Chemical Equilibrium: Concepts, Constants, and Calculations

Chapter 14: Chemical Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a closed system are equal, resulting in constant concentrations of reactants and products over time. This dynamic state is fundamental to many chemical processes.

• Dynamic Equilibrium: Both forward and reverse reactions continue to occur, but there is no net change in the concentrations of reactants and products.

• Macroscopic Constancy: Observable properties (concentration, pressure) remain constant at equilibrium.

• Reversibility: Equilibrium can be approached from either direction (starting with reactants or products).

• Example: The reaction $2 ext{HI} (g) \rightleftharpoons ext{H}_2 (g) + ext{I}_2 (g)$ reaches equilibrium when the rate of HI decomposition equals the rate of HI formation.

PhET Simulations: Reversible Reactions

Simulations can help visualize how changes in conditions (such as temperature) affect equilibrium. For example, increasing temperature can shift the equilibrium position, changing the relative amounts of reactants and products.

• Le Châtelier's Principle: If a system at equilibrium is disturbed, it will shift to counteract the disturbance.

• Example: In a simulation where A (green) and B (red) interconvert, increasing temperature may favor the endothermic direction, altering the ratio of A to B.

The Equilibrium Constant

The equilibrium constant quantifies the ratio of product and reactant concentrations at equilibrium for a given reaction at a specific temperature.

• Law of Mass Action: For a general reaction $aA + bB \rightleftharpoons cC + dD$, the equilibrium constant expression is:

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

• Units: The units of $K_c$ depend on the reaction stoichiometry, but $K$ itself is often treated as dimensionless in practice.

• Magnitude: A large $K$ indicates product-favored equilibrium; a small $K$ indicates reactant-favored equilibrium.

• Temperature Dependence: $K$ is constant at a given temperature but changes if the temperature changes.

• Example: For $2 ext{HI} (g) \rightleftharpoons ext{H}_2 (g) + ext{I}_2 (g)$, $K_c = \frac{[ ext{H}_2][ ext{I}_2]}{[ ext{HI}]^2}$

Homogeneous and Heterogeneous Equilibria

Equilibria are classified based on the physical states of the reactants and products.

• Homogeneous Equilibrium: All reactants and products are in the same phase (usually gas or aqueous).

• Heterogeneous Equilibrium: Reactants and products are in different phases (e.g., solids, liquids, gases).

• Note: Pure solids and liquids are omitted from the equilibrium constant expression.

• Example: For $\text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g)$, $K = [\text{CO}_2]$

Equilibrium Constants in Terms of Pressure: $K_p$

For gaseous reactions, equilibrium can also be expressed in terms of partial pressures.

• Expression: For $aA (g) + bB (g) \rightleftharpoons cC (g) + dD (g)$:

$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

• Relationship to $K_c$: $K_p$ and $K_c$ are related by the equation:

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

• Where $R$ is the gas constant (0.08314 bar·L·mol-1·K-1), $T$ is temperature in Kelvin, and $\Delta n$ is the change in moles of gas ($\Delta n =$ moles of gaseous products $-$ moles of gaseous reactants).

• Example: For $2 ext{HI} (g) \rightleftharpoons ext{H}_2 (g) + ext{I}_2 (g)$, $\Delta n = 2 - 2 = 0$, so $K_p = K_c$.

Writing Equilibrium Expressions

General guidelines for constructing equilibrium constant expressions:

• Products over reactants; each concentration or pressure is raised to the power of its coefficient in the balanced equation.

• Exclude pure solids and liquids from the expression.

• Use $K_c$ for concentrations (mol/L), $K_p$ for partial pressures (bar or atm).

• Example: For $\text{N}_2 (g) + 3\text{H}_2 (g) \rightleftharpoons 2\text{NH}_3 (g)$:

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ $K_p = \frac{(P_{\text{NH}_3})^2}{P_{\text{N}_2}(P_{\text{H}_2})^3}$

Summary Table: $K_c$, $K_p$, and Their Relationship

Symbol	Expression	Units	When Used
$K_c$	$\frac{[\text{products}]}{[\text{reactants}]}$	varies (often mol/L)	Concentration (aqueous or gas)
$K_p$	$\frac{(P_{\text{products}})}{(P_{\text{reactants}})}$	varies (often bar or atm)	Partial pressures (gases)
$K_p = K_c (RT)^{\Delta n}$

Worked Examples

• Example 1: For $\text{N}_2\text{O}_4 (g) \rightleftharpoons 2\text{NO}_2 (g)$, $K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$

• Example 2: For $\text{Li}_2\text{O} (s) + \text{H}_2\text{O} (g) \rightleftharpoons 2\text{LiOH} (s)$, $K = \frac{1}{P_{\text{H}_2\text{O}}}$ (solids omitted)

Interpreting the Magnitude of $K$

• Very large $K$ ($K \gg 1$): Reaction proceeds nearly to completion; products are favored at equilibrium.

• Very small $K$ ($K \ll 1$): Reaction hardly proceeds; reactants are favored at equilibrium.

• $K \approx 1$: Significant amounts of both reactants and products are present at equilibrium.

Sample Data Table: Initial and Equilibrium Concentrations

The following table illustrates how initial partial pressures and equilibrium values are used to calculate $K_p$ for the reaction $\text{H}_2 (g) + \text{I}_2 (g) \rightleftharpoons 2\text{HI} (g)$ at 640 K.

Initial $P_{\text{H}_2}$ (bar)	Initial $P_{\text{I}_2}$ (bar)	Initial $P_{\text{HI}}$ (bar)	Equilibrium $P_{\text{H}_2}$ (bar)	Equilibrium $P_{\text{I}_2}$ (bar)	Equilibrium $P_{\text{HI}}$ (bar)	$K_p$
2.000	2.000	0.000	0.936	0.936	3.124	67
1.000	1.000	0.000	0.760	0.760	1.480	67
1.000	1.000	1.000	0.295	0.295	2.410	67

Additional info: Table values are representative; actual values may vary depending on the specific experiment.

Concept Check: Predicting Equilibrium Mixtures

Given different values of $K$ and initial concentrations, you can predict the composition of equilibrium mixtures. For example, if $K$ is large, expect more products; if $K$ is small, expect more reactants.

• Example: If $K = 2$ for $A (g) \rightleftharpoons B (g)$ and the system starts with 8 B and 4 A, the equilibrium will shift to favor B, but some A will remain.