Chemical Equilibrium Lecture Notes

The Reaction Quotient

• The size of the equilibrium constant K and the time required to reach equilibrium are not directly related.
• Three different initial conditions are given for the reaction A = 2B:
  • a) [A]i = 3.0 M and [B]i = 1.0 M
  • b) [A]i = 1.0 M and [B]i = 4.0 M
  • c) [A]i = 2.0 M and [B]i = 2.0 M
• The initial concentrations are used to determine the direction of the shift toward equilibrium.
• The equilibrium constant K is defined as: K = \frac{[B]e^2}{[A]e} = 2.0
• The reaction quotient Q is defined as: Q = \frac{[B]i^2}{[A]i}
  • If Q < K, the reaction shifts towards the products (right).
  • If Q > K, the reaction shifts towards the reactants (left).
  • If Q = K, the reaction is at equilibrium and will not shift.
• Calculations for each initial condition:
  • a) Q = \frac{(1.0)^2}{3.0} = 0.33. Since Q < K, the reaction shifts right (towards products).
  • b) Q = \frac{(4.0)^2}{1.0} = 16. Since Q > K, the reaction shifts left (towards reactants).
  • c) Q = \frac{(2.0)^2}{2.0} = 2.0. Since Q = K, the reaction is at equilibrium and will not shift.
• For a reactant-heavy system, the reaction shifts right.
• For a product-heavy system, the reaction shifts left.

Ethyl Acetate Synthesis

• Ethyl acetate is synthesized in a non-reacting solvent according to the reaction:
  CH3CO2H + C2H5OH \rightleftharpoons CH3CO2C2H5 + H_2O
• The equilibrium constant for this reaction is K = 2.2. Defined as: K = \frac{[CH3CO2C2H5][H2O]}{[CH3CO2H][C2H_5OH]}
• The direction the reaction must shift to reach equilibrium given initial concentrations is determined by comparing Q and K.
• a) Given: [CH3CO2C2H5] = 0.22 M, [H2O] = 0.10 M, [CH3CO2H] = 0.010 M, [C2H_5OH] = 0.010 M
  Q = \frac{(0.22)(0.10)}{(0.010)(0.010)} = 220
  Since Q > K, the reaction will shift towards the reactants (left).
• b) Given: [CH3CO2C2H5] = 0.88 M, [H2O] = 0.12 M, [CH3CO2H] = 0.044 M, [C2H_5OH] = 6.0 M
  Q = \frac{(0.88)(0.12)}{(0.044)(6.0)} = 0.40
  Since Q < K, the reaction will shift towards the products (right).

Basic Chemical Equilibrium Problem

• Reaction: A \rightleftharpoons B
• At equilibrium, [A] = 10. M and [B] = 15. M.
• What will the equilibrium concentrations be if [A]i = 60. M and [B]i = 60. M?
• Calculate the equilibrium constant:
  K = \frac{[B]e}{[A]e} = \frac{15}{10} = 1.5
• Calculate the reaction quotient with the initial concentrations:
  Q = \frac{[B]i}{[A]i} = \frac{60}{60} = 1.0
• Since Q < K, the reaction shifts to the right.
• Set up an ICE table:
  • Initial: [A] = 60, [B] = 60
  • Change: [A] = -x, [B] = +x
  • Equilibrium: [A] = 60 - x, [B] = 60 + x
• Solve for x:
  K = 1.5 = \frac{60 + x}{60 - x}
  1.5(60 - x) = 60 + x
  90 - 1.5x = 60 + x
  2.5x = 30
  x = 12
• Calculate the equilibrium concentrations:
  [A]e = 60 - 12 = 48 M [B]e = 60 + 12 = 72 M

