Study Guide for Chemical Equilibrium and Le Chatelier's Principle

Chemical Equilibrium: Constants and Expressions

• Definition of Equilibrium Constant Expression ($K_{eq}$): The equilibrium constant expression for a chemical reaction represents the ratio of the molar concentrations of products to reactants, with each raised to the power of its stoichiometric coefficient as defined by the balanced equation.

• Phases in Equilibrium Expressions:     * Only species in the gaseous ($g$) or aqueous ($aq$) phases are included in the $K_{eq}$ expression.     * Pure solids ($s$) and pure liquids ($l$) are omitted because their concentrations remain constant regardless of the amount present.

• Calculations of $K_{eq}$ Examples:     * Reaction 1: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$         * Expression: $K_{eq} = \frac{[NH_3]^2}{[N_2][H_2]^3}$         * Given Concentrations: $[NH_3] = 0.01002\,M$, $[N_2] = 0.0200\,M$, $[H_2] = 0.0200\,M$         * Calculation: $K_{eq} = \frac{(0.0100)^2}{(0.0200) \times (0.0200)^3} = \frac{0.0001}{0.00000016} = 625$     * Reaction 2: $2KClO_3(s) ightleftharpoons 2KCl(s) + 3O_2(g)$         * Expression: $K_{eq} = [O_2]^3$ (Solids are excluded).         * Given: $[O_2] = 0.0500\,M$         * Calculation: $K_{eq} = (0.0500)^3 = 0.000125$     * Reaction 3: $H_2O(l) ightleftharpoons H^+(aq) + OH^-(aq)$         * Expression: $K_{eq} = [H^+][OH^-]$ (Liquid water is excluded).         * Given: $[H^+] = 1 \times 10^{-8}\,M$, $[OH^-] = 1 \times 10^{-6}\,M$         * Calculation: $K_{eq} = (1 \times 10^{-8}) \times (1 \times 10^{-6}) = 1 \times 10^{-14}$     * Reaction 4: $2CO(g) + O_2(g) ightleftharpoons 2CO_2(g)$         * Expression: $K_{eq} = \frac{[CO_2]^2}{[CO]^2[O_2]}$         * Given: $[CO] = 2.0\,M$, $[O_2] = 1.5\,M$, $[CO_2] = 3.0\,M$         * Calculation: $K_{eq} = \frac{(3.0)^2}{(2.0)^2 \times 1.5} = \frac{9}{6} = 1.5$     * Reaction 5: $Li_2CO_3(s) ightleftharpoons 2Li^+(aq) + CO_3^{2-}(aq)$         * Expression: $K_{eq} = [Li^+]^2[CO_3^{2-}]$         * Given: $[Li^+] = 0.20\,M$, $[CO_3^{2-}] = 0.10\,M$         * Calculation: $K_{eq} = (0.20)^2 \times 0.10 = 0.004$

Reaction Quotient ($Q$) and Predicting Reaction Direction

• The Reaction Quotient ($Q$): $Q$ is calculated using the same expression as $K_{eq}$ but uses the current (non-equilibrium) concentrations of reactants and products.

• Comparing $Q$ and $K_{eq}$:     * If Q < K_{eq}: The reaction shifts to the right (toward products) to reach equilibrium.     * If Q > K_{eq}: The reaction shifts to the left (toward reactants) to reach equilibrium.     * If $Q = K_{eq}$: The system is at equilibrium.

• Specific Calculation Examples:     * Ammonia Synthesis ($500\,^{\circ}C$): $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$, $K_{eq} = 0.080$.         * Current values: $[NH_3] = 0.0596\,M$, $[N_2] = 0.600\,M$, $[H_2] = 0.420\,M$.         * $Q = \frac{(0.0596)^2}{(0.600) \times (0.420)^3} = 0.080$.         * Since $Q = K_{eq}$, the system is at equilibrium.     * Antimony Pentachloride Decomposition ($448\,^{\circ}C$): $SbCl_5(g) ightleftharpoons SbCl_3(g) + Cl_2(g)$, $K_{eq} = 0.0251$.         * Current values: $[SbCl_5] = 0.095\,M$, $[SbCl_3] = 0.020\,M$, $[Cl_2] = 0.050\,M$.         * $Q = \frac{(0.020) \times (0.050)}{0.095} = 0.0105$.         * Since Q < K_{eq}, the reaction shifts right.     * Hydrofluoric Acid Decomposition ($1000\,^{\circ}C$): $2HF(g) ightleftharpoons H_2(g) + F_2(g)$, $K_{eq} = 1.0 \times 10^{-13}$.         * Current values: $[HF] = 23.0\,M$, $[H_2] = 0.540\,M$, $[F_2] = 0.380\,M$.         * $Q = \frac{(0.540) \times (0.380)}{(23.0)^2} = 3.88 \times 10^{-4}$.         * Since Q > K_{eq}, the reaction shifts left.     * Sulfur Trioxide Formation ($1227\,^{\circ}C$): $2SO_2(g) + O_2(g) ightleftharpoons 2SO_3(g)$, $K_{eq} = 0.15$.         * Current values: $[SO_2] = 0.344\,M$, $[O_2] = 0.172\,M$, $[SO_3] = 0.056\,M$.         * $Q = \frac{(0.056)^2}{(0.344)^2 \times (0.172)} = 0.154$.         * Since $Q \approx K_{eq}$, the system is essentially at equilibrium.

