7.1. Equilibrium Constant

Course-wide Conventions & Notation

Overview and Learning Objectives¶

Chapter 6 treated one kind of equilibrium — the coexistence of two phases of a single substance — and showed that the condition for equilibrium reduces to equality of chemical potentials, $\mu_\alpha = \mu_\beta$. Chemical equilibrium is the multi-species generalization: instead of two phases of one substance, we now have several chemical species interconverting through a reaction, and the analogous condition (derived below) is

In this section we translate that abstract condition into the quantitative tools students have seen in general chemistry: the extent of reaction $\xi$, the reaction quotient $Q_p$, the equilibrium constant $K$, and the temperature dependence given by the van’t Hoff equation. The key ideas are:

• A reaction at constant $T,P$ proceeds in the direction that decreases the Gibbs free energy of the system.

• We track progress with the extent of reaction $\xi$.

• At constant $T$ and $P$, equilibrium occurs when the Gibbs free energy is minimized, i.e.

  $$\left(\frac{\partial G}{\partial \xi}\right)_{T,P}=0.$$

  (2)

This minimization gives the chemical equilibrium condition $\Delta_r G=0$, and (for ideal gases) the familiar relationship

Learning objectives:

• Use the extent of reaction $\xi$ to relate changes in species amounts via $dn_i=\nu_i\,d\xi$.

• Derive $\Delta_r G=\sum_i \nu_i\mu_i$ and show $(\partial G/\partial\xi)_{T,P}=\Delta_r G$.

• Derive $\Delta_r G=\Delta_r G^\circ + RT\ln Q_p$ for ideal gases and define $Q_p$ as a dimensionless product of pressure ratios.

• Use $\Delta_r G^\circ=-RT\ln K$ and the comparison of $Q_p$ and $K$ to predict reaction direction.

• Apply the van’t Hoff equation to predict how $K$ changes with temperature.

Core Ideas and Derivations¶

7.1.1 Extent of reaction $\xi$¶

Consider a general reaction written with stoichiometric coefficients:

A convenient way to describe composition changes is the extent of reaction $\xi$ (units of moles). If $n_{i,0}$ are initial amounts, then as the reaction proceeds,

Differentiating gives the compact relation

if we adopt the signed stoichiometry convention: $\nu_i<0$ for reactants and $\nu_i>0$ for products.

7.1.2 Gibbs free energy and the equilibrium condition¶

Treat the system Gibbs energy as a function of $T,P$, and composition:

Its total differential is

The partial derivative $\left(\partial G/\partial n_i\right)_{T,P,n_{j\ne i}}$ was introduced in § 6.1.2 as the chemical potential $\mu_i$, so

At constant $T$ and $P$,

Using Eq. (6),

This motivates the definition of the Gibbs free energy of reaction:

and therefore

Spontaneity and equilibrium (constant $T,P$)¶

• $\Delta_r G < 0$: reaction proceeds forward (toward products).

• $\Delta_r G = 0$: equilibrium.

• $\Delta_r G > 0$: reaction proceeds backward (toward reactants).

At equilibrium,

This is the multi-species analog of the phase-equilibrium condition $\mu_\alpha = \mu_\beta$ from § 6.1.2. Instead of transferring matter between two phases, we now transfer it among several chemical species; equilibrium is again the stationary point of $G$ at fixed $T,P$.

7.1.3 Reaction quotient $Q_p$ and the equilibrium constant $K$¶

To connect $\Delta_r G$ to measurable composition variables, we need an expression for the chemical potentials.

Ideal-gas chemical potential¶

Section 5.3 showed that for one mole of an ideal gas at constant $T$, integrating $dG = V\,dP$ from $P^\circ$ to $P$ gives $G(T,P) - G^\circ(T) = RT\ln(P/P^\circ)$. For a single species in an ideal-gas mixture, the same argument with $P_i$ in place of $P$ yields

where $P^{\circ}$ is the standard-state pressure (1 bar, per the notation conventions) and $\mu_i^{\circ}(T)$ is the standard chemical potential.

