Chemistry Lecture Notes - Dynamic Equilibrium, Kc, and Kp

Chapter 16 - Part 1

CHE 112 - Spring 2026

Book Sections 16.1 - 16.3

Dynamic Equilibrium

• **Dynamic equilibrium: **

  • Discusses reaction rates (refer to Chapter 14).

  • Constant concentration of reactants and products.

• Key Characteristics:

  • Reversible reactions.

  • Rate of the forward reaction equals the rate of the reverse reaction.

    • Example: $N_2O_4(g) <br>ightleftharpoons 2NO_2(g)$

  • At equilibrium:

    • Concentrations of reactants and products are constant.

    • The concentrations are not necessarily equal to each other.

Equilibrium Constant Expression

• Concentration Notation:

  • Square brackets [ ] denote concentration.

• Definition:

  • Equilibrium constant expression - Always a ratio of concentrations of products over reactants at equilibrium.

• Variables:

  • Lowercase letters represent coefficients (e.g. a, b, c, d).

  • Uppercase letters represent species (e.g. A, B, C, D).

• General Equation:

  • $aA + bB ightleftharpoons cC + dD$

    • $K_c = rac{[C]^c[D]^d}{[A]^a[B]^b}$

  • $K_c$ - equilibrium constant (unitless).

Equilibrium Expressions

• Inclusion Criteria:

  • Only include chemicals in aqueous and gaseous phases in the equilibrium expression.

  • Solids (s) and liquids (l) do not appear in the equilibrium expression and are incorporated into the value of K.

• Example Reaction:

  • $Ca^{2+}(aq) + 2 HCl_{(aq)} ightleftharpoons CaCl_2(s) + CO_2(g) + H_2O(l)$

    • $K_c = rac{[CO_2]}{[Ca^{2+}][HCl]^2}$

    • Note: solid CaCl2 and water are excluded from K value.

Writing Equilibrium Expressions

• Exercises: