1.3: Dynamic Equilibrium

• Dynamic Equilibrium is explored in Chem 26 Lecture Notes 03 by TJ Sotelo from the Institute of Chemistry, College of Science, University of the Philippines Diliman

Learning Outcomes

• Grasp the principles of chemical equilibrium
• Be able to derive Keq in terms of Kc and Kp
• Be able to evaluate the extent of a reaction based on the magnitude of Keq and Q
• Learn to apply Le Chatelier’s Principle to equilibrium systems
• Learn to determine the effects of changes to equilibrium systems
• Know the effect of a catalyst to the equilibrium system

Dynamic Equilibrium

• Dynamic Equilibrium involves opposing processes taking place at equal rates
• Example reactions:
  • H2O(l) ⇌ H2O(g)
  • NaCl(s) ⇌ NaCl(aq)
  • CO(g) + 2 H2(g) ⇌ CH3OH(g)
  • I2(H2O)
  • I2(CCl4)

Equilibrium Constant Expression

• The oxidation-reduction reaction of copper (II) and tin (II) in aqueous solution is reversible
• Example reactions:
• 2 Cu²⁺(aq) + Sn²⁺(aq) ⇄ 2 Cu⁺ (aq) + Sn⁴⁺(aq)
• 2 Cu⁺(aq) + Sn⁴⁺(aq) → 2 Cu²⁺(aq) + Sn²⁺(aq)
• Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction

General Expression for K

• For a reaction of the form: aA + bB ... → gG + hH ...
• The equilibrium constant Kc = [G]ᵍ[H]ʰ ... / [A]ᵃ[B]ᵇ ...
• Keq = (aG)ᵍ(aH)ʰ ... / (aA)ᵃ(aB)ᵇ ... ≈ (1/c°) ᐁⁿ[G]ᵍ[H]ʰ ... / [A]ᵐ[B]ⁿ ... = (1/c°) ᐁⁿ Kc
• where ∆n = {g + h ...} - {a + b ...}
• (1/c°) ᐁⁿ ensures that K is a dimensionless quantity

Activity

• Activity is a thermodynamic concept introduced by Lewis
• Activity is a dimensionless ratio referring to a chosen reference state
• Activity helps account for non-ideal behaviors in solutions and gases
• Activity represents an effective concentration in a system
• aB = γB[B] / c°B
• γB = activity coefficient (≈ 1 in an ideal solution)
• c°B = standard reference state (1 mol L⁻¹)
• aB = [B]
• A similar expression exists for gases
• aB = γBPB / P°B
• P°B = standard reference state (1 bar)
• aB = PB

Relationships Involving Equilibrium Constants

• Reversing an equation causes inversion of K
• Multiplying coefficients by a common factor raises the equilibrium constant to a corresponding power
• Dividing the coefficients by a common factor causes equilibrium constants to be taken to that root

Evaluation of the Equilibrium Constant

• The synthesis of NH3 from its elements at 298 K has K = 5.80 x 10⁵
• N2(g) + 3H2(g) ⇌ 2NH3(g)
• Determine the value of K at 298 K for the following reaction:
• NH3(g) ⇌ 1/2N2(g) + 3/2H2(g)
• A: For 2NH3(g) ⇌ N2(g) + 3H2(g); K’ = 1 / 5.80 x 10⁵
• A: For NH3(g) ⇌ ½ N2(g) + 3/2H2(g); K’ = √(1 / 5.80 x 10⁵) = 1.30 x 10⁻³
• Methanol is a gasoline additive that increases its octane value
• Methanol can be synthesized according to the following reaction:
• CO(g) + 2 H2 (g) ⇌ CH3OH (g); K = 9.23 x 10⁻³
• The equilibrium constant, K’, for the reverse reaction is 1.08 x 10²
• The equilibrium constant, K”, if two moles of methanol were formed is 8.52 x 10⁻⁵
• 2 CO(g) + 4 H2 (g) ⇌ 2 CH3OH (g)

Combining Equilibrium Constant Expressions

• To determine Kc for the reaction N₂O(g) + ½O₂(g) ⇌ 2NO(g)
• Using the information from these reactions:
• N₂(g) + ½O₂(g) ⇌ N₂O(g); Kc₁ = [N₂O] / [N₂][O₂]⁰⁵ = 5.4 x 10⁻¹⁹
• N₂(g) + O₂(g) ⇌ 2NO(g); Kc₂ = [NO]² / [N₂][O₂] = 4.6 x 10⁻³¹
• Kc = ([NO]² / [N₂][O₂]) / ([N₂O] / [N₂][O₂]⁰⁵) = Kc₂ / Kc₁ = 8.5 x 10⁻¹³

