$ to a single number you can look up in a table: the standard reaction Gibbs energy, $\Delta_rG^\circ$ . The relation $\Delta_rG^\circ = -RT\ln K$ is arguably the most-used equation in chemical thermodynamics. It converts a calorimetric or spectroscopic quantity into a prediction about concentrations, pressures, and yields — for acid dissociation, protein folding, industrial ammonia synthesis, and battery voltages alike. You will see how the chemical potential of an ideal gas leads to a general expression for $\Delta_rG$ at any composition. Then you will watch that expression collapse, at equilibrium, into the compact law relating $\Delta_rG^\circ$ and $K$ . Two worked numerical examples follow, along with the traps that catch most students on exams.

Key takeaways

The reaction Gibbs energy at any composition is $\Delta_rG = \Delta_rG^\circ + RT\ln Q$ , where $Q$ is the reaction quotient built from actual partial pressures or activities.
At equilibrium $\Delta_rG = 0$ , so $\Delta_rG^\circ = -RT\ln K$ ; $K$ is the special value $Q$ takes when the reaction stops moving.
$\Delta_rG^\circ$ is evaluated at standard state (1 bar, specified $T$ , usually 298 K) but $K$ itself is temperature-dependent, not composition-dependent.
A large negative $\Delta_rG^\circ$ means $K \gg 1$ (products favored); a large positive $\Delta_rG^\circ$ means $K \ll 1$ (reactants favored).
The sign of $\Delta_rG$ (not $\Delta_rG^\circ$ ) tells you which way a reaction runs from any given starting mixture, because it depends on $Q$ through the actual composition.
Key idea: equilibrium is the composition at which the falling free-energy curve of the reaction mixture reaches its minimum, and $-RT\ln K$ is just the depth of that minimum measured from the standard state.

The core insight

Because each species' chemical potential rises logarithmically with its partial pressure (or activity), the reaction Gibbs energy of a mixture is the standard value plus $RT\ln Q$ ; setting $\Delta_rG = 0$ at the composition where Gibbs energy is minimized converts this single equation into the thermodynamic definition of the equilibrium constant, $\Delta_rG^\circ = -RT\ln K$ .

The physical picture

Consider a general gas-phase reaction,

aA + bB \rightleftharpoons cC + dD

running at constant temperature and pressure in a closed vessel. At every instant the mixture has a total Gibbs energy $G$ that depends on how far the reaction has progressed, measured by the extent of reaction $\xi$ . The slope of $G$ against $\xi$ is the reaction Gibbs energy,

\Delta_rG = \left(\frac{\partial G}{\partial \xi}\right)_{T,p}

As the reaction proceeds from pure reactants toward pure products, $\Delta_rG$ starts negative (the mixture's Gibbs energy is falling — the reaction runs forward), passes through zero at some intermediate composition, and becomes positive beyond that point (Gibbs energy would rise, so the reaction resists going further). The zero-crossing is equilibrium: not "no reaction happening," but the composition where forward and reverse tendencies exactly balance. Each species contributes to $G$ through its chemical potential, which for an ideal gas depends on its partial pressure through

\mu_J = \mu_J^\circ + RT\ln(p_J/p^\circ)

This single logarithmic dependence is the entire engine behind everything that follows. Because $\mu_J$ is not fixed — it depends on how much of $J$ is actually present — the reaction Gibbs energy itself must depend on composition. Summing the weighted chemical potentials of products minus reactants (weighted by stoichiometric coefficients) produces a composition-dependent term that turns out to be exactly $RT\ln Q$ , where $Q$ is the reaction quotient built from the same partial-pressure ratios. The second figure below shows why the equilibrium point is special: it is precisely where the total- $G$ -versus- $\xi$ curve has zero slope, i.e., where $\Delta_rG = 0$ .

Figure

Reaction Gibbs energy falls to zero at the equilibrium extent of reaction

Figure

Standard reaction Gibbs energy sets K, but ΔrG tracks the running quotient Q

The core relations

\Delta_rG = \Delta_rG^\circ + RT\ln Q

where $Q$ is the reaction quotient of activities or partial pressures at the current, possibly non-equilibrium, composition.

