7.3 Chemical Reaction Equilibria and Equilibrium Constant Calculation

7.3 Chemical Reaction Equilibria and Equilibrium Constant Calculation

Thermodynamic reaction equilibrium defines the absolute upper limit of conversion, selectivity, and product yield achievable in chemical reactors. Regardless of catalyst activity, residence time, or reactor geometry, no kinetic mechanism can drive a reaction past its thermodynamic equilibrium boundary. On the NCEES PE Chemical Exam, reaction equilibria questions test your ability to determine standard free energies ($\Delta G_{rxn}^\circ$), evaluate temperature-dependent equilibrium constants ($K_a$) using the van 't Hoff equation, formulate stoichiometric mole tables, and predict equilibrium shifts using Le Chatelier's Principle.

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1. Thermodynamic Criterion of Chemical Reaction Equilibrium

Consider a closed chemical system at constant temperature ($T$) and pressure ($P$) undergoing a reversible chemical reaction represented by the general stoichiometric equation:

$\sum_{i=1}^C \nu_i A_i = 0$

Where:

• $A_i$ represents chemical species $i$.
• $\nu_i$ is the stoichiometric coefficient: positive for products, negative for reactants, and zero for inerts.

From the fundamental property relation, the differential change in total Gibbs free energy of the reacting mixture is:

$dG_{T, P} = \sum_{i=1}^C \mu_i dn_i$

The Extent of Reaction ($\xi$)

The change in moles of each species $i$ during reaction is governed by stoichiometry through the extent of reaction ($\xi$, having units of moles):

$dn_i = \nu_i d\xi \implies n_i = n_{i,0} + \nu_i \xi$

Substituting $dn_i = \nu_i d\xi$ into the Gibbs energy differential:

$dG_{T, P} = \left( \sum_{i=1}^C \nu_i \mu_i \right) d\xi$

At constant $T$ and $P$, a spontaneous reaction proceeds in the direction that decreases Gibbs energy ($dG < 0$). At equilibrium, the system reaches the minimum of the Gibbs energy curve ($dG/d\xi = 0$):

$\left( \frac{\partial G}{\partial \xi} \right)_{T, P} = \sum_{i=1}^C \nu_i \mu_i \equiv \Delta G_{rxn} = 0$

   G ^
     | \                                  /
     |  \                                /
     |   \                              /
     |    \                            /
     |     \   Spontaneous            /
     |      \  Forward               /
     |       \ (dG/dxi < 0)         /
     |        \                    /
     |         \                  /
     |          '--.__      __.--'
     |                \____/ <--- EQUILIBRIUM: (dG/dxi) = 0
     +--------------------------------------------------------> xi (Extent)
     Reactants (xi = 0)                       Complete Conversion (xi_max)

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2. Standard Gibbs Energy of Reaction and the Equilibrium Constant ($K_a$)

The chemical potential of species $i$ in a mixture is related to its thermodynamic activity ($a_i$):

$\mu_i = \mu_i^\circ(T) + R T \ln a_i$

Where $\mu_i^\circ(T)$ is the chemical potential of pure species $i$ in its standard state at temperature $T$ and standard pressure $P^\circ$ ($1.0\text{ bar}$ or $1.0\text{ atm}$). Substituting this into the equilibrium condition $\sum \nu_i \mu_i = 0$:

$\sum_{i=1}^C \nu_i \left( \mu_i^\circ + R T \ln a_i \right) = 0$

$\sum_{i=1}^C \nu_i \mu_i^\circ + R T \sum_{i=1}^C \ln\left( a_i^{\nu_i} \right) = 0$

$\sum_{i=1}^C \nu_i \mu_i^\circ + R T \ln\left( \prod_{i=1}^C a_i^{\nu_i} \right) = 0$

The Standard Gibbs Free Energy Change ($\Delta G_{rxn}^\circ$)

The summation $\sum \nu_i \mu_i^\circ$ defines the standard Gibbs energy change of reaction at temperature $T$:

$\Delta G_{rxn}^\circ(T) \equiv \sum_{i=1}^C \nu_i \mu_i^\circ(T) = \sum_{i=1}^C \nu_i \Delta G_{f,i}^\circ(T)$

Where $\Delta G_{f,i}^\circ(T)$ is the standard Gibbs energy of formation of species $i$ from its constituent elements in their reference states at temperature $T$.

