9.1 Chemical Equilibrium: Equilibrium Constants, Le Chatelier's Principle, and the van 't Hoff Equation

Chemical Equilibrium: Equilibrium Constants, Le Chatelier's Principle, and the van 't Hoff Equation

Chemical equilibrium is the macroscopic state in which the forward and reverse rates of a chemical process are exactly balanced, resulting in time-invariant concentrations of reactants and products. On the PRC Chemist Licensure Examination, chemical equilibrium is tested from both a thermodynamic standpoint—linking free energy, chemical potentials, and activities—and a practical problem-solving perspective involving gas-phase equilibria, temperature shifts, and heterogeneous systems.

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1. Thermodynamic Formulation of Chemical Equilibrium

Consider a general closed-system chemical reaction operating at constant temperature ($T$) and pressure ($P$):

$\sum_{i} \nu_i A_i = 0$

where $\nu_i$ denotes the stoichiometric coefficient of species $A_i$ (defined as positive for products and negative for reactants).

The infinitesimal change in Gibbs free energy ($dG$) at constant $T$ and $P$ as the reaction advances by an extent of reaction $d\xi$ is:

$dG = \sum_{i} \mu_i dn_i = \left( \sum_{i} \nu_i \mu_i \right) d\xi$

The differential Gibbs free energy of reaction ($\Delta_r G$) is defined as the partial derivative of $G$ with respect to the extent of reaction $\xi$:

$\Delta_r G = \left( \frac{\partial G}{\partial \xi} \right)_{T, P} = \sum_{i} \nu_i \mu_i$

Chemical Potential and Activity Expansion

For each reacting component $i$, its chemical potential $\mu_i$ is expressed in terms of its standard chemical potential $\mu_i^\circ$ and its thermodynamic activity $a_i$:

$\mu_i = \mu_i^\circ + RT \ln a_i$

Substituting this activity relationship into the expression for $\Delta_r G$ yields:

$\Delta_r G = \sum_{i} \nu_i \left( \mu_i^\circ + RT \ln a_i \right) = \sum_{i} \nu_i \mu_i^\circ + RT \sum_{i} \ln (a_i^{\nu_i})$

Recognizing that $\Delta_r G^\circ = \sum \nu_i \mu_i^\circ$ is the standard reaction Gibbs free energy, and defining the reaction quotient $Q$ as the activity product:

$Q = \prod_{i} a_i^{\nu_i}$

we obtain the central master equation of chemical thermodynamics:

$\Delta_r G = \Delta_r G^\circ + RT \ln Q$

The Equilibrium Criterion

At dynamic chemical equilibrium, the system reaches an extremum (minimum) in Gibbs free energy at constant $T$ and $P$, meaning the driving force vanishes:

$\Delta_r G = 0$

Setting $\Delta_r G = 0$ in the master equation, the reaction quotient $Q$ achieves its unique equilibrium value, termed the thermodynamic equilibrium constant $K$:

$0 = \Delta_r G^\circ + RT \ln K \implies \Delta_r G^\circ = -RT \ln K$

Equivalently, solving for $K$:

$K = \exp\left( -\frac{\Delta_r G^\circ}{RT} \right)$

Exam Distinction: Never confuse $\Delta_r G$ with $\Delta_r G^\circ$. $\Delta_r G^\circ$ is a fixed thermodynamic property at a given temperature that reflects the difference between pure standard-state products and pure standard-state reactants ($1\text{ bar}$, ideal behavior). In contrast, $\Delta_r G$ is the instantaneous slope $(\partial G / \partial \xi)_{T,P}$ at the specific actual composition of the mixture. At equilibrium, $\Delta_r G = 0$, but $\Delta_r G^\circ$ is generally not zero.

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2. Equilibrium Constant Representations: $K_p$, $K_c$, and $K_x$

In homogeneous gas reactions, activities can be expressed using partial pressures, molar concentrations, or mole fractions.