Q = \prod_J a_J^{\,\nu_J} \quad\text{(ideal gas: } a_J \to p_J/p^\circ\text{)}

where $\nu_J$ is positive for products and negative for reactants.

\Delta_rG^\circ = -RT\ln K

where $K$ is the thermodynamic equilibrium constant, the value $Q$ takes when $\Delta_rG = 0$ .

K = \exp\!\left(\frac{-\Delta_rG^\circ}{RT}\right)

where a more negative $\Delta_rG^\circ$ gives an exponentially larger $K$ .

Derivation AFrom μ = μ° + RT ln p to ΔrG° = −RT ln K

Goal. Show that the chemical potential's logarithmic pressure dependence forces $\Delta_rG = \Delta_rG^\circ + RT\ln Q$ , and that setting $\Delta_rG=0$ at equilibrium yields $\Delta_rG^\circ = -RT\ln K$ .

Step 1 — Write G as a function of extent of reaction. For the reaction $aA + bB \rightleftharpoons cC + dD$ , let $\xi$ measure how far the reaction has proceeded. As $\xi$ advances by $d\xi$ , the amounts change by $dn_J = \nu_J\, d\xi$ for each species $J$ (with $\nu_J>0$ for products, $\nu_J<0$ for reactants). The total differential of Gibbs energy at constant $T,p$ is

dG = \sum_J \mu_J\, dn_J = \left(\sum_J \nu_J \mu_J\right) d\xi

so by definition

\Delta_rG \equiv \left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \sum_J \nu_J \mu_J = (c\mu_C + d\mu_D) - (a\mu_A + b\mu_B)

Step 2 — Insert the ideal-gas chemical potential. Each chemical potential depends on its own partial pressure through

\mu_J = \mu_J^\circ + RT\ln(p_J/p^\circ)

Substituting for all four species,

\Delta_rG = \Big[c\mu_C^\circ + d\mu_D^\circ - a\mu_A^\circ - b\mu_B^\circ\Big] + RT\Big[c\ln(p_C/p^\circ) + d\ln(p_D/p^\circ) - a\ln(p_A/p^\circ) - b\ln(p_B/p^\circ)\Big]

The bracketed standard-state term is exactly the definition of $\Delta_rG^\circ$ , the reaction Gibbs energy when every species sits at its standard partial pressure of 1 bar.

Step 3 — Collapse the logarithms with log laws. Using $c\ln x - a\ln y = \ln(x^c/y^a)$ repeatedly, the composition-dependent bracket becomes a single logarithm of a ratio of powers of partial pressures,

RT\ln\!\left[\frac{(p_C/p^\circ)^c (p_D/p^\circ)^d}{(p_A/p^\circ)^a (p_B/p^\circ)^b}\right] = RT\ln Q

This ratio is defined as the reaction quotient $Q$ . Hence

\Delta_rG = \Delta_rG^\circ + RT\ln Q

This holds at any composition — equilibrium or not — because no equilibrium assumption entered Steps 1–3. It is a statement about instantaneous slopes of $G(\xi)$ .

Step 4 — Impose the equilibrium condition. Equilibrium at constant $T,p$ is the condition of minimum Gibbs energy (established in the prerequisite treatment of equilibrium), meaning

\left(\frac{\partial G}{\partial \xi}\right)_{T,p} = 0 \quad\Longrightarrow\quad \Delta_rG = 0 \text{ at } \xi = \xi_{eq}

Substituting this into the Step 3 result, evaluated with the equilibrium partial pressures,

0 = \Delta_rG^\circ + RT\ln Q_{eq}

The quotient evaluated specifically at equilibrium partial pressures is given its own name and symbol, $K$ :

Q_{eq} \equiv K

Step 5 — Solve for the standard reaction Gibbs energy. Rearranging,

\boxed{\Delta_rG^\circ = -RT\ln K}

Because $\Delta_rG^\circ$ depends only on $T$ (it is built from standard-state chemical potentials, which carry no pressure or concentration dependence), $K$ itself can depend only on temperature — never on the starting composition, total pressure, or presence of a catalyst. That composition-independence is what makes $K$ useful as a fixed benchmark against which any $Q$ can be compared.