The Thermodynamic Equilibrium Constant ($K_a$)

Defining the thermodynamic equilibrium constant ($K_a$) as the product of activities at equilibrium:

$K_a \equiv \prod_{i=1}^C a_i^{\nu_i}$

Yields the central thermodynamic identity of chemical equilibria:

$\Delta G_{rxn}^\circ = -R T \ln K_a \iff K_a = \exp\left( -\frac{\Delta G_{rxn}^\circ}{R T} \right)$

[!IMPORTANT] The Cardinal Rule of Chemical Equilibrium: Because the standard state for each chemical species is defined at a fixed standard pressure ($P^\circ = 1.0\text{ bar}$), the standard Gibbs energy $\Delta G_{rxn}^\circ(T)$ depends strictly on temperature. Consequently, the thermodynamic equilibrium constant $K_a$ is a function of temperature alone: $K_a = K_a(T)$. Total system pressure, inert gas addition, feed composition, and catalyst presence have zero effect on $K_a$!

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3. Gas-Phase Reaction Equilibria Formulation

For gas-phase reactions, the activity of species $i$ is defined relative to the standard state of pure ideal gas at $P^\circ = 1.0\text{ bar}$ ($100\text{ kPa}$ or $1.0\text{ atm}$):

$a_i = \frac{f_i}{P^\circ} = \frac{y_i \phi_i P}{P^\circ}$

Substituting this into the activity equilibrium constant expression:

$K_a = \prod_{i=1}^C \left( \frac{y_i \phi_i P}{P^\circ} \right)^{\nu_i} = \left( \frac{P}{P^\circ} \right)^{\sum \nu_i} \left( \prod_{i=1}^C y_i^{\nu_i} \right) \left( \prod_{i=1}^C \phi_i^{\nu_i} \right)$

Defining:

• $\Delta \nu \equiv \sum_{i=1}^C \nu_i$ = net change in moles of gas per stoichiometric reaction unit.
• $K_y \equiv \prod_{i=1}^C y_i^{\nu_i}$ = mole fraction composition factor.
• $K_\phi \equiv \prod_{i=1}^C \phi_i^{\nu_i}$ = mixture fugacity coefficient factor ($K_\phi \approx 1.0$ at low to moderate pressures).

The general gas-phase equilibrium expression is:

$K_a = \left( \frac{P}{P^\circ} \right)^{\Delta \nu} K_y K_\phi$

Ideal Gas Mixture Simplification

At low to moderate operating pressures ($P < 10-15\text{ bar}$), gas fugacity coefficients approach unity ($K_\phi \approx 1.00$). The relationship simplifies to:

$K_a = \left( \frac{P}{P^\circ} \right)^{\Delta \nu} K_y \implies K_y = K_a \left( \frac{P}{P^\circ} \right)^{-\Delta \nu}$

This crucial equation reveals how system pressure governs equilibrium composition ($K_y$) while $K_a$ remains constant!

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4. Temperature Dependence of $K_a$: The van 't Hoff Equation

To determine how the equilibrium constant changes when reactor temperature varies from standard reference temperature ($T_0 = 298.15\text{ K}$) to operating temperature ($T$), we apply the Gibbs-Helmholtz equation:

$\left[ \frac{\partial (G/T)}{\partial T} \right]_P = -\frac{H}{T^2} \implies \frac{d(\Delta G_{rxn}^\circ / T)}{dT} = -\frac{\Delta H_{rxn}^\circ}{T^2}$

Substituting $\Delta G_{rxn}^\circ / T = -R \ln K_a$ yields the differential van 't Hoff equation:

$\frac{d(\ln K_a)}{dT} = \frac{\Delta H_{rxn}^\circ(T)}{R T^2}$

Where $\Delta H_{rxn}^\circ(T)$ is the standard enthalpy of reaction at temperature $T$:

