Dynamic Equilibrium – Comprehensive Study Notes

Dynamic Equilibrium – Study Notes

Learning outcomes (summary)

  • Explain the principles of chemical equilibrium.
  • Derive Keq in terms of Kc and Kp.
  • Evaluate the extent of a reaction based on the magnitude of Keq and Q.
  • Apply Le Chatelier’s Principle to equilibrium systems.
  • Determine the effect of changes in the equilibrium system.
  • Determine the effect of a catalyst on the equilibrium system.

Dynamic equilibrium (conceptual overview)

  • Dynamic equilibrium: opposing processes occur at equal rates, so macroscopic properties are constant though microscopic processes continue.
  • Illustrative examples (from lecture):
    • I2 in water and I2 in CCl4; H2O(l) ⇄ H2O(g); NaCl(s) ⇄ NaCl(aq); CO(g) + 2 H2(g) ⇄ CH3OH(g).
  • Key idea: balance between forward and reverse reactions leads to constant concentrations of reactants and products at equilibrium.

Equilibrium constant expression (Kc)

  • For a general reaction: a \,A + b \,B \rightleftharpoons g \,G + h \,H
    • Equilibrium constant expression: K_c = \frac{[G]^g [H]^h}{[A]^a [B]^b}
    • For stoichiometric coefficients, exponents match coefficients (g, h, a, b).
  • The thermodynamic equilibrium constant is dimensionless: it uses activities, not just concentrations.
  • Net change in moles of gas during the reaction is: \Delta n = (g + h) - (a + b)
  • The forward and reverse rates are equal at equilibrium (rate law balance).

Activities and non-ideality

  • Activity is a dimensionless quantity that reduces to an effective concentration under non-ideal conditions.
  • General definitions:
    • For solutes in solution: ai = \gammai \dfrac{[i]}{[i]^{\circ}} where ( [i]^{\circ} = 1\,\text{M} ) is the standard reference.
    • For pure solutes: activity of a pure solid or liquid is taken as 1 (excluded from K expressions).
    • For gases: ai = \phii \dfrac{Pi}{P^{\circ}} where (\phii) is the fugacity coefficient and (P^{\circ} = 1\,\text{bar} ) (or 1 atm) is the standard pressure depending on convention.
  • Practical point: the use of activities makes the equilibrium constant dimensionless and accounts for non-ideal behavior.

Equilibrium involving gases: Kp and Kc

  • For reactions involving gases: the relationship between Kp and Kc is
    • Kp = Kc \,(R T)^{\Delta n} where \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}.
  • Example: for the gas-phase reaction
    • 2\mathrm{SO}2(g) + \mathrm{O}2(g) \rightleftharpoons 2\mathrm{SO}_3(g)
    • Reactant moles (gas): 3; product moles (gas): 2; \Delta n = 2 - 3 = -1 hence Kp = Kc \,(R T)^{-1} = \frac{K_c}{R T}.
  • General derivation from concentrations to pressures uses partial pressures: Pi = Xi P_{tot} and the ideal gas relation to relate to Kc.

Equilibria with pure liquids and solids

  • In a heterogeneous equilibrium, pure solids and liquids do not appear in the equilibrium expression because their activities are 1.
  • Example: for the decomposition/formation of calcium carbonate
    • \mathrm{CaCO}3(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}2(g)
    • Equilibrium expression in terms of concentrations: Kc = [\mathrm{CO}2] (since solid activities are 1, only CO2 appears in the expression).
    • In terms of partial pressures: Kp = P{\mathrm{CO}_2} (same idea).
  • Additional note: the presence of pure solids or liquids in a reaction mixture cannot alter the position of equilibrium by changing their concentrations.

Magnitude of the equilibrium constant

  • If K >> 1, products are favored at equilibrium.
  • If K << 1, reactants are favored at equilibrium.
  • If K \approx 1, neither side dominates; substantial amounts of both reactants and products are present.

Connection between Gibbs energy and equilibrium (ΔG and Keq)

  • General relation: \Delta G = \Delta G^{\circ} + RT \ln Q where Q is the reaction quotient.
  • At equilibrium, Q = K{eq} and hence \Delta G = 0 = \Delta G^{\circ} + RT \ln K{eq} \quad \Rightarrow\quad \Delta G^{\circ} = -RT \ln K_{eq}
  • Interpretation: the standard Gibbs energy change is linked to the position of equilibrium via Keq.