Worked examples

Example 1 — Foundational: finding K from a tabulated Δ_rG°. For the synthesis of ammonia,

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

tables give $\Delta_fG^\circ(\text{NH}_3, g) = -16.4\ \text{kJ mol}^{-1}$ at 298 K, with $\Delta_fG^\circ$ of the elements zero by convention. Step 1: compute $\Delta_rG^\circ$ .

\Delta_rG^\circ = 2(-16.4\ \text{kJ mol}^{-1}) - [1(0) + 3(0)] = -32.8\ \text{kJ mol}^{-1}

Step 2: apply $\Delta_rG^\circ = -RT\ln K$ , solved for $K$ .

\ln K = \frac{-\Delta_rG^\circ}{RT} = \frac{32{,}800\ \text{J mol}^{-1}}{(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})} = \frac{32{,}800}{2478} = 13.24

Step 3: exponentiate.

K = e^{13.24} = 5.6\times10^{5}

Sanity check: $\Delta_rG^\circ$ is comfortably negative (about $-33$ kJ/mol), so $K$ should be large — and $5.6\times10^5$ is indeed $\gg 1$ , consistent with ammonia being strongly favored thermodynamically at 298 K (even though the rate is slow without a catalyst — a separate kinetic issue).

Example 2 — Applied: predicting which way a reaction runs from a non-equilibrium mixture. Suppose a reactor is charged with N₂, H₂, and NH₃ such that at 298 K the partial pressures are

p_{N_2} = 2.0\ \text{bar}, \quad p_{H_2} = 4.0\ \text{bar}, \quad p_{NH_3} = 6.0\ \text{bar}

Using $K = 5.6\times10^5$ from Example 1, determine the direction of spontaneous change. Step 1: build $Q$ from the same expression as $K$ but with actual pressures (each divided by $p^\circ = 1$ bar, so numerically we can use the bar values directly).

Q = \frac{p_{NH_3}^2}{p_{N_2}\, p_{H_2}^3} = \frac{(6.0)^2}{(2.0)(4.0)^3} = \frac{36}{(2.0)(64)} = \frac{36}{128} = 0.281

Step 2: compute $\Delta_rG$ at this composition using $\Delta_rG = \Delta_rG^\circ + RT\ln Q$ .

RT\ln Q = (8.314)(298)\ln(0.281) = (2478\ \text{J mol}^{-1})(-1.269) = -3145\ \text{J mol}^{-1} = -3.14\ \text{kJ mol}^{-1}

\Delta_rG = -32.8\ \text{kJ mol}^{-1} + (-3.14\ \text{kJ mol}^{-1}) = -35.9\ \text{kJ mol}^{-1}

Step 3: interpret the sign. Since $\Delta_rG < 0$ , the forward reaction (more NH₃ formation) is still spontaneous — the mixture has not yet reached equilibrium and needs still more product to arrive at $Q = K = 5.6\times10^5$ . Sanity check: $Q = 0.281$ is far below $K = 5.6\times10^5$ , so the reaction quotient must increase toward $K$ ; since $Q$ grows as NH₃ increases, the forward direction is correctly predicted by both the $Q$ -vs- $K$ comparison and the negative $\Delta_rG$ — the two criteria agree, as they must.

Limiting cases & common traps

Situation	What happens	Why
$Q < K$	$\Delta_rG < 0$ , reaction runs forward	mixture is "reactant-heavy" relative to equilibrium
$Q > K$	$\Delta_rG > 0$ , reaction runs in reverse	mixture is "product-heavy" relative to equilibrium
$Q = K$	$\Delta_rG = 0$ , no net change	this is the definition of equilibrium
$\Delta_rG^\circ \ll 0$	$K \gg 1$	products dominate at equilibrium
$\Delta_rG^\circ \gg 0$	$K \ll 1$	reactants dominate at equilibrium
$\Delta_rG^\circ = 0$	$K = 1$	comparable amounts of reactants and products at equilibrium

Other traps: forgetting that $Q$ and $K$ must use the same standard state and units (partial pressures divided by $p^\circ = 1$ bar, so both are dimensionless); assuming $K$ depends on total pressure or on adding an inert gas (it does not, though mole fractions at equilibrium may shift); and forgetting that only $\Delta_rG^\circ$ , not $\Delta_rG$ , connects directly to $K$ through a temperature-only exponential — feeding this relation lightly into the later Van 't Hoff analysis of how $K$ itself varies with $T$ .