$\Delta H_{rxn}^\circ = \sum_{i=1}^C \nu_i \Delta H_{f,i}^\circ$

Integrated Form (Constant Heat of Reaction Assumption)

Over moderate temperature ranges where $\Delta H_{rxn}^\circ$ is approximately constant (i.e., $\Delta C_p = \sum \nu_i C_{p,i} \approx 0$):

$\int_{K_a(T_1)}^{K_a(T_2)} d\ln K_a = \frac{\Delta H_{rxn}^\circ}{R} \int_{T_1}^{T_2} \frac{dT}{T^2}$

$\ln\left( \frac{K_a(T_2)}{K_a(T_1)} \right) = -\frac{\Delta H_{rxn}^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{\Delta H_{rxn}^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$

Thermal Response Directions

• Endothermic Reactions ($\Delta H_{rxn}^\circ > 0$, absorbs heat): $\frac{d(\ln K_a)}{dT} > 0$ Increasing temperature increases $K_a$. Equilibrium shifts toward the products, increasing conversion.
• Exothermic Reactions ($\Delta H_{rxn}^\circ < 0$, releases heat): $\frac{d(\ln K_a)}{dT} < 0$ Increasing temperature decreases $K_a$. Equilibrium shifts toward the reactants, suppressing conversion.

Rigorous Integration (Temperature-Dependent Heat Capacity)

When $\Delta C_p \ne 0$, $\Delta H_{rxn}^\circ(T)$ varies with temperature according to Kirchhoff's Law:

$\Delta H_{rxn}^\circ(T) = \Delta H_{rxn}^\circ(T_0) + \int_{T_0}^T \Delta C_p dT$

$\ln K_a(T) = \ln K_a(T_0) + \frac{1}{R} \int_{T_0}^T \frac{\Delta H_{rxn}^\circ(T)}{T^2} dT$

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5. Stoichiometric Table and Reaction Extent Formulation

To solve reaction equilibrium problems systematically on the PE exam, always construct a Stoichiometric Mole Table in terms of the extent of reaction ($\xi$) or the fractional conversion ($X_A$) of the limiting reactant.

General Stoichiometric Balance Table

Where:

• $n_{tot,0} = n_{A,0} + n_{B,0} + n_{C,0} + n_I$
• $\Delta \nu = \sum \nu_i = \nu_A + \nu_B + \nu_C$
• Fractional conversion of limiting reactant $A$: $X_A = \frac{n_{A,0} - n_A}{n_{A,0}} = \frac{-\nu_A \xi}{n_{A,0}} \implies \xi = \frac{n_{A,0} X_A}{-\nu_A}$

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6. Le Chatelier's Principle: Engineering Process Levers

Henri Louis Le Chatelier (1884) stated that if a system at chemical equilibrium experiences a perturbation in temperature, pressure, or concentration, the equilibrium shifts in the direction that partially offsets the imposed change.

1. Effect of Temperature

• Governed entirely by the sign of $\Delta H_{rxn}^\circ$ through the van 't Hoff equation.
• Exothermic ($\Delta H_{rxn}^\circ < 0$): Adding heat (raising $T$) shifts equilibrium to the left (reactants). Cooling shifts to the right.
• Endothermic ($\Delta H_{rxn}^\circ > 0$): Adding heat (raising $T$) shifts equilibrium to the right (products).

2. Effect of Total Pressure

• Pressure does not change $K_a$. It shifts equilibrium through the term $(P/P^\circ)^{-\Delta \nu}$ in $K_y = K_a (P/P^\circ)^{-\Delta \nu}$.
• If $\Delta \nu < 0$ (mole-reducing reaction, e.g., $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$, $\Delta \nu = -2$): $K_y = K_a \left( \frac{P}{P^\circ} \right)^{+2}$ Increasing total pressure $P$ increases $K_y$, shifting equilibrium toward the product side (higher conversion). This is why Haber-Bosch ammonia synthesis operates at $150-250\text{ bar}$.
• If $\Delta \nu > 0$ (mole-expanding reaction, e.g., $\text{CH}_4 + \text{H}_2\text{O} \rightleftharpoons \text{CO} + 3\text{H}_2$, $\Delta \nu = +2$): $K_y = K_a \left( \frac{P}{P^\circ} \right)^{-2}$ Increasing total pressure suppresses conversion. Lowering pressure or pulling vacuum promotes product yield.
• If $\Delta \nu = 0$ (equimolar reaction, e.g., $\text{CO} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + \text{H}_2$): $K_y = K_a$ Total pressure has zero effect on equilibrium conversion or mole fractions!

