7.4 Chemical Equilibrium, Heats of Reaction, and Mixing

Chemical Reaction Equilibrium

Chemical reaction equilibrium calculations determine the maximum possible conversion (thermodynamic limit) of a chemical reaction at given operating conditions. The fundamental criterion for chemical equilibrium is that the total Gibbs free energy of the system is minimized at constant temperature and pressure. For a general reaction $\sum \nu_i A_i = 0$ (where stoichiometric coefficients $\nu_i$ are positive for products and negative for reactants), the standard Gibbs free energy change of reaction ($\Delta G_{rxn}^\circ$) is calculated from the standard Gibbs free energies of formation ($\Delta G_{f,i}^\circ$) found in reference tables:

$$\Delta G_{rxn}^\circ = \sum \nu_i \Delta G_{f,i}^\circ$$

The equilibrium constant ($K_{eq}$) is directly related to the standard Gibbs free energy change by:

$$\Delta G_{rxn}^\circ = -RT \ln K_{eq} \implies K_{eq} = \exp\left(-\frac{\Delta G_{rxn}^\circ}{RT}\right)$$

where $R$ is the universal gas constant and $T$ is the absolute temperature. The equilibrium constant is defined in terms of the activities ($a_i$) of the reacting species:

$$K_{eq} = \prod (a_i)^{\nu_i}$$

For an ideal gas mixture, the activity of species $i$ is $a_i = y_i P / P^\circ$, where $P^\circ$ is the standard-state pressure ($1 \text{ bar}$ or $100 \text{ kPa}$). This gives:

$$K_{eq} = \left(\frac{P}{P^\circ}\right)^{\sum \nu_i} \prod (y_i)^{\nu_i} = \left(\frac{P}{P^\circ}\right)^{\Delta \nu} K_y$$

where $\Delta \nu = \sum \nu_i$ is the change in the total number of moles of gas. Note that if $\Delta \nu > 0$, increasing the system pressure $P$ will decrease the equilibrium mole fraction constant $K_y$, thereby decreasing the equilibrium conversion (Le Chatelier's Principle).

Temperature Effects: The van 't Hoff Equation

The standard Gibbs free energy change, and therefore $K_{eq}$, is a function of temperature. The effect of temperature on the equilibrium constant is quantitatively described by the van 't Hoff equation:

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

where $\Delta H_{rxn}^\circ$ is the standard enthalpy of reaction. Assuming $\Delta H_{rxn}^\circ$ is approximately constant over a moderate temperature range, integrating the van 't Hoff equation from $T_1$ to $T_2$ yields:

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

This relationship reveals that:

• For endothermic reactions ($\Delta H_{rxn}^\circ > 0$), $K_{eq}$ increases as temperature increases, shifting the equilibrium toward the products.
• For exothermic reactions ($\Delta H_{rxn}^\circ < 0$), $K_{eq}$ decreases as temperature increases, shifting the equilibrium toward the reactants.

Heats of Reaction and Temperature Dependence

The standard heat of reaction ($\Delta H_{rxn}^\circ$) is calculated from standard heats of formation ($\Delta H_{f,i}^\circ$):

$$\Delta H_{rxn}^\circ = \sum \nu_i \Delta H_{f,i}^\circ$$

To determine the heat of reaction at a temperature other than the standard state temperature ($T^\circ = 298.15 \text{ K}$), we apply Kirchhoff's Law:

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

where $\Delta C_p = \sum \nu_i C_{p,i}(T)$ is the change in heat capacity upon reaction. If the heat capacities are functions of temperature (typically expressed as polynomials $C_{p,i} = a_i + b_i T + c_i T^2 + d_i T^3$), $\Delta C_p$ must be integrated accordingly.

Thermodynamics of Mixing

When pure substances are mixed to form a solution, the thermodynamic properties change. For an ideal solution, the molecules of different species are similar in size and chemical nature, resulting in zero heat of mixing ($\Delta H_{mix} = 0$) and zero volume change on mixing ($\Delta V_{mix} = 0$). The entropy and Gibbs free energy of mixing for an ideal solution are:

$$\Delta S_{mix}^{\text{ideal}} = -R \sum x_i \ln x_i > 0$$ $$\Delta G_{mix}^{\text{ideal}} = RT \sum x_i \ln x_i < 0$$

For non-ideal solutions, we define excess properties ($M^E$) as the difference between the actual solution property ($M$) and that of an ideal solution ($M^{\text{ideal}}$) at the same temperature, pressure, and composition. The heat of mixing is equal to the excess enthalpy of the solution:

$$\Delta H_{mix} = H^E = H - \sum x_i H_i$$

Partial Molar Properties

In a mixture, the contribution of a single mole of component $i$ to the total property $M$ (such as volume, enthalpy, or Gibbs free energy) is described by its partial molar property ($\bar{M}_i$):

$$\bar{M}_i = \left(\frac{\partial (n M)}{\partial n_i}\right)_{T, P, n_{j \ne i}}$$

The chemical potential $\mu_i$ is simply the partial molar Gibbs free energy ($\bar{G}_i$).

The total property of a solution is calculated from the partial molar properties via the summability relation:

$$nM = \sum n_i \bar{M}_i \implies M = \sum x_i \bar{M}_i$$

The changes in partial molar properties are constrained by the Gibbs-Duhem equation. At constant temperature and pressure, the Gibbs-Duhem relation is:

$$\sum x_i d\bar{M}_i = 0$$

For a binary mixture, this simplifies to:

$$x_1 d\bar{M}_1 + x_2 d\bar{M}_2 = 0 \implies x_1 \frac{d\bar{M}_1}{dx_1} + x_2 \frac{d\bar{M}_2}{dx_1} = 0$$

This relationship shows that the partial molar properties of components in a mixture cannot change independently; if one increases, the other must decrease.

Worked Example: van 't Hoff Temperature Correction

For the water-gas shift reaction ($CO + H_2O \rightleftharpoons CO_2 + H_2$), the equilibrium constant is $K_{eq} = 22.4$ at $600\text{ K}$. The standard enthalpy of reaction is constant over the temperature range at $\Delta H_{rxn}^\circ = -41.2\text{ kJ/mol} = -41200\text{ J/mol}$. Calculate the equilibrium constant at $800\text{ K}$. The universal gas constant is $R = 8.314\text{ J/(mol}\cdot\text{K)}$.

Step 1: Write down the integrated van 't Hoff equation.

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

Step 2: Substitute the known values.

$$T_1 = 600 \text{ K}, \quad K_{eq}(T_1) = 22.4$$ $$T_2 = 800 \text{ K}$$ $$\ln\left(\frac{K_{eq}(800\text{ K})}{22.4}\right) = -\frac{-41200 \text{ J/mol}}{8.314 \text{ J/(mol}\cdot\text{K)}} \left(\frac{1}{800 \text{ K}} - \frac{1}{600 \text{ K}}\right)$$

Step 3: Solve the arithmetic.

$$\frac{1}{800} - \frac{1}{600} = 0.00125 - 0.001667 = -0.0004167 \text{ K}^{-1}$$ $$\ln\left(\frac{K_{eq}(800\text{ K})}{22.4}\right) = 4955.5 \times (-0.0004167) \approx -2.065$$

Step 4: Exponentiate to find $K_{eq}(800\text{ K})$.

$$\frac{K_{eq}(800\text{ K})}{22.4} = \exp(-2.065) \approx 0.1268$$ $$K_{eq}(800\text{ K}) = 22.4 \times 0.1268 \approx 2.84$$

As expected for an exothermic reaction, raising the temperature from $600\text{ K}$ to $800\text{ K}$ significantly reduces the equilibrium constant from $22.4$ to $2.84$, meaning that the equilibrium composition shifts toward the reactants, reducing the maximum conversion of carbon monoxide.