Predicting the direction of net change (using Qc)

  • Reaction: general A + b B ⇌ g G + h H, with concentrations at time t.
  • Reaction quotient: Q_c = \frac{[G]^g [H]^h}{[A]^a [B]^b} (same form as Kc, but with current concentrations).
  • At equilibrium, Qc = Kc.
  • For a specific example: \mathrm{CO(g)} + 2\mathrm{H}2(g) \rightleftharpoons \mathrm{CH}3\mathrm{OH}(g)
    • Qc = \frac{[\mathrm{CH}3\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2} (products in numerator).
  • Decision rule (qualitative):
    • If Qc < Kc, the reaction will move toward the products (forward) to reach equilibrium.
    • If Qc > Kc, the reaction will move toward the reactants (reverse) to reach equilibrium.

Examples (practice problems)

  • Example 1: ammonia synthesis
    • Reaction: \mathrm{N}2(g) + 3\mathrm{H}2(g) \rightleftharpoons 2\mathrm{NH}_3(g)
    • At 298 K, K_c = 5.80 \times 10^{5}.
    • Consider the reverse: 2\mathrm{NH}3(g) \rightleftharpoons \mathrm{N}2(g) + 3\mathrm{H}_2(g)
    • The relation: K'c = \frac{1}{Kc} = \frac{1}{5.80 \times 10^{5}} = 1.72 \times 10^{-6}
    • If a problem asks for the equilibrium constant for the reverse direction, take the reciprocal of the forward value.
  • Example 2: methanol synthesis and reverse
    • Reaction: \mathrm{CO(g)} + 2\mathrm{H}2(g) \rightleftharpoons \mathrm{CH}3\mathrm{OH}(g) with K = 9.23 \times 10^{-3}
    • Reverse reaction: \mathrm{CH}3\mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g) + 2\mathrm{H}2(g)
    • K' = \frac{1}{K} = \frac{1}{9.23 \times 10^{-3}} = 1.08 \times 10^{2}
  • For the reaction producing two methanol molecules:
    • 2\mathrm{CO}(g) + 4\mathrm{H}2(g) \rightleftharpoons 2\mathrm{CH}3\mathrm{OH}(g)
    • The equilibrium constant for this multiply-by-2 process is K'' = K^{2} = (9.23 \times 10^{-3})^{2} = 8.52 \times 10^{-5}

Combining equilibrium expressions

  • Sometimes multiple equilibria combine to give a net reaction; their constants multiply.
  • Example framework (as shown in lecture):
    • If you have two equilibria with given Kc values, the overall Kc for the sum/difference of those reactions is the product or quotient of the individual Kc values, depending on the direction of each equilibrium.

Equilibrium constant for gases (derivation sketch)

  • In gas-phase, concentrations can be expressed as mole fractions and volumes, or more directly via partial pressures.
  • Substituting: [X] = \dfrac{nX}{V}, \quad PX = X_X P^{\circ} where X is a species.
  • For a representative reaction: a A + b B \rightleftharpoons g G + h H with only gases, the relationship becomes
    • Kp = \dfrac{PG^{g} PH^{h}}{PA^{a} P_B^{b}}
  • The general link to Kc: Kp = Kc (R T)^{\Delta n}, \quad \Delta n = (g+h) - (a+b).

Examples and practice problems (gas-focused)

  • Example 1 (from slides): Calculate KP for the reaction
    • 2\mathrm{NH}3(g) \rightleftharpoons \mathrm{N}2(g) + 3\mathrm{H}_2(g)
    • Given forward Kc = 2.8 \times 10^{-9} at 298 K.
    • For the reverse reaction, K'c = \dfrac{1}{Kc} = \dfrac{1}{2.8 \times 10^{-9}} = 3.57 \times 10^{8} (example, adjust per exact direction in problem).
  • Example 2: Pressure-volume considerations for gas mixtures (conceptual)
    • Raising pressure at constant temperature shifts toward fewer moles of gas (Le Chatelier’s principle).
    • Decreasing volume has the opposite effect.

Le Chatelier’s Principle – altering conditions (conceptual)

  • Temperature, pressure, and concentration changes shift equilibrium to counter the change:
    • If you add a reactant or remove a product, the system shifts to form more product (toward products).
    • If you add a product or remove a reactant, the system shifts to form more reactant (toward reactants).
  • Example (organic oxide formation):
    • Reaction: 2\mathrm{SO}2(g) + \mathrm{O}2(g) \rightleftharpoons 2\mathrm{SO}_3(g)
    • If SO3 is added (increase product concentration), the system shifts to the left (toward reactants) to reduce the added SO3.
  • Practical quantitative note (Qc vs Kc): the comparison of reaction quotient and equilibrium constant dictates direction of shift.