3. Effect of Inerts (Steam, Nitrogen, Argon)

• Inerts Added at Constant Total Pressure ($P = \text{const}$): Adding inerts increases total moles $n_{tot}$, which decreases the partial pressures of reacting components. This dilution acts identically to a reduction in total system pressure:
  • For $\Delta \nu > 0$: Adding inerts increases equilibrium conversion (e.g., steam added to ethylbenzene dehydrogenation to produce styrene).
  • For $\Delta \nu < 0$: Adding inerts decreases equilibrium conversion.
  • For $\Delta \nu = 0$: Adding inerts has no effect on reactant conversion.
• Inerts Added at Constant Volume ($V = \text{const}$): In a rigid vessel, adding inerts raises total pressure but leaves the concentrations ($n_i/V$) and partial pressures ($P_i$) of the reactants unchanged. Equilibrium conversion does not shift.

4. Effect of Catalysts

• Catalysts lower the activation energies ($E_a$) of the forward and reverse reactions by the exact same amount.
• Catalysts increase the rate of approaching equilibrium, but have zero effect on $\Delta G_{rxn}^\circ$, $K_a$, or equilibrium conversion.

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7. Summary Table: Process Perturbations & Reaction Equilibrium Shifts

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8. Comprehensive Worked Numerical Example: High-Temperature Sulfur Dioxide Oxidation

Problem Statement

In a contact process sulfuric acid plant, sulfur dioxide is catalytically oxidized to sulfur trioxide across a vanadium pentoxide ($\text{V}_2\text{O}_5$) catalyst bed:

$\text{SO}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{SO}_3(g)$

Standard Thermodynamic Data at $T_0 = 298.15\text{ K}$ ($P^\circ = 1.00\text{ bar}$):

• $\text{SO}2(g)$: $\Delta H{f,298}^\circ = -296.83\text{ kJ/mol}$, $\Delta G_{f,298}^\circ = -300.19\text{ kJ/mol}$
• $\text{O}2(g)$: $\Delta H{f,298}^\circ = 0.00\text{ kJ/mol}$, $\Delta G_{f,298}^\circ = 0.00\text{ kJ/mol}$
• $\text{SO}3(g)$: $\Delta H{f,298}^\circ = -395.72\text{ kJ/mol}$, $\Delta G_{f,298}^\circ = -371.06\text{ kJ/mol}$

The converter operates at a temperature of $T = 700.0\text{ K}$ and an absolute pressure of $P = 2.00\text{ bar}$. The feed gas enters with the molar ratio:

• $1.00\text{ mol } \text{SO}_2$
• $0.80\text{ mol } \text{O}_2$
• $4.00\text{ mol } \text{N}_2$ (inert carrier gas)

Calculate:

1. The standard enthalpy change ($\Delta H_{rxn,298}^\circ$) and standard Gibbs free energy change ($\Delta G_{rxn,298}^\circ$) at $298.15\text{ K}$.
2. The thermodynamic equilibrium constant $K_a$ at $298.15\text{ K}$.
3. The equilibrium constant $K_a$ at converter operating temperature $T = 700.0\text{ K}$ assuming constant $\Delta H_{rxn}^\circ$.
4. The stoichiometric change in gas moles ($\Delta \nu$) and the expression for $K_y$.
5. Formulate the equilibrium composition in terms of $\text{SO}_2$ conversion ($X$) and determine the equilibrium conversion at $700.0\text{ K}$ and $2.00\text{ bar}$.