Equilibrium Constant for Gases

• In concentration, another substitution can be used for the reaction:
• 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
• Concentration can be expressed as (nx / V) hence
• [SO₃] = nSO₃/ V = PSO₃ / RT
• [O₂] = nO₂/ V = PO₂ / RT
• [SO₂] = nSO₂/ V = PSO₂ / RT
• αₓ = [X] / C° = (Px / RT )/ C°

Equilibrium Constant for Gases (2)

• From the previous slide:
• Kp = α²SO₃ / α²SO₂αO₂ = P°(P²SO₃ / P²SO₂PO₂)
• Kp = α²SO₃ / α²SO₂αO₂ = P°(([SO₃]RT)² / ([SO₂]RT)²([O₂]RT))
• Kp = (P°/ RT) ([SO₃]² / ([SO₂]²[O₂])) = Kc/ RT
• In general terms:
• aA + bB … → gG + hH … -∆n = {g + h …} - {a + b …} -Kp = Kc(RT)∆n

Equilibria Involving Pure Liquids and Solids (1)

• Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component
• For the reaction: C(s) + H₂O(g) ⇌ CO(g) + H₂(g):
• Kc = [CO][H₂] / [H₂O]² = PCOPH₂/ PH₂O² (RT)⁻¹
• Changes in pure solid or pure liquid concentrations should not affect the position of equilibrium

Equilibria Involving Pure Liquids and Solids (2)

• Decomposition of CaCO₃(s) yields CaO(s) and CO₂(g), which exerts equilibrium partial pressure.
• CaCO₃(s) ⇌ CaO(s) + CO₂(g)
• Introducing more solid CaCO₃(s) or CaO(s) will not affect the partial pressure exerted by CO₂(g)
• Kc = [CO₂]
• Kp = PCO₂ = Kc(RT)

Significance of the Magnitude of the Equilibrium Constant

• If K » 1, products formation is favored
• If K « 1, reactants formation is favored
• Equilibrium Constants of Some Common Reactions:
• 2 H₂(g) + O₂(g) ⇌ 2 H₂O(l); Kp = 1.4 × 10⁸³ at 298 K
• CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.9 × 10⁻²³ at 298 K; 1.0 at ~1200 K
• 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g); Kp = 3.4 at 1000 K
• C(s) + H₂O(g) ⇌ CO(g) + H₂(g); Kp = 1.6 × 10⁻²¹ at 298 K; 10.0 at ~1100 K

Predicting the Direction of Net Change with the Reaction Quotient

• For the reaction CO(g) + 2H2(g) ⇌ CH3OH(g), a qualitative determination of initial conditions as equilibrium is approached is needed
• Determining equilibrium shift is key
• Equilibrium can be approached various ways
• Qc = [G]ᵍ[H]ʰ / [A]ᵐ[B]ⁿ
• At equilibrium Qc = Kc, where Q is called the reaction quotient

Predicting the Direction of Net Change with the Reaction Quotient (2)

• Consider the reaction: Reactants ⇌ Products
• If Q < Kc, approaches the right
• initial reaction proceeds to the right
• pure reactants will approach equilibrium to the right
• If Q = Kc, equilibrium is reached
• If Q > Kc, the reaction proceeds to the left

Altering Equilibrium Conditions

• Le Châtelier's Principle states: When an equilibrium system is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.

Effect of Changing the Pressure

• Adding a gaseous reactant or product changes Pgas and changes the equilibrium in the same way that concentration changes equilibrium
• Adding an inert gas changes the total pressure but the relative partial pressures are unchanged, so there is no change in equilibrium
• At high P and low V, the direction shifts to the direction that produces the lower moles of gas

Effect of Volume Change

• When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas
• When the volume is increased, a net change occurs in the direction that produces more moles of gas

Effect of Temperature

• Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction
• Lowering the temperature causes a shift in the direction of the exothermic reaction

Effect of Adding a Catalyst

• A catalyst changes the mechanism of a reaction to one with a lower activation energy
• A catalyst has no effect on the condition of equilibrium in a reversible reaction
• A catalyst affects the rate at which equilibrium is attained