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Completion of Chemical Equilibrium Module 14

Overview of Chemical Equilibrium

• The focus of Module 14 is on quantitative calculations and Le Chatelier's principle.

• Objective: To predict the state of a given reaction—at equilibrium, moving towards products (right), or moving towards reactants (left).

Reaction Quotient (Q)

• Definition: The reaction quotient (Q) is calculated using the expression of the equilibrium constant with experimental concentrations at a given time (t).

  • Q is calculated using the formula:
    $Q = rac{[ ext{Products}]}{[ ext{Reactants}]}$

Determining the Direction of the Reaction

• If $Q = K$ (equilibrium constant), then the system is at equilibrium.

• If $Q < K$, the reaction will move to the right (towards products) to reach equilibrium.

• If $Q > K$, the reaction will move to the left (towards reactants) to reach equilibrium, indicating an excess of products.

Example: Haber Process

• The reaction is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

• Equilibrium Constant (K): 0.159 at 453°C

• Given amounts in a 1 L container:

  • 0.683 moles N₂ → concentration = $0.683 ext{ M}$

  • 0.744 moles H₂ → concentration = $0.744 ext{ M}$

  • 8.8 moles NH₃ → concentration = $8.8 ext{ M}$

Calculating Q

• The formula for calculating Q for this reaction:
  Q=rac[extNH3]2[extN2][extH2​]3

• Plugging values into the expression:
  $Q = rac{(8.8)^2}{(0.683)(0.744)^3}$

• Calculation yields: $Q ext{ (calculated)} = 275$

• Since $Q > K$ (275 > 0.159), the reaction will shift left, towards reactants, until equilibrium is reached where $Q = K$.

Le Chatelier's Principle

• Definition: Le Chatelier's principle states that if a system at equilibrium is disturbed by changes in temperature, pressure, or concentration, the system will adjust to counteract the disturbance and restore a new equilibrium.

Factors Affecting Equilibrium

1. Change in Concentration

   • Adding/removing products or reactants changes their concentrations, shifting equilibrium.

2. Change in Pressure

   • Increasing pressure favors the side with fewer moles of gas ions.

   • Example: Increasing internal pressure with nitrogen gas does not affect the equilibrium of carbon monoxide and carbon dioxide.

3. Change in Temperature

   • A reaction requiring heat will shift left if the temperature decreases (removing heat).

Example Scenario

• Consider the reaction: C(s) + CO₂(g) ⇌ 2CO(g) + heat.

  • If CO₂ is removed, equilibrium will shift left (to produce more CO₂).

  • If temperature decreases, the reaction will also shift left (to consume heat).

  • If the volume of the vessel is increased, equilibrium shifts to produce more CO (right) since it leads to more gas moles.

  • Adding solid carbon does not affect the equilibrium as solids do not appear in K expressions.

Calculations with Equilibrium Constants

Acetic Acid Example

• Reaction: CH₃COOH (aq) + H₂O ⇌ CH₃COO⁻ (aq) + H₃O⁺ (aq)

• Given: $K_a = 1.8 imes 10^{-5}$, [CH₃COOH] = 0.5 M, [CH₃COO⁻] = 0.0056 M.

• Finding [H₃O⁺]:

  • Write expression for Ka:
    $K_a = rac{[CH₃COO^-][H₃O^+]}{[CH₃COOH]}$

• Substitute known values:
  $1.8 imes 10^{-5} = rac{(0.0056)([H₃O^+])}{0.5}$

• Solve for [H₃O⁺]:
  $[H₃O^+] = rac{(1.8 imes 10^{-5})(0.5)}{0.0056} = 1.6 imes 10^{-3} ext{ M}$

Phosphorus Trichloride Example

• Reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)

• Given: $K_c = 49$ at 230°C; starting with 4 moles each of PCl₃ and Cl₂ in a 4L vessel.

• Initial concentrations:

  • $[PCl₃] = [Cl₂] = rac{4 ext{ moles}}{4 ext{ L}} = 1 ext{ M}$

• Constructing ICE table:

  • Initial concentrations:

  • $[PCl₃] = 1 M, [Cl₂] = 1 M, [PCl₅] = 0$

  • Changes at equilibrium:

  • $[PCl₃] = 1 - x, [Cl₂] = 1 - x, [PCl₅] = x$

• Kc expression: $K_c = rac{[PCl₅]}{[PCl₃][Cl₂]}$

  • Substituting values:
    $49 = rac{x}{(1-x)(1-x)}$

  • Rearranging gives a quadratic equation:
    $49 - 99x + 49x^2 = 0$

• Solve for x using the quadratic formula: $x = rac{99 ext{ ± } ext{sqrt}(99^2 - 4(49)(49))}{2(49)}$

  • Solutions yield two values, but only the valid one ($x = 0.867$) can be used due to initial concentration limits.

• Equilibrium concentrations:

  • $[PCl₅] ext{ at equilibrium } = x = 0.867 ext{ M}$

  • $[PCl₃] = 1 - 0.867 = 0.13 ext{ M}$

Complex Stoichiometry Example

• Reaction: 4PCl₃ → P + 6Cl₂

• Initial concentration: 1 M of PCl₃, x is equilibrium concentration for P.

• Construct ICE table:

  • Changes:

  • Initial: $1 ext{ M}$ (PCl₃), 0 (P), 0 (Cl₂)

  • Change: $-4x$ (PCl₃), $+x$ (P), $+6x$ (Cl₂)

  • At equilibrium: $[PCl₃] = 1 - 4x, [P] = x, [Cl₂] = 6x$

• K expression:
  $K_c = rac{[Cl₂]^6 [P]}{[PCl₃]^4}$

• Substituting in equilibrium concentrations produces the correct relation for evaluation in future questions.

Conclusion

• Completed Module 14 on chemical equilibrium, emphasizing the importance of understanding reaction quotients, Le Chatelier's principle, and the application of ICE tables in equilibrium calculations. Students encouraged to practice additional questions and activities in the textbook.