Thermodynamic Activity of Gases

For an ideal gas component $i$, activity is defined relative to the standard pressure $P^\circ = 1\text{ bar}$ (or $1\text{ atm}$ in classical systems):

$a_i = \frac{P_i}{P^\circ}$

Because $P^\circ = 1\text{ bar}$, the thermodynamic equilibrium constant $K_p^\circ$ is strictly dimensionless:

$K_p^\circ = \prod_{i} \left( \frac{P_i}{P^\circ} \right)^{\nu_i}$

In routine practice, chemists write $K_p = \prod P_i^{\nu_i}$, with the understanding that pressure values are referenced to $1\text{ bar}$.

Relationship Between $K_p$ and $K_c$

From the ideal gas equation of state, $P_i V = n_i RT$, the partial pressure is related to molar concentration $c_i = n_i / V$ by:

$P_i = c_i RT$

Substituting this into the definition of $K_p$:

$K_p = \prod_{i} (c_i RT)^{\nu_i} = \left( \prod_{i} c_i^{\nu_i} \right) (RT)^{\sum \nu_i}$

Defining $\Delta n_g = \sum \nu_{i, gas} = \sum \nu_{products, gas} - \sum \nu_{reactants, gas}$, we derive:

$K_p = K_c (RT)^{\Delta n_g}$

• Units check: When pressures are in $\text{bar}$ and concentrations in $\text{mol/L}$, use $R = 0.083145\text{ L}\cdot\text{bar}/(\text{mol}\cdot\text{K})$. When pressures are in $\text{atm}$, use $R = 0.082057\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$.
• Special Case: If $\Delta n_g = 0$, then $(RT)^0 = 1$, and $K_p = K_c$.

Relationship to Mole Fractions ($K_x$)

By Dalton's law of partial pressures, $P_i = x_i P_{tot}$, where $x_i$ is the mole fraction and $P_{tot}$ is the total system pressure:

$K_p = \prod_{i} (x_i P_{tot})^{\nu_i} = \left( \prod_{i} x_i^{\nu_i} \right) P_{tot}^{\Delta n_g} = K_x P_{tot}^{\Delta n_g}$

Rearranging for the mole fraction equilibrium constant $K_x$:

$K_x = K_p P_{tot}^{-\Delta n_g}$

Unlike $K_p$, which depends only on temperature for ideal gases, $K_x$ explicitly depends on the total system pressure whenever $\Delta n_g \neq 0$.

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3. Temperature Dependence: The van 't Hoff Equation

Because standard reaction Gibbs free energy is related to enthalpy and entropy by $\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$, substituting $-RT \ln K = \Delta_r G^\circ$ gives:

$\ln K = -\frac{\Delta_r H^\circ}{RT} + \frac{\Delta_r S^\circ}{R}$

Differentiating with respect to temperature at constant pressure, and utilizing the Gibbs-Helmholtz relation $\left[\frac{\partial(\Delta G^\circ / T)}{\partial T}\right]_P = -\frac{\Delta H^\circ}{T^2}$, yields the differential van 't Hoff isobar equation:

$\frac{d\ln K}{dT} = \frac{\Delta_r H^\circ}{RT^2}$

Transforming to the reciprocal temperature coordinate $1/T$, since $d(1/T) = -\frac{1}{T^2} dT$:

$\frac{d\ln K}{d(1/T)} = -\frac{\Delta_r H^\circ}{R}$

Integrated Two-Point van 't Hoff Equation

Assuming that the standard enthalpy of reaction $\Delta_r H^\circ$ is approximately constant over the temperature interval from $T_1$ to $T_2$:

$\int_{K_1}^{K_2} d\ln K = \frac{\Delta_r H^\circ}{R} \int_{T_1}^{T_2} \frac{dT}{T^2}$

$\ln\left( \frac{K_2}{K_1} \right) = -\frac{\Delta_r H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{\Delta_r H^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$

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4. Le Chatelier's Principle: System Perturbations

Le Chatelier's principle states that if a system at dynamic equilibrium is subjected to an external disturbance (change in concentration, pressure, volume, or temperature), the system adjusts in a direction that partially offsets the imposed change.

Summary of Perturbations and Thermodynamic Responses

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5. Heterogeneous Equilibria

When reactions involve more than one phase (such as solids and gases), the activities of pure condensed phases (liquids and solids) are defined as unity ($a = 1$) because their molar concentrations and vapor pressures are constant at a given temperature.

