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) $[A]<em>i = 3.0 M$ and $[B]</em>i = 1.0 M$
  • b) $[A]<em>i = 1.0 M$ and $[B]</em>i = 4.0 M$
  • c) $[A]<em>i = 2.0 M$ and $[B]</em>i = 2.0 M$
• The initial concentrations are used to determine the direction of the shift toward equilibrium.
• The equilibrium constant $K$ is defined as: $K = \frac{[B]<em>e^2}{[A]</em>e} = 2.0$
• The reaction quotient $Q$ is defined as: $Q = \frac{[B]<em>i^2}{[A]</em>i}$
  • 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:
  $CH<em>3CO</em>2H + C<em>2H</em>5OH \rightleftharpoons CH<em>3CO</em>2C<em>2H</em>5 + H_2O$
• The equilibrium constant for this reaction is $K = 2.2$. Defined as: $K = \frac{[CH<em>3CO</em>2C<em>2H</em>5][H<em>2O]}{[CH</em>3CO<em>2H][C</em>2H_5OH]}$
• The direction the reaction must shift to reach equilibrium given initial concentrations is determined by comparing $Q$ and $K$.
• a) Given: $[CH<em>3CO</em>2C<em>2H</em>5] = 0.22 M$, $[H<em>2O] = 0.10 M$, $[CH</em>3CO<em>2H] = 0.010 M$, $[C</em>2H_5OH] = 0.010 M$
  $Q = \frac{(0.22)(0.10)}{(0.010)(0.010)} = 220$
  Since Q > K, the reaction will shift towards the reactants (left).
• b) Given: $[CH<em>3CO</em>2C<em>2H</em>5] = 0.88 M$, $[H<em>2O] = 0.12 M$, $[CH</em>3CO<em>2H] = 0.044 M$, $[C</em>2H_5OH] = 6.0 M$
  $Q = \frac{(0.88)(0.12)}{(0.044)(6.0)} = 0.40$
  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 $[A]<em>i = 60. M$ and $[B]</em>i = 60. M$?
• Calculate the equilibrium constant:
  $K = \frac{[B]<em>e}{[A]</em>e} = \frac{15}{10} = 1.5$
• Calculate the reaction quotient with the initial concentrations:
  $Q = \frac{[B]<em>i}{[A]</em>i} = \frac{60}{60} = 1.0$
• 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]<em>e = 60 - 12 = 48 M$$[B]</em>e = 60 + 12 = 72 M$

More Complicated Chemical Equilibrium Problem

• Reaction: $H<em>2(g) + F</em>2(g) \rightleftharpoons 2HF(g)$
• 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, $[H<em>2]</em>i = [F<em>2]</em>i = [HF]_i = 2.0 M$
• Reaction quotient: $Q = 1.0$
• Since Q < K, the reaction shifts right.
• ICE table:
  | | $H<em>2$ | $F</em>2$ | $2HF$ |
  |----------|-------|-------|--------|
  | 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:
  $K = \frac{[HF]^2}{[H<em>2][F</em>2]} = \frac{(2.0 + 2x)^2}{(2.0 - x)(2.0 - x)} = \frac{(2.0 + 2x)^2}{(2.0 - x)^2}$
• 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:
  $[H<em>2]</em>e = [F<em>2]</em>e = 2.0 - x = 2.0 - 1.53 = 0.47 M$
  $[HF]_e = 2.0 + 2x = 2.0 + 2(1.53) = 5.1 M$

Chemical Equilibria of Reactions With Multiple Steps

• Consider two reactions:
  $aA + bB \rightleftharpoons cC$ with equilibrium constant $K<em>1$$cC \rightleftharpoons dD + eE$ with equilibrium constant $K</em>2$
• Equilibrium constants:
  $K<em>1 = \frac{[C]^c}{[A]^a[B]^b}$$K</em>2 = \frac{[D]^d[E]^e}{[C]^c}$
• Overall reaction: $aA + bB \rightleftharpoons dD + eE$
• The equilibrium constant for the overall reaction, $K<em>3$, is the product of the individual equilibrium constants: $K</em>3 = \frac{[D]^d[E]^e}{[A]^a[B]^b} = K<em>1 \times K</em>2 = \frac{[C]^c}{[A]^a[B]^b} \times \frac{[D]^d[E]^e}{[C]^c}$
• 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:
  $K<em>{overall} = K</em>1^{-1} \times K<em>2 \times K</em>3^{-1} \times K<em>4 = \frac{K</em>2 \times K<em>4}{K</em>1 \times K_3} = \frac{(6.2 \times 10^{-10}) \times (1.2 \times 10^{-6})}{(1.8 \times 10^{-2}) \times (1.3 \times 10^{6})} = 3.2 \times 10^{-20}$