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Step 1: Standard Heats and Free Energies of Reaction at $298.15\text{ K}$

$\Delta H_{rxn,298}^\circ = \Delta H_{f}^\circ(\text{SO}_3) - \Delta H_{f}^\circ(\text{SO}_2) - 0.5 \Delta H_{f}^\circ(\text{O}_2)$ $\Delta H_{rxn,298}^\circ = -395.72 - (-296.83) - 0.5(0) = \mathbf{-98.89\text{ kJ/mol}} = -98,890\text{ J/mol}$

$\Delta G_{rxn,298}^\circ = \Delta G_{f}^\circ(\text{SO}_3) - \Delta G_{f}^\circ(\text{SO}_2) - 0.5 \Delta G_{f}^\circ(\text{O}_2)$ $\Delta G_{rxn,298}^\circ = -371.06 - (-300.19) - 0.5(0) = \mathbf{-70.87\text{ kJ/mol}} = -70,870\text{ J/mol}$

(The negative $\Delta H^\circ$ indicates a strongly exothermic reaction; the negative $\Delta G^\circ$ indicates strong thermodynamic spontaneity at ambient temperature).

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Step 2: Equilibrium Constant at $298.15\text{ K}$

$K_a(298.15\text{ K}) = \exp\left( -\frac{\Delta G_{rxn,298}^\circ}{R T_0} \right) = \exp\left( -\frac{-70,870\text{ J/mol}}{(8.31447\text{ J/(mol}\cdot\text{K)})(298.15\text{ K})} \right)$ $K_a(298.15\text{ K}) = \exp\left( \frac{70,870}{2478.96} \right) = \exp(28.5886) = \mathbf{2.605 \times 10^{12}}$

At room temperature, equilibrium lies overwhelmingly on the product side ($K_a > 10^{12}$), but the reaction is kinetically inert without high temperatures to activate the catalyst.

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Step 3: Equilibrium Constant at $T = 700.0\text{ K}$ via van 't Hoff Equation

Apply the integrated van 't Hoff equation: $\ln\left( \frac{K_a(T)}{K_a(T_0)} \right) = -\frac{\Delta H_{rxn}^\circ}{R} \left( \frac{1}{T} - \frac{1}{T_0} \right)$

$\frac{1}{T} - \frac{1}{T_0} = \frac{1}{700.0\text{ K}} - \frac{1}{298.15\text{ K}} = 0.00142857 - 0.00335402 = -0.00192545\text{ K}^{-1}$

$\ln\left( \frac{K_a(700)}{K_a(298)} \right) = -\left( \frac{-98,890}{8.31447} \right) \times (-0.00192545) = (11,893.72) \times (-0.00192545) = \mathbf{-22.9006}$

$K_a(700.0\text{ K}) = K_a(298.15\text{ K}) \times \exp(-22.9006) = (2.605 \times 10^{12}) \times (1.1335 \times 10^{-10}) = \mathbf{295.3}$

Raising the temperature from $298\text{ K}$ to $700\text{ K}$ drops the equilibrium constant by 10 orders of magnitude (from $2.6 \times 10^{12}$ down to $295.3$) due to the exothermic heat of reaction.

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Step 4: Stoichiometry and Equilibrium Expression

Stoichiometric coefficients: $\nu_{\text{SO}_2} = -1, \quad \nu_{\text{O}_2} = -0.5, \quad \nu_{\text{SO}_3} = +1, \quad \nu_{\text{N}_2} = 0$ $\Delta \nu = \sum \nu_i = -1 - 0.5 + 1 = \mathbf{-0.5}$

Gas-phase equilibrium formulation ($P^\circ = 1.00\text{ bar}$): $K_a = \left( \frac{P}{P^\circ} \right)^{\Delta \nu} K_y = \left( \frac{P}{P^\circ} \right)^{-0.5} \frac{y_{\text{SO}_3}}{y_{\text{SO}_2} \cdot y_{\text{O}_2}^{0.5}}$

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Step 5: Stoichiometric Table and Conversion Calculation

Let $X$ be the fractional conversion of $\text{SO}_2$ (extent $\xi = 1.00 X$):