The chemist behind it

The equilibrium constant traces back to Cato Guldberg and Peter Waage's law of mass action (1864), but the thermodynamic derivation connecting free energy to $K$ is the achievement of Josiah Willard Gibbs and, in more accessible chemical form, of the American physical chemist Gilbert N. Lewis, who introduced fugacity and activity to make the ideal-gas relations rigorous for real systems.

Origins · 1 minGilbert Newton Lewis and how this was discovered

Gilbert Newton Lewis

1875–1946 · Formalized chemical potential, activity, and fugacity, turning Gibbs's abstract thermodynamics into working tools for chemists

quiz[A reaction mixture has Q greater than K. Which direction does it spontaneously proceed]

Forward, toward more products
Reverse, toward more reactants ✓
No net change occurs
Direction depends on the sign of ΔrG°

quiz[The relation ΔrG = ΔrG° + RT ln Q was derived using which key fact]

The equilibrium constant is always dimensionless
The chemical potential of an ideal gas varies logarithmically with its partial pressure ✓
Gibbs energy is conserved in a closed system
Enthalpy and entropy are both temperature-independent

Ask the tutor: How does the Van 't Hoff equation let you predict K at a new temperature once you know Δ_rH° and Δ_rG° at 298 K?

Why is K dimensionless if partial pressures have units of pressure

Each partial pressure in $Q$ or $K$ is always divided by the standard pressure $p^\circ = 1$ bar before being raised to a power. This ratio $p_J/p^\circ$ is dimensionless, so the whole product defining $Q$ or $K$ is dimensionless too. Textbooks often write $p_J$ alone as shorthand, but that only works if $p_J$ is understood to be measured in bar. Carrying $p^\circ$ explicitly avoids confusion and keeps $\ln K$ mathematically valid, since logarithms of dimensioned quantities are undefined.

Does the equilibrium constant depend on the starting mixture

No. $K$ is fixed once temperature is fixed, because it is defined through $\Delta_rG^\circ$ , which involves only standard-state chemical potentials with no reference to actual amounts present. Different starting mixtures reach different equilibrium partial pressures, but those pressures always combine, through the same quotient expression, to the same numerical $K$ . This composition-independence is exactly what makes $K$ a useful predictive constant rather than just another description of one particular experiment.

How does temperature affect Delta_rG° and K

Both change with temperature, but through different mechanisms. $\Delta_rG^\circ = \Delta_rH^\circ - T\Delta_rS^\circ$ varies roughly linearly with $T$ if enthalpy and entropy are treated as constant over a modest range. Because $K$ depends exponentially on $\Delta_rG^\circ/T$ , even a small enthalpy term produces a strong temperature sensitivity in $K$ . This coupling is exactly what the Van 't Hoff equation quantifies, expressing $d(\ln K)/dT$ directly in terms of $\Delta_rH^\circ$ .

Can this relation be used for reactions involving solids, liquids, or solutions

Yes, with activities replacing partial pressures. Pure solids and liquids are assigned activity 1 in their standard states, so they drop out of $Q$ and $K$ entirely, which is why, for example, $CaCO_3(s)$ does not appear in the equilibrium expression for its decomposition. For solutes, activity is approximated by molar concentration divided by the standard concentration $c^\circ = 1\ \text{mol dm}^{-3}$ , mirroring the gas-phase treatment exactly.

What is the difference between Q and K in practice

$Q$ is calculable at any instant, using whatever partial pressures or concentrations happen to be present, whether or not the system is at equilibrium. $K$ is the single specific value $Q$ takes only when the system has reached equilibrium at a given temperature. Comparing a freshly measured $Q$ to the known $K$ tells you immediately, via the sign of $RT\ln(Q/K)$ , which direction the reaction must shift to reach balance.