Derivation of $\Delta_r G = \Delta_r G^{\circ} + RT\ln Q_p$¶

Substitute Eq. (15) into Eq. (12):

Define the standard Gibbs energy of reaction

and combine the logarithms to define the reaction quotient $Q_p$:

Therefore,

Equilibrium: $Q_p = K$¶

At equilibrium, $\Delta_r G=0$, so

with

For an ideal-gas reaction this $K$ is the pressure-based equilibrium constant (and equals the limit of $Q_p$ at equilibrium).

Using $Q_p$ vs $K$ to predict direction¶

Since $\Delta_r G = RT\ln(Q_p/K)$:

• If $Q_p<K$, then $\Delta_r G<0$ and the reaction proceeds toward products.

• If $Q_p>K$, then $\Delta_r G>0$ and the reaction proceeds toward reactants.

• If $Q_p=K$, the system is at equilibrium.

7.1.4 Worked example: gas-phase dimerization of $\mathrm{NO_2}$¶

Consider the equilibrium

($\mathrm{NO_2}$ is brown; $\mathrm{N_2O_4}$ is colorless.)

Thermodynamics and $K$ at 298.15 K¶

At $298.15\,\mathrm{K}$ (values from § 5.3-style tabulated data):

giving

Equilibrium composition at a specified total pressure¶

Suppose we start with 2 mol of $\mathrm{NO_2}$ and 0 mol of $\mathrm{N_2O_4}$. Build an ICE table in terms of $\xi$:

• Initial: $n_{\mathrm{NO_2},0}=2$, $n_{\mathrm{N_2O_4},0}=0$

• Change: $\Delta n_{\mathrm{NO_2}}=-2\xi$, $\Delta n_{\mathrm{N_2O_4}}=+\xi$

• Equilibrium: $n_{\mathrm{NO_2}}=2-2\xi$, $n_{\mathrm{N_2O_4}}=\xi$, $n_{\text{tot}}=2-\xi$

Mole fractions (ideal-gas mixture):

At total pressure $P$, partial pressures are $P_i=y_i P$, and

For $P=P^{\circ}$ and $K=6.74$, numerical root-finding gives

so

$G(\xi)$ has a minimum at $\xi_{\mathrm{eq}}$¶

The equilibrium condition $\Delta_r G = 0$ is the statement that $G(\xi)$ has a stationary point at equilibrium. For this system, with mole fractions $y_i(\xi)$ and $P=P^\circ$, the Gibbs free energy relative to pure reactants is

The first term is the “bookkeeping” shift in standard chemical potentials; the second is the ideal-mixing entropy that prevents the reaction from running to completion. Plotting:

import numpy as np
import matplotlib.pyplot as plt

R = 8.314462618          # J mol^-1 K^-1
T = 298.15
dG_std = -4.74e3         # J per mole of reaction
K_eq = np.exp(-dG_std / (R*T))

def G_of_xi(xi):
    n_NO2 = 2 - 2*xi
    n_N2O4 = xi
    n_tot = 2 - xi
    mix = 0.0
    if n_NO2 > 0:
        mix += n_NO2 * np.log(n_NO2 / n_tot)
    if n_N2O4 > 0:
        mix += n_N2O4 * np.log(n_N2O4 / n_tot)
    return xi*dG_std + R*T*mix   # J (per initial 2 mol NO2)

xis = np.linspace(1e-4, 1 - 1e-4, 400)
G_vals = np.array([G_of_xi(x) for x in xis]) / 1e3   # kJ

