Equilibrium Review Guide

Definition of Equilibrium: Equilibrium in a chemical reaction occurs when the rate of the forward reaction equals the rate of the reverse reaction.
- Behavior of Concentrations: At equilibrium, the concentrations of reactants and products remain constant over time.
- Does the Reaction Stop? No, the reaction does NOT stop at equilibrium; instead, both forward and reverse reactions continue to occur at equal rates, maintaining constant concentrations.

Equilibrium Reaction Example: For the reaction 2 NOCl (g) + heat ⇌ Cl2 (g) + NO (g)

Change 1:
- Observation: No sharp increase or decrease in concentration.
- Possible Change: This change indicates a shift due to a change in heat or temperature, rather than pressure since the number of moles of gas on both sides is equal (2 moles).
- Detailed Explanation: The concentrations of NO and Cl2 increase while NOCl decreases, indicating the forward reaction is occurring due to added heat, shifting towards the products.
Change 2:
- Observation: There is a sharp increase in concentrations of all species (NOCl, Cl2, and NO).
- Possible Change: This indicates an increase in the concentrations of reactants and products, prompting the reverse reaction to occur after the increase.
Change 3:
- Observation: A sharp decrease in Cl2 concentration.
- Possible Change: The removal of Cl2 caused the forward reaction to offset the loss, as indicated by the graph where NOCl decreases and Cl2 and NO increase.
Equilibrium Points: Draw vertical lines to indicate the four points on the graph where the system was at equilibrium before and after changes.

For the reaction 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g):
- Equilibrium concentrations:
- [SO2] = 0.20 M
- [O2] = 0.20 M
- [SO3] = 0.40 M

Keq Expression: The equilibrium constant (K) is expressed as:
$K_{eq} = \frac{[products]}{[reactants]}$

Example Calculation 1:
- For the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
  - Given: [PCl5] = 0.058, [PCl3] = 0.042, [Cl2] = 0.042
  - Calculation:
    $K = \frac{[0.042][0.042]}{[0.058]} = 0.0304$
- Interpretation: Since K < 1, reactants are favored at equilibrium.
Example Calculation 2:
- For the reaction 2NO (g) + 2H2(g) ⇌ N2(g) + 2H2O(g)
  - Given concentrations:
  - [NO] = 0.062
  - [H2] = 0.012
  - [N2] = 0.019
  - [H2O] = 0.138
  - Calculation:
    $K = \frac{[0.019][0.138]^2}{[0.062]^2[0.012]^2} = 654$
- Interpretation: Since K > 1, products are favored at equilibrium.
Finding Concentration Example:
- Calculate [H2] given:
  - [NO] = 0.102, [H2O] = 0.329, [N2] = 0.138
  - $K = 654 = \frac{[0.138][0.329]^2}{[0.102]^2[x]^2}$
  - Solving gives: $x = 0.047 M = [H2]$
Further Equilibrium Calculations:
- For N2(g) + 3H2(g) ⇌ 2NH3(g):
  - Given: K = 3.5 x 10^{-5}, [N2]= 0.50 M, [H2] = 1.00 M
  - Concentration of NH3 at equilibrium:
  - $K = 3.5 x 10^{-5} = \frac{x^2}{[0.50][1.00]^3}$
  - Solving gives: $x = 0.0042 M = [NH3]$
Final Equilibrium Calculation:
- For CO(g) + H2O(g) ⇌ CO2(g) + H2(g):
  - Given K = 2.4 x 10^{-3}, [CO] = 0.10 M, [H2O] = 0.20 M, [H2] = 0.33 M
  - Concentration of CO2 at equilibrium:
  - $K = 2.4 x 10^{-3} = \frac{x}{[0.33][0.10][0.20]}$
  - Solving gives: $x = 0.000145 M = [CO2]$

Reaction Example: 2CH4 (g) ⇌ C2H2 (g) + 3H2(g), Keq = 0.154
- Determine direction for starting concentrations of 2.60M CH4, 1.05M C2H2, and 5.00M H2.
Reaction Example: H2(g) + I2(g) ⇌ 2HI(g)
- Given Keq = 50.2 and initial concentrations [H2]=0.3M, [I2]=0.04M, [HI]=0.26M. Determine shift direction.
K Value Interpretation: If the value of K = 8.1 for the reaction of H2 + O2 ⇌ H2O,
- Interpretation: Since K > 1, products are favored.

Example Q Calculation 1:
- For the first reaction: Q = 19.42
  - $Q = \frac{[1.05][5.00]^3}{[2.60]^2}$
  - Direction: Since Q > K, the reaction will shift left to produce more reactants to reach equilibrium.
Example Q Calculation 2:
- For the second reaction: Q = 5.63
  - $Q = \frac{[0.26]^2}{[0.3][0.04]}$
  - Direction: Since Q < K, the reaction will shift right to produce more products until equilibrium is reached.

Equilibrium Definition: A reaction is at equilibrium when reactant and product concentrations are constant, with equal rates of forward and reverse reactions.
Le Chatelier's Principle: If an external change (stress) is applied to a system at equilibrium, the equilibrium will shift in a direction that counteracts this stress, reestablishing equilibrium.

Three Factors:
- Changes in concentration
- Changes in temperature
- Changes in pressure or volume
Effect of Increased Reactant Concentration: If the concentration of PCl5 is increased:
- Change Direction: Reaction shifts to the right (toward products) to reestablish equilibrium.
- Effect on Cl2 Amount: Amount of Cl2 increases as the system adjusts.
Effect of Increased Pressure:
- Direction of Shift: System shifts toward fewer moles of gas.
  - Reactant side has 1 mole, product side has 2 moles.
  - Therefore, the reaction will shift left (toward reactants) due to increased pressure.
Effect of Adding N2:
- Direction of Shift: Reaction shifts left (toward reactants) if N2 (a product) concentration is increased.
Heat in Exothermic Reaction:
- In an exothermic reaction, heat is treated as a product; thus, it should be written on the right side of the equation.
- Effect of Adding Heat: Adding heat will shift the reaction left (towards reactants) to counter the increase in product concentration.
Effect of Increasing Volume:
- Increasing the volume of the container decreases pressure, causing the equilibrium to shift towards the side with more moles of gas.
- If reactant side has 4 moles and product side has 3 moles, this will shift equilibrium left, decreasing water H2O production.