Effect of changing pressure and volume (gases)

  • Increasing pressure (or decreasing volume) shifts equilibrium toward the side with fewer moles of gas (fewer gas moles).
  • If you add an inert gas at constant volume and temperature, the total pressure changes but partial pressures remain unchanged; equilibrium position is unaffected.
  • Mathematical intuition: for reaction aA + bB ⇌ gG + hH, the term (ngas products − ngas reactants) drives the pressure/volume response via Δn.
  • Example relationship for a specific reaction: for 2S O2(g) + O2(g) ⇌ 2SO3(g), Δn = -1; increasing pressure favors SO3 formation.

Effect of volume change (quantitative relation)

  • Volume change affects concentrations; the equilibrium constant in terms of concentrations, Kc, remains the same at a given T, but the actual concentrations adjust according to stoichiometry.
  • Volume reduction increases overall concentrations; the net shift is toward the side with fewer moles of gas (depending on Δn).

Effect of temperature (endothermic vs exothermic)

  • Increasing temperature:
    • Shifts equilibrium in the direction of the endothermic reaction (absorbs heat).
  • Decreasing temperature:
    • Shifts toward the exothermic reaction (releases heat).
  • Practical implication: temperature controls the position of equilibrium for reactions with ΔH ≠ 0.

Effect of a catalyst

  • A catalyst provides an alternate mechanism with lower activation energy.
  • Catalysts do not change the position of equilibrium (ratio of products to reactants at equilibrium remains the same).
  • They affect the rate at which equilibrium is attained (increase the rates of both forward and reverse reactions equally).

Example multi-part problem (practice set)

  • Problem (general framework): A complex reaction system is given with a solid Ni(s) catalyst and a gas mixture; impact of temperature and volume on equilibrium is asked.
  • Typical tasks:
    • (a) Write the equilibrium constant expressions in Kp and Kc for the balanced reaction.
    • (b) Determine whether raising temperature shifts toward products or reactants (based on ΔH).
    • (c) Determine the effect on the rate when the Ni catalyst is removed.
    • (d) If the system volume changes from 1000 mL to 500 mL, discuss how equilibrium would shift (based on gas moles and Le Chatelier’s principle).

Quick reference: useful relations to memorize

  • General equilibrium constant expression: K_c = \frac{\prod [\text{products}]^{\text{coefficients}}}{\prod [\text{reactants}]^{\text{coefficients}}}
  • For gases: Kp = Kc (R T)^{\Delta n}, \quad \Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})
  • Relationship to Gibbs energy: \Delta G^{\circ} = -RT \ln K_{eq}
  • Reaction quotient: Q_c = \frac{\prod [\text{products}]^{\text{coefficients}}}{\prod [\text{reactants}]^{\text{coefficients}}} (evaluated with current concentrations)
  • At equilibrium: Qc = Kc
  • Activities formalism (soln and gas): ai = \gammai \dfrac{[i]}{[i]^{\circ}} \quad\text{or}\quad ai = \phii \dfrac{P_i}{P^{\circ}} for gases

Quick worked examples (summaries)

  • Example A: CO + 2 H2 ⇄ CH3OH
    • Given: K for forward reaction at a temperature: K = 9.23 \times 10^{-3}
    • Reverse constant: K' = \dfrac{1}{K} = 1.08 \times 10^{2}
    • For doubling stoichiometric coefficients (e.g., 2 CO + 4 H2 ⇄ 2 CH3OH): K'' = K^2 = (9.23 \times 10^{-3})^2 = 8.52 \times 10^{-5}
  • Example B: Pure solids in gas equilibria
    • For a solid–gas decomposition, the partial pressure of CO2 establishes the gas-phase equilibrium; the solid activities do not appear in K expressions.
  • Example C: ΔG and Keq
    • If the system is at equilibrium, ΔG = 0, thus ΔG° = -RT ln K_eq.

Summary of key concepts

  • Equilibrium is dynamic, with equal forward and reverse rates.
  • Keq expresses the ratio of product activities to reactant activities, raised to stoichiometric powers.
  • Kp and Kc are related by the Δn term for gaseous reactions: Kp = Kc (R T)^{\Delta n}.
  • Activities account for non-ideality; for pure solids/liquids, activities are 1 and do not appear in equilibrium expressions.
  • The magnitude of Keq indicates the favored side at equilibrium; ΔG° is tied to Keq via the relation \Delta G^{\circ} = -RT \ln K_{eq}.
  • The reaction quotient Q allows prediction of the direction of net change toward equilibrium.
  • Le Chatelier’s Principle explains how changes in temperature, pressure/volume, or concentrations shift equilibria.
  • A catalyst changes the rate to reach equilibrium but not the position of equilibrium.

Notes:

  • All numerical references and formulas are presented as given, with standard conventions for Kc, Kp, and Qc. When performing problems, ensure consistent units and, for gas-phase reactions, carefully determine Δn to relate Kp and Kc.