Le Chatelier's Principle

• Fundamental Statement: When a system at equilibrium is subjected to a stress (change in concentration, pressure, or temperature), the system will shift its equilibrium position to relieve the stress.

• Effect of Stress on Reaction Shifts ($N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g) + 22.0\,kcal$ - Exothermic):     * Add Reactant ($N_2$ or $H_2$): Shift right; reactant concentrations decrease over time to reach new equilibrium; product concentration increases; $K$ remains same.     * Add Product ($NH_3$): Shift left; product concentration decreases; reactants increase; $K$ remains same.     * Remove Reactant ($N_2$ or $H_2$): Shift left; reactants increase to replace loss; $K$ remains same.     * Remove Product ($NH_3$): Shift right; product increases to replace loss; $K$ remains same.     * Increase Temperature: Shift left (toward endothermic side); reactants increase; products decrease; $K$ decreases.     * Decrease Temperature: Shift right (toward exothermic side); products increase; $K$ increases.     * Increase Pressure: Shift right (toward fewer moles of gas: 4 moles reactant vs 2 moles product); $K$ remains same.     * Decrease Pressure: Shift left (toward more moles of gas); $K$ remains same.

• Effect of Stress on Endothermic Reaction ($12.6\,kcal + H_2(g) + I_2(g) ightleftharpoons 2HI(g)$):     * Increase Temperature: Shift right (consuming added heat); $K$ increases.     * Decrease Temperature: Shift left; $K$ decreases.     * Pressure Change: No change in shift because both sides have 2 moles of gas.

• Solubility and Le Chatelier ($NaOH(s) ightleftharpoons Na^+(aq) + OH^-(aq) + 10.6\,kcal$):     * Add $NaOH(s)$: No shift (solid concentrations do not change).     * Add $NaCl$ (Source of $Na^+$): Common ion effect shifts equilibrium left; amount of solid $NaOH$ increases.     * Add $H^+$ (Removes $OH^-$ via neutralization): Shift right to replace $OH^-$; solid $NaOH$ decreases.

Equilibrium Problem Set and Applications

• Pressure Effects Specifics:     * $2H_2S(g) ightleftharpoons 2H_2(g) + S_2(g)$: Increase in Pressure shifts Left (3 moles gas vs 2 moles gas).     * $2N_2O_5(g) ightleftharpoons 4NO_2(g) + O_2(g)$: Increase in Pressure shifts Left.     * $4NO(g) + 2O_2(g) ightleftharpoons 2N_2O_4(g)$: Increase in Pressure shifts Right.     * $SO_2(g) + NO_2(g) ightleftharpoons SO_3(g) + NO(g)$: Increase in Pressure causes No Change (2 moles gas on both sides).

• Weather Indicators (Cobalt Chloride):     * Reaction: $CoCl_2 \cdot 6H_2O(s) \text{ (Pink)} ightleftharpoons CoCl_2(s) \text{ (Blue)} + 6H_2O(g)$.     * Interpretation: High humidity (high $H_2O(g)$) shifts the equilibrium left, producing the pink hydrate. Therefore, pink indicates moist air. Dry air (low $H_2O(g)$) shifts right, producing the blue anhydrous solid.

• Photosynthesis:     * Reaction: $6CO_2(g) + 6H_2O(l) ightleftharpoons C_6H_{12}O_6(s) + 6O_2(g)$, $\Delta H^{\circ} = +2801.69\,kJ/mol$.     * Removing half of the $C_6H_{12}O_6(s)$. Result: No Change (solid concentrations are irrelevant to equilibrium position).     * Decreasing Temperature: Shift Left (since reaction is endothermic).     * Adding Catalyst: No Change in equilibrium position; it only speeds up the rate to reach equilibrium.

• Calculating $K_{eq}$ from Molar Quantities:     * Reaction: $A(g) + B(g) ightleftharpoons C(g) + D(g)$.     * Equilibrium Moles in 1.00 L: $A = 0.40$, $B = 0.40$, $C = 1.60$, $D = 1.60$.     * Calculation: $K_{eq} = \frac{(1.60) \times (1.60)}{(0.40) \times (0.40)} = \frac{2.56}{0.16} = 16$.