# Analytical xi_eq from K = xi(2-xi)/(2-2xi)^2
from scipy.optimize import brentq
xi_eq = brentq(lambda x: x*(2-x)/(2-2*x)**2 - K_eq, 0.01, 0.99)

fig, ax = plt.subplots(figsize=(6.0, 4.0))
ax.plot(xis, G_vals, lw=2)
ax.axvline(xi_eq, color="k", ls="--", lw=1)
ax.plot([xi_eq], [G_of_xi(xi_eq)/1e3], "o", color="tab:red", zorder=5)
ax.annotate(fr"$\xi_{{\mathrm{{eq}}}} \approx {xi_eq:.2f}$",
            xy=(xi_eq, G_of_xi(xi_eq)/1e3),
            xytext=(xi_eq - 0.35, G_of_xi(xi_eq)/1e3 + 1.0),
            arrowprops=dict(arrowstyle="->", lw=1))
ax.set_xlabel(r"Extent of reaction $\xi$ (mol)")
ax.set_ylabel(r"$G(\xi) - G(0)$ (kJ)")
ax.set_title(r"$2\,\mathrm{NO_2}(g) \rightleftharpoons \mathrm{N_2O_4}(g)$, " 
             r"$T=298.15$ K, $P=P^\circ$")
fig.subplots_adjust(left=0.14, right=0.96, top=0.90, bottom=0.14)
plt.show()

The minimum of $G(\xi)$ lies precisely at the $\xi_{eq}$ computed from $K$. Note that the curve is asymmetric: the left branch is driven by $\Delta_r G^\circ < 0$ (pushing $\xi$ upward from pure reactants), while the right branch is pulled back up by the $RT\sum n_i \ln y_i$ mixing term as the mixture becomes dominated by $\mathrm{N_2O_4}$. The “tug-of-war” between these two terms is the reason chemical equilibrium is generically an interior minimum rather than either pure reactants or pure products.

Interactive visualization¶

7.1.5 Temperature dependence: the van’t Hoff equation¶

The temperature dependence of $K$ follows from the Gibbs–Helmholtz relation:

Using $\Delta_r G^{\circ}=-RT\ln K$ gives the van’t Hoff equation:

An equivalent integrated form between $T_1$ and $T_2$ (assuming $\Delta_r H^{\circ}$ is approximately constant over the range) is

Example trend for $2\,\mathrm{NO_2}\rightleftharpoons\mathrm{N_2O_4}$¶

At $P=P^{\circ}$:

Because $\Delta_r H^{\circ}<0$ for dimerization, increasing temperature drives $K$ down and shifts equilibrium toward $\mathrm{NO_2}$ (less $\mathrm{N_2O_4}$). This is the quantitative form of Le Châtelier’s principle for an exothermic reaction.

7.1.6 Computing $K$ from tabulated thermochemistry¶

In § 7.1.3 we derived the key link between equilibrium and thermochemistry:

So if you can compute $\Delta_r G^{\circ}$ at a temperature of interest, you can immediately compute the equilibrium constant. This subsection is a practical workflow for getting $\Delta_r G^{\circ}$ from tabulated thermochemical data.

Step 0: Write the reaction (with phases) and choose a standard state¶

1. Balance the reaction and include phases, e.g. $\mathrm{NH_3(g)}$ vs $\mathrm{NH_3(\ell)}$.

2. Use a consistent standard state across all species (most modern tables use $P^\circ=1\ \mathrm{bar}$ at $T=298.15\ \mathrm{K}$; always check the table header).

Step 1: Pull species data at the same temperature¶

For each species $i$ in the balanced equation, collect either:

• Option A (most common at 298.15 K): $\Delta_f H_i^\circ$ and $S_i^\circ$ (and optionally $\Delta_f G_i^\circ$).

• Option B (temperature-dependent, e.g. $T\neq 298.15\,\mathrm{K}$): $H_i^\circ(T)$ and $S_i^\circ(T)$ (or directly $G_i^\circ(T)$ if provided).

Where to get the data:

• NIST Chemistry WebBook: fast way to grab $\Delta_f H^\circ(298.15\ \mathrm{K})$ and $S^\circ(298.15\ \mathrm{K})$ for many species (check the phase!).

