5.1 Chemical Equilibrium, Equilibrium Constants & Le Chatelier's Principle

5.1 Chemical Equilibrium, Equilibrium Constants & Le Chatelier's Principle

Reversible reactions progress toward dynamic chemical equilibrium, an energetic minimum where forward and reverse reaction rates balance.

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Dynamic Equilibrium and the Law of Mass Action

Chemical equilibrium is dynamic: forward and reverse reactions proceed continuously at identical rates ($r_f = r_r$). Consequently, macroscopic system parameters (concentrations, partial pressures, color) remain invariant. Thermodynamically, equilibrium represents the state of minimum Gibbs free energy where:

$$\Delta G = 0$$

For a reversible reaction $aA + bB \rightleftharpoons cC + dD$, the law of mass action defines the equilibrium constant ($K_c$):

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

CRITICAL MCAT RULE: Pure solids ($s$) and pure liquids ($l$, including solvent water) are omitted from equilibrium expressions because their thermodynamic activity equals 1 (molar density is constant). Only aqueous solutes and gases appear in $K_{\text{eq}}$.

For gas-phase equilibria, partial pressures define $K_p$:

$$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} = K_c (RT)^{\Delta n_{\text{gas}}}$$

where $\Delta n_{\text{gas}} = (c + d)_{\text{gas}} - (a + b)_{\text{gas}}$, $R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$, and $T$ is absolute temperature. If $\Delta n_{\text{gas}} = 0$, then $K_p = K_c$.

Manipulating balanced equations alters $K_{\text{eq}}$ algebraically:

• Reversing reaction: $K_{\text{rev}} = 1 / K_{\text{fwd}}$.
• Multiplying coefficients by $n$: $K' = (K_{\text{orig}})^n$.
• Summing consecutive reactions: $K_{\text{net}} = K_1 \times K_2$.

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Reaction Quotient ($Q$) and Predicting Directionality

The reaction quotient $Q$ evaluates instantaneous non-equilibrium concentrations using the mass-action expression. Comparing $Q$ to $K_{\text{eq}}$ predicts reaction direction:

• $Q < K_{\text{eq}}$: Products are deficient; the forward reaction proceeds spontaneously (shifts right $\rightarrow$, $\Delta G < 0$).
• $Q = K_{\text{eq}}$: Dynamic equilibrium; forward and reverse rates balance (no shift, $\Delta G = 0$).
• $Q > K_{\text{eq}}$: Products are in excess; the reverse reaction proceeds spontaneously (shifts left $\leftarrow$, $\Delta G > 0$ for forward direction).

Thermodynamically, $\Delta G = RT\ln(Q / K_{\text{eq}})$. When $Q < K_{\text{eq}}$, $\Delta G < 0$, driving spontaneous product formation.

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Le Chatelier's Principle and System Perturbations

When a stress is applied to an equilibrated system, the equilibrium position shifts to partially offset the disturbance.

1. Concentration Perturbations

Adding reactants or removing products lowers $Q$ below $K_{\text{eq}}$, shifting the reaction right ($\rightarrow$). Adding products or removing reactants raises $Q$ above $K_{\text{eq}}$, shifting left ($\leftarrow$). The numerical value of $K_{\text{eq}}$ remains unchanged.

2. Pressure and Volume (Gases)

According to Boyle's law ($P \propto 1/V$):

• Decreasing Volume (Increasing Pressure): Shifts toward the side with fewer moles of gas ($\sum n_{\text{gas}}$) to relieve pressure.
• Increasing Volume (Decreasing Pressure): Shifts toward the side with more moles of gas to restore pressure.
• If $\Delta n_{\text{gas}} = 0$: Volume and pressure changes cause zero shift.

3. Inert Gas Additions

• Constant Volume: Adding an unreactive gas (He, Ar) increases total pressure without changing reactant or product partial pressures. Thus, $Q = K_{\text{eq}}$ and no shift occurs.
• Constant Total Pressure: The container expands ($V \uparrow$), decreasing individual gas partial pressures. The reaction shifts toward the side with more moles of gas.