• Thermal Decomposition of Calcium Carbonate: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ Because $a(CaCO_3, s) = 1$ and $a(CaO, s) = 1$: $K_p = P_{CO_2}$ Equilibrium is established at a given temperature if and only if the partial pressure of carbon dioxide equals the decomposition pressure $K_p(T)$.

• Ammonium Hydrosulfide Decomposition: $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$ $K_p = P_{NH_3} P_{H_2S}$ If solid $NH_4HS$ is placed into an evacuated container, stoichiometry requires $P_{NH_3} = P_{H_2S} = \frac{1}{2} P_{tot}$. Thus: $K_p = \left( \frac{P_{tot}}{2} \right)^2 = \frac{P_{tot}^2}{4} \implies P_{tot} = 2\sqrt{K_p}$

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6. Worked Examination Calculation

Problem: Consider the industrial synthesis of methanol from synthesis gas:

$CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$

At $298.15\text{ K}$, the standard thermodynamic values are:

• $\Delta_r H^\circ = -90.70\text{ kJ/mol} = -90700\text{ J/mol}$
• $\Delta_r S^\circ = -221.0\text{ J}/(\text{mol}\cdot\text{K})$

Task:

1. Calculate $\Delta_r G^\circ$ and $K_p$ at $298.15\text{ K}$.
2. Calculate $K_c$ at $298.15\text{ K}$.
3. Calculate $K_p$ at $500.0\text{ K}$ assuming $\Delta_r H^\circ$ is constant.

Step 1: Standard Gibbs Free Energy and $K_p$ at $298.15\text{ K}$

$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ = -90700\text{ J/mol} - (298.15\text{ K})(-221.0\text{ J}/(\text{mol}\cdot\text{K}))$ $\Delta_r G^\circ = -90700 + 65891 = -24809\text{ J/mol} = -24.81\text{ kJ/mol}$

Using $\Delta_r G^\circ = -RT \ln K_p$: $\ln K_p = -\frac{\Delta_r G^\circ}{RT} = -\frac{-24809\text{ J/mol}}{(8.3145\text{ J}/(\text{mol}\cdot\text{K}))(298.15\text{ K})} = \frac{24809}{2479.0} = 10.008$ $K_p = e^{10.008} = 2.22 \times 10^4\text{ bar}^{-2}$

Step 2: Concentration Equilibrium Constant $K_c$

$\Delta n_g = 1 - (1 + 2) = -2$ $K_p = K_c (RT)^{\Delta n_g} \implies K_c = K_p (RT)^{-\Delta n_g} = K_p (RT)^2$ Using $R = 0.083145\text{ L}\cdot\text{bar}/(\text{mol}\cdot\text{K})$: $RT = 0.083145 \times 298.15 = 24.790\text{ L}\cdot\text{bar/mol}$ $K_c = (2.22 \times 10^4) \times (24.790)^2 = (2.22 \times 10^4) \times 614.54 = 1.36 \times 10^7\text{ M}^{-2}$

Step 3: High-Temperature Equilibrium Constant $K_p$ at $500.0\text{ K}$

Using the integrated van 't Hoff equation: $\ln\left( \frac{K_{500}}{K_{298}} \right) = \frac{\Delta_r H^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) = \frac{-90700}{8.3145} \left( \frac{500 - 298.15}{298.15 \times 500} \right)$ $\ln\left( \frac{K_{500}}{K_{298}} \right) = -10908.7 \times \left( \frac{201.85}{149075} \right) = -10908.7 \times 0.0013540 = -14.77$ $\frac{K_{500}}{K_{298}} = e^{-14.77} = 3.85 \times 10^{-7}$ $K_{500} = (2.22 \times 10^4) \times (3.85 \times 10^{-7}) = 8.55 \times 10^{-3}\text{ bar}^{-2}$

Chemical Insight: Because methanol synthesis is strongly exothermic ($\Delta_r H^\circ < 0$), raising the temperature to $500\text{ K}$ causes $K_p$ to collapse by over six orders of magnitude (from $22200$ down to $0.00855$). This demonstrates why industrial methanol reactors must operate at high pressure ($50\text{--}100\text{ bar}$) to force high equilibrium conversion despite the elevated temperatures required for catalyst kinetic activity.