• NIST–JANAF Thermochemical Tables: best when you need $H^\circ(T)$ and $S^\circ(T)$ over a range of temperatures (not just 298.15 K).

• ATcT (Active Thermochemical Tables): best when you care about high-accuracy formation thermochemistry (often includes uncertainties). In practice, many workflows use ATcT for $\Delta_f H^\circ$ and JANAF for $S^\circ(T)$.

Step 2: Convert species data into reaction values¶

Once you have consistent species data, compute reaction properties using stoichiometric coefficients $\nu_i$ (positive for products, negative for reactants):

If your data are enthalpies of formation, this becomes the familiar “products minus reactants” form:

Step 3: Compute $\Delta_r G^\circ$¶

If you have $\Delta_r H^\circ$ and $\Delta_r S^\circ$ at the same temperature, then

Step 4: Compute $K$¶

Finally,

For an ideal-gas reaction, this $K$ corresponds to a pressure-based equilibrium constant with the dimensionless ratio $(P_i/P^\circ)$ inside $Q_p$.

Worked example: ammonia formation at 298.15 K¶

For the Haber–Bosch reaction (as written in § 5.3),

Using the reaction values quoted in § 5.3 at 298.15 K,

gives

Then

Interpretation. $K\gg 1$ at room temperature, so the equilibrium strongly favors $\mathrm{NH_3}$ at the standard state. (Section 5.3 explains why temperature and pressure “knobs” still matter for industrial operation, where the 500 °C kinetic sweet spot drives $K$ down dramatically.)

Common pitfalls (and quick fixes)¶

• Wrong phase: $\mathrm{H_2O(\ell)}$ and $\mathrm{H_2O(g)}$ (or $\mathrm{NH_3(\ell)}$ vs $\mathrm{NH_3(g)}$) have very different $S^\circ$ and $\Delta_f H^\circ$.

• Mixed standard states: make sure all species use the same $P^\circ$ convention and the same temperature.

• Unit mismatches: always reconcile kJ vs J and “per mole of reaction” vs “per mole of species.”

• Sign mistakes: products minus reactants (or use $\nu_i$ with the sign convention consistently).

Concept Checks¶

1. Why is $\Delta_r G=0$ the correct equilibrium condition at constant $T,P$? What would change if the reaction instead ran at constant $T,V$? (The constant-$T,V$ case is developed in § 7.2.)

2. Why must $Q_p$ and $K$ be dimensionless? What role does $P^\circ$ play?

3. How does comparing $Q_p$ to $K$ predict the direction of spontaneous change?

4. Inspecting Eq. (32): which term in $G(\xi)$ would vanish if the mixture were not ideal (i.e., if there were no entropy of mixing)? Why does this entropy term prevent the reaction from running to completion even when $\Delta_r G^\circ \ll 0$?

5. What changes in the derivation of $\Delta_r G = \Delta_r G^\circ + RT\ln Q_p$ if the mixture is non-ideal and activities replace partial-pressure ratios?

Key Takeaways¶

• Extent of reaction $\xi$ provides a compact way to track composition changes via stoichiometry: $dn_i = \nu_i\,d\xi$.

• At constant $T,P$, equilibrium corresponds to minimizing $G$, giving $\Delta_r G = \sum_i\nu_i\mu_i = 0$ — the multi-species analog of the phase-equilibrium condition of § 6.1.2.

• For ideal gases, $\Delta_r G = \Delta_r G^\circ + RT\ln Q_p$ and $\Delta_r G^\circ = -RT\ln K$.

• The sign of $\Delta_r G$ (equivalently the comparison $Q_p$ vs $K$) determines spontaneous direction.

• The temperature dependence of $K$ is governed by van’t Hoff: $\partial\ln K/\partial T = \Delta_r H^\circ/(RT^2)$.

• Thermochemical tables (NIST WebBook, JANAF, ATcT) provide the $\Delta_f H^\circ$ and $S^\circ$ values needed to compute $\Delta_r G^\circ$ and hence $K$.