More Complicated Chemical Equilibrium Problem

• Reaction: H2(g) + F2(g) \rightleftharpoons 2HF(g)
• Equilibrium constant: K = 1.15 \times 10^2 (Products favored)
• Initial conditions: 1.0 L flask charged with 2.0 moles of each compound. Therefore, [H2]i = [F2]i = [HF]_i = 2.0 M
• Reaction quotient: Q = 1.0
• Since Q < K, the reaction shifts right.
• ICE table:
  | | H2 | F2 | 2HF |
  |----------|-------|-------|--------|
  | Initial | 2.0 M | 2.0 M | 2.0 M |
  | Change | -x | -x | +2x |
  | Equilibrium| 2.0-x | 2.0-x | 2.0+2x |
• Equilibrium expression:
  K = \frac{[HF]^2}{[H2][F2]} = \frac{(2.0 + 2x)^2}{(2.0 - x)(2.0 - x)} = \frac{(2.0 + 2x)^2}{(2.0 - x)^2}
• Since K = 1.15 \times 10^2, take the square root of both sides:
  \sqrt{K} = \sqrt{1.15 \times 10^2} = 10.72
  10.72 = \frac{2.0 + 2x}{2.0 - x}
• Solve for x:
  10.72(2.0 - x) = 2.0 + 2x
  21.44 - 10.72x = 2.0 + 2x
  12.72x = 19.44
  x = 1.53
• Calculate equilibrium concentrations:
  [H2]e = [F2]e = 2.0 - x = 2.0 - 1.53 = 0.47 M
  [HF]_e = 2.0 + 2x = 2.0 + 2(1.53) = 5.1 M

Chemical Equilibria of Reactions With Multiple Steps

• Consider two reactions:
  aA + bB \rightleftharpoons cC with equilibrium constant K1 cC \rightleftharpoons dD + eE with equilibrium constant K2
• Equilibrium constants:
  K1 = \frac{[C]^c}{[A]^a[B]^b} K2 = \frac{[D]^d[E]^e}{[C]^c}
• Overall reaction: aA + bB \rightleftharpoons dD + eE
• The equilibrium constant for the overall reaction, K3, is the product of the individual equilibrium constants: K3 = \frac{[D]^d[E]^e}{[A]^a[B]^b} = K1 \times K2 = \frac{[C]^c}{[A]^a[B]^b} \times \frac{[D]^d[E]^e}{[C]^c}
• When adding multiple reactions to derive an overall reaction, the equilibrium constant for the overall reaction is the product of the individual equilibrium constants.

Multistep Equilibrium Problem

• Very large and very small equilibrium constants are difficult to measure directly, so they are often obtained through the combination of reactions.
• Overall reaction:
  CuCN(s) \rightleftharpoons Cu^+(aq) + CN^-(aq) \quad K = ?
• Given the following reactions:
  1. CuCl(s) + HCN(aq) \rightleftharpoons CuCN(s) + HCl(aq) \quad K_1 = 1.8 \times 10^{-2}
  2. HCN(aq) \rightleftharpoons H^+(aq) + CN^-(aq) \quad K_2 = 6.2 \times 10^{-10}
  3. HCl(aq) \rightleftharpoons H^+(aq) + Cl^-(aq) \quad K_3 = 1.3 \times 10^{6}
  4. CuCl(s) \rightleftharpoons Cu^+(aq) + Cl^-(aq) \quad K_4 = 1.2 \times 10^{-6}
• To obtain the overall reaction, we need to manipulate these reactions:
  • Reverse reaction 1: CuCN(s) + HCl(aq) \rightleftharpoons CuCl(s) + HCN(aq) \quad K_1^{-1}
  • Keep reaction 2: HCN(aq) \rightleftharpoons H^+(aq) + CN^-(aq) \quad K_2
  • Reverse reaction 3: H^+(aq) + Cl^-(aq) \rightleftharpoons HCl(aq) \quad K_3^{-1}
  • Keep reaction 4: CuCl(s) \rightleftharpoons Cu^+(aq) + Cl^-(aq) \quad K_4
• Adding the reactions together, we get the desired overall reaction:
  CuCN(s) \rightleftharpoons Cu^+(aq) + CN^-(aq)
• The equilibrium constant for the overall reaction is:
  K{overall} = K1^{-1} \times K2 \times K3^{-1} \times K4 = \frac{K2 \times K4}{K1 \times K_3} = \frac{(6.2 \times 10^{-10}) \times (1.2 \times 10^{-6})}{(1.8 \times 10^{-2}) \times (1.3 \times 10^{6})} = 3.2 \times 10^{-20}