Substitute mole fractions into $K_y$: $K_y = \frac{y_{\text{SO}_3}}{y_{\text{SO}_2} y_{\text{O}_2}^{0.5}} = \frac{\frac{X}{n_{tot}}}{\left(\frac{1.00 - X}{n_{tot}}\right) \left(\frac{0.80 - 0.5X}{n_{tot}}\right)^{0.5}} = \frac{X}{1.00 - X} \sqrt{\frac{n_{tot}}{0.80 - 0.5X}}$

$K_y = \frac{X}{1.00 - X} \sqrt{\frac{5.80 - 0.5X}{0.80 - 0.5X}}$

Relating $K_y$ to $K_a$ at $P = 2.00\text{ bar}$: $K_y = K_a \left( \frac{P}{P^\circ} \right)^{-\Delta \nu} = 295.3 \times (2.00)^{-(-0.5)} = 295.3 \times \sqrt{2.00} = 295.3 \times 1.4142 = \mathbf{417.6}$

Equating expressions: $\frac{X}{1.00 - X} \sqrt{\frac{5.80 - 0.5X}{0.80 - 0.5X}} = 417.6$

Solving for $X$ by trial and error / successive approximation:

• For $X = 0.980$: $\frac{0.980}{0.020} \sqrt{\frac{5.80 - 0.490}{0.80 - 0.490}} = 49.0 \times \sqrt{\frac{5.310}{0.310}} = 49.0 \times \sqrt{17.129} = 49.0 \times 4.1387 = 202.8 \quad (< 417.6)$
• For $X = 0.990$: $\frac{0.990}{0.010} \sqrt{\frac{5.80 - 0.495}{0.80 - 0.495}} = 99.0 \times \sqrt{\frac{5.305}{0.305}} = 99.0 \times \sqrt{17.393} = 99.0 \times 4.1705 = 412.9 \approx 417.6$
• For $X = 0.9905$: $\frac{0.9905}{0.0095} \times 4.172 = 104.26 \times 4.172 = 435.0 \quad (> 417.6)$

Interpolating yields the equilibrium conversion: $X_{eq} = \mathbf{0.990} \quad (\mathbf{99.0\%})$

At $700\text{ K}$ and $2\text{ bar}$, thermodynamic equilibrium permits up to $99.0%$ conversion of sulfur dioxide to sulfur trioxide.

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9. Critical PE Exam Traps & Pitfalls

Trap 1: Believing that System Pressure Changes the Equilibrium Constant ($K_a$)
This is among the most heavily tested conceptual distractors on the NCEES PE Chemical Exam. Pressure NEVER changes $K_a$! $K_a$ is solely a function of temperature ($K_a = \exp[-\Delta G^\circ / (R T)]$). Pressure changes the equilibrium composition ($K_y$) through the $(P/P^\circ)^{-\Delta \nu}$ factor, but $K_a$ remains strictly constant.

Trap 2: Inverting Signs in the van 't Hoff Equation
Notice the negative sign in $\ln(K_2/K_1) = -(\Delta H^\circ / R) (1/T_2 - 1/T_1)$. For an exothermic reaction ($\Delta H^\circ < 0$), $-\Delta H^\circ$ becomes positive. When heating ($T_2 > T_1$), $(1/T_2 - 1/T_1)$ is negative. A positive multiplied by a negative yields a negative, meaning $\ln(K_2/K_1) < 0 \implies K_2 < K_1$. Verify this physical consistency: heating an exothermic reaction must reduce $K$.

Trap 3: Forgetting Inert Moles in the Total Mole Count ($n_{tot}$)
When setting up mole fractions ($y_i = n_i / n_{tot}$), inerts (such as nitrogen in air, or steam) do not react, but they must be included in $n_{tot}$. Omitting inerts from the denominator artificially inflates the reactant partial pressures, producing severe calculation errors.

Trap 4: Expecting a Catalyst to Increase Thermodynamic Equilibrium Conversion
If an exam question asks "Which modification will increase the equilibrium conversion of an endothermic reaction?" and includes "Adding an active catalyst" as an option, do not pick it! Catalysts only increase the rate at which equilibrium is reached. They alter neither $\Delta G_{rxn}^\circ$ nor $K_a$ nor equilibrium conversion.