Chemical Equilibrium, Reaction Quotient, and Gibbs Free Energy

Note that $K$ is a unitless value.

Interpreting the Equilibrium Constant ($K$)

The magnitude of $K$ tells us the "position" of the equilibrium:

• If $K \gg 1$: The equilibrium favors the products (the forward reaction is favored).
• If $K \ll 1$: The equilibrium favors the reactants (the reverse reaction is favored).
• If $K = 1$: Neither direction is significantly favored.

Equilibrium Constants for Gases ($K_c$ and $K_p$)

When dealing with gases, the concentration of a substance is proportional to its partial pressure. Therefore, we can express the equilibrium constant in terms of molarity ($K_c$) or partial pressure ($K_p$):

$$K_{\mathrm{c}} = \frac{[\mathrm{C}]^{c} [\mathrm{D}]^{d}}{[\mathrm{A}]^{a} [\mathrm{B}]^{b}} \quad \text{and} \quad K_{\mathrm{p}} = \frac{P_{\mathrm{C}}^{c} P_{\mathrm{D}}^{d}}{P_{\mathrm{A}}^{a} P_{\mathrm{B}}^{b}}$$

In $K_p$ expressions, partial pressures must be in atmospheres (atm). The two constants are related by the following equation:

$$K_{\mathrm{p}} = K_{\mathrm{c}}(RT)^{\Delta n}$$

where $\Delta n$ is the difference between the moles of gaseous products and gaseous reactants, $R$ is the gas constant ($0.08206 \text{ L}\cdot\text{atm}/\text{mol}\cdot\text{K}$), and $T$ is the temperature in Kelvin.

Important Note: Pure solids and pure liquids are excluded from equilibrium constant expressions because their concentrations do not change significantly during the reaction.

The Reaction Quotient ($Q$) and the Direction of Change

If a reaction mixture is not at equilibrium, we can still use the same mathematical expression used for $K$, but we call it the reaction quotient ($Q$). By comparing $Q$ to $K$, we can predict which direction the reaction will shift to reach equilibrium:

• If $Q < K$: The ratio of products to reactants is less than the equilibrium ratio. The reaction will proceed in the forward direction to produce more products.
• If $Q > K$: The ratio of products to reactants is greater than the equilibrium ratio. The reaction will proceed in the reverse direction to produce more reactants.
• If $Q = K$: The system is already at equilibrium.

Solving for Equilibrium: The ICE Table

To find equilibrium concentrations when starting from initial amounts, chemists use an ICE table (Initial, Change, Equilibrium):

1. I (Initial): List the starting concentrations.
2. C (Change): Use stoichiometry to represent the change in each species as $+x$ or $-x$.
3. E (Equilibrium): Sum the initial and change values to find the equilibrium concentrations.

Gibbs Free Energy ($\Delta G$) and Spontaneity

Thermodynamics allows us to predict whether a process will occur spontaneously (without ongoing outside intervention). This is determined by the Gibbs Free Energy ($\Delta G$).

Standard vs. Nonstandard Conditions

The Standard Gibbs Free Energy ($\Delta G^{\circ}$) refers to the free energy change when all reactants and products are in their standard states (1 M concentration for solutions, 1 atm for gases, and a specified temperature, usually 298 K).

However, most real-world reactions occur under nonstandard conditions. The relationship between the actual free energy change ($\Delta G$) and the standard free energy change ($\Delta G^{\circ}$) is:

$$\Delta G = \Delta G^{\circ} + RT \ln Q$$

where $R$ is the gas constant ($8.314 \text{ J}/\text{mol}\cdot\text{K}$) and $Q$ is the reaction quotient.

The Relationship Between $\Delta G^{\circ}$ and $K$

At equilibrium, the system has no driving force to change, meaning $\Delta G = 0$. By substituting this into the equation above, we derive a fundamental link between thermodynamics and equilibrium:

$$0 = \Delta G^{\circ} + RT \ln K \implies \Delta G^{\circ} = -RT \ln K$$

This relationship allows us to predict the equilibrium constant if we know the standard free energy, or vice versa:

Exam Tip: When calculating $\Delta G$ using $\Delta G = \Delta G^{\circ} + RT \ln Q$, ensure your units for $\Delta G^{\circ}$ and $R$ are consistent (e.g., convert kJ to J or vice versa).

Worked Example: Calculating $\Delta G$ under Nonstandard Conditions

Problem: Calculate $\Delta G$ at 298 K for the reaction: $2 \text{ NO}_{(g)} + \text{O}_{2(g)} \rightarrow 2 \text{ NO}_{2(g)}$ Given $\Delta G^{\circ} = -71.2 \text{ kJ}$, with partial pressures: $P_{\text{NO}} = 0.100 \text{ atm}$, $P_{\text{O}_2} = 0.100 \text{ atm}$, and $P_{\text{NO}_2} = 2.00 \text{ atm}$.

Step 1: Calculate $Q$

$$Q = \frac{(P_{\text{NO}_2})^2}{(P_{\text{NO}})^2 (P_{\text{O}_2})} = \frac{(2.00)^2}{(0.100)^2 (0.100)} = \frac{4}{0.001} = 4000$$

Step 2: Calculate $\Delta G$ Convert $R$ to $\text{kJ}/\text{mol}\cdot\text{K}$: $R = 8.314 \times 10^{-3} \text{ kJ}/\text{mol}\cdot\text{K}$.

$$\begin{aligned} \Delta G &= \Delta G^{\circ} + RT \ln Q \\ \Delta G &= (-71.2 \text{ kJ}) + (8.314 \times 10^{-3} \text{ kJ}/\text{mol}\cdot\text{K})(298 \text{ K}) \ln(4000) \\ \Delta G &= -71.2 \text{ kJ} + 20.5 \text{ kJ} \\ \Delta G &= -50.7 \text{ kJ} \end{aligned}$$

Since $\Delta G$ is negative, the reaction is spontaneous under these specific conditions.