4. Temperature (The Sole Stress Altering $K_{\text{eq}}$)

Temperature is the only perturbation that alters the numerical value of $K_{\text{eq}}$:

• Endothermic ($\Delta H^\circ > 0$): Heat behaves as a reactant. Heating shifts right and increases $K_{\text{eq}}$; cooling shifts left and decreases $K_{\text{eq}}$.
• Exothermic ($\Delta H^\circ < 0$): Heat behaves as a product. Heating shifts left and decreases $K_{\text{eq}}$; cooling shifts right and increases $K_{\text{eq}}$.
• Quantified by the van 't Hoff equation: $\ln(K_2 / K_1) = -(\Delta H^\circ / R)(1/T_2 - 1/T_1)$.

5. Catalysts

Catalysts lower activation energy ($E_a$) equally for forward and reverse directions, accelerating $k_f$ and $k_r$ identically. Catalysts allow the mixture to reach equilibrium faster without altering equilibrium concentrations, yields, or $K_{\text{eq}}$.

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Le Chatelier Perturbation Summary Table

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Solubility Product Constant ($K_{\text{sp}}$) and Common Ion Effect

The dissolution of sparingly soluble ionic salts constitutes a heterogeneous equilibrium:

$$M_m X_x(s) \rightleftharpoons m M^{n+}(aq) + x X^{y-}(aq) \implies K_{\text{sp}} = [M^{n+}]^m [X^{y-}]^x$$

Molar solubility ($s$) represents moles of solid dissolved per liter of saturated solution:

• 1:1 Salt ($AgCl$): $K_{\text{sp}} = s^2 \implies s = \sqrt{K_{\text{sp}}}$
• 1:2 or 2:1 Salt ($PbI_2, CaF_2$): $K_{\text{sp}} = 4s^3 \implies s = (K_{\text{sp}}/4)^{1/3}$
• 1:3 Salt ($Al(OH)_3$): $K_{\text{sp}} = 27s^4 \implies s = (K_{\text{sp}}/27)^{1/4}$
• 2:3 Salt ($Ca_3(PO_4)_2$): $K_{\text{sp}} = 108s^5 \implies s = (K_{\text{sp}}/108)^{1/5}$

Comparing $Q_{\text{sp}}$ to $K_{\text{sp}}$ predicts precipitation:

• $Q_{\text{sp}} < K_{\text{sp}}$: Unsaturated; no precipitate forms.
• $Q_{\text{sp}} = K_{\text{sp}}$: Saturated; dynamic equilibrium.
• $Q_{\text{sp}} > K_{\text{sp}}$: Supersaturated; spontaneous precipitation occurs until $Q_{\text{sp}} = K_{\text{sp}}$.

Common Ion Effect and Selective Precipitation

Adding a soluble salt sharing an ion with a sparingly soluble salt increases $[M^{n+}]$ or $[X^{y-}]$, driving $Q_{\text{sp}} > K_{\text{sp}}$. The system shifts left, precipitating salt and reducing molar solubility without changing $K_{\text{sp}}$.

For example, $\text{BaSO}_4$ ($K_{\text{sp}} = 1.1 \times 10^{-10}$) has a solubility of $1.05 \times 10^{-5}\text{ M}$ in pure water ($s = \sqrt{K_{\text{sp}}}$). In $0.10\text{ M } \text{Na}_2\text{SO}_4$, $[\text{SO}_4^{2-}] \approx 0.10\text{ M}$, yielding $[\text{Ba}^{2+}] = 1.1 \times 10^{-10} / 0.10 = 1.1 \times 10^{-9}\text{ M}$—a 10,000-fold reduction.

Selective precipitation separates mixed cations by adding a counter-anion dropwise. The compound with the lower required precipitant concentration reaches its $K_{\text{sp}}$ threshold first ($Q_{\text{sp}} > K_{\text{sp}}$) and precipitates before the more soluble species forms solid.