Chemical Equilibrium: Principles, Calculations, and Applications

Chemical Equilibrium

Introduction to Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry describing the state in which the concentrations of reactants and products remain constant over time because the forward and reverse reactions occur at equal rates. This balance is dynamic, not static, meaning that reactions continue to occur, but there is no net change in the system.

• Equilibrium State: Achieved when the rate of the forward reaction equals the rate of the reverse reaction.

• Dynamic Nature: Both reactants and products are continuously formed and consumed.

• Example: The dissolution and precipitation of a salt in water until no further net change is observed.

Chemical Equilibria Defined

Dynamic Equilibrium in Reactions

As a chemical system approaches equilibrium, both the forward and reverse reactions are active. At equilibrium, the rates of these reactions are equal, and the concentrations of all species remain constant.

• Constant Concentrations: The amounts of reactants and products do not change once equilibrium is reached.

• Example: For the reaction $N_2O_4(g) \rightleftharpoons 2 NO_2(g)$, the concentrations of $N_2O_4$ and $NO_2$ become constant at equilibrium.

The Equilibrium Constant

Defining the Equilibrium Constant ($K$)

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

• General Form: For a reaction $aA + bB \rightleftharpoons dD + eE$:

Equilibrium Expression: $K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b}$

• For Gases: The equilibrium constant can also be expressed in terms of partial pressures:

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

• Relationship to Rate Constants: At equilibrium, $k_{forward}[N_2O_4] = k_{reverse}[NO_2]^2$

Equilibrium Constants: Concentration or Pressure

Using $K_c$ and $K_p$

Equilibrium constants can be based on concentrations ($K_c$) or partial pressures ($K_p$), especially for reactions involving gases. The ideal gas law ($PV = nRT$) relates these quantities.

• Conversion: $P = MRT$ (where $M$ is molarity)

• For gases: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas.

The Haber Bosch Process

Industrial Application of Equilibrium

The Haber process synthesizes ammonia from nitrogen and hydrogen gases:

$N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)$

• Equilibrium Expression: $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

• Stoichiometry: The exponents in the equilibrium expression correspond to the coefficients in the balanced equation.

Achieving Equilibrium

Pathways to Equilibrium

Regardless of whether a reaction starts with only reactants or only products, the system will reach the same equilibrium state, with the relative amounts determined by stoichiometry.

• Example: $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ achieves the same equilibrium composition from either direction.

Experimental Observations of $K$

Constancy of the Equilibrium Ratio

Experimental data show that the ratio of product to reactant concentrations at equilibrium is constant at a given temperature, regardless of initial concentrations.

Experiment	Initial [NO2]	Initial [N2O4]	Equilibrium [NO2]	Equilibrium [N2O4]	Kc
1	0.0400	0.0000	0.0320	0.0040	0.256
2	0.0000	0.0400	0.0120	0.0340	0.256

Using Equilibrium Constants

Magnitude and Direction

• If $K \gg 1$: Products are favored; equilibrium lies to the right.

• If $K \ll 1$: Reactants are favored; equilibrium lies to the left.

The Direction of the Equation and $K$

Reversing and Scaling Reactions

• Reverse Reaction: The equilibrium constant for the reverse reaction is the reciprocal of the forward reaction's constant.

$K_{reverse} = \frac{1}{K_{forward}}$

• Scaling: If the coefficients in a reaction are multiplied by $n$, raise $K$ to the $n$th power.

Combining Equilibrium Expressions

Multiple Equilibria

• When reactions are added, the equilibrium constant for the overall reaction is the product of the individual constants.

$K_3 = K_1 \times K_2$

Heterogeneous Equilibria

Homogeneous vs. Heterogeneous

• Homogeneous Equilibria: All reactants and products are in the same phase.

• Heterogeneous Equilibria: Reactants and products are in different phases.

• Pure Solids and Liquids: Their concentrations are omitted from the equilibrium expression.

Example: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ $K_c = [CO_2]$

Calculating Equilibrium Constants

Stepwise Calculation

1. Tabulate all known initial and equilibrium concentrations.

2. Calculate the change in concentrations.

3. Use stoichiometry to find changes for all species.

4. Find equilibrium concentrations.

5. Substitute into the equilibrium expression to solve for $K$.

ICE Tables and Calculation Example

Using ICE (Initial, Change, Equilibrium) Tables

• Organize data for equilibrium calculations.

Reagent	[H2], M	[I2], M	[HI], M
Initial	1.000 × 10-3	2.000 × 10-3	0
Change	-9.35 × 10-4	-9.35 × 10-4	+1.87 × 10-3
Equilibrium	6.5 × 10-5	1.065 × 10-3	1.87 × 10-3

Equilibrium Expression: $K_c = \frac{[HI]^2}{[H_2][I_2]}$

Reaction Quotient ($Q$) and Applications of $K$

Determining Reaction Direction

• Reaction Quotient ($Q$): Calculated like $K$, but with current (not necessarily equilibrium) concentrations.

• Comparing $Q$ and $K$:

  • If $Q < K$, the reaction proceeds toward products.

  • If $Q = K$, the system is at equilibrium.

  • If $Q > K$, the reaction proceeds toward reactants.

Calculating Equilibrium Concentrations

Using $K$ to Find Concentrations

• Set up an ICE table with initial concentrations and changes based on stoichiometry.

• Express equilibrium concentrations in terms of $x$ and solve using the equilibrium expression.

• Quadratic equations may be required for nontrivial cases:

$ax^2 + bx + c = 0$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Le Châtelier’s Principle

Predicting Shifts in Equilibrium

If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the system will shift to counteract the disturbance and re-establish equilibrium.

• Concentration: Adding a component shifts equilibrium to consume it; removing shifts to produce it.

• Pressure/Volume: Increasing pressure (decreasing volume) favors the side with fewer gas molecules; decreasing pressure favors more gas molecules.

• Temperature:

  • Endothermic: Heat is a reactant; increasing temperature shifts toward products ($K$ increases).

  • Exothermic: Heat is a product; increasing temperature shifts toward reactants ($K$ decreases).

The Effect of Catalysis

Catalysts and Equilibrium

• Catalysts: Increase the rate of both forward and reverse reactions equally.

• Effect: Equilibrium is reached faster, but the equilibrium position and composition remain unchanged.

• Activation Energy: Lowered by catalysts, allowing equilibrium to be established at lower temperatures.

Heterogeneous Equilibria: Solubility

Solubility-Product Constant ($K_{sp}$)

• Ionic Compounds: Dissociate completely in solution; equilibrium involves the solid and its ions.

• Expression: For $Ba_3(PO_4)_2(s) \rightleftharpoons 3 Ba^{2+}(aq) + 2 PO_4^{3-}(aq)$:

$K_{sp} = [Ba^{2+}]^3 [PO_4^{3-}]^2$

Solubility and $K_{sp}$

Distinction and Units

• $K_{sp}$: Not the same as solubility; it is the equilibrium constant for the dissolution process.

• Solubility: The amount of substance that dissolves to form a saturated solution, typically in g/L or mol/L (M).

Converting between Solubility and $K_{sp}$

Calculation Pathways

• Use the empirical formula to relate molar solubility to ion concentrations and $K_{sp}$.

• Given $K_{sp}$, calculate solubility; given solubility, calculate $K_{sp}$.

Example: For $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2 F^-(aq)$, $K_{sp} = [Ca^{2+}][F^-]^2$

Additional info: The process involves setting up an ICE table for the dissolution and solving for the ion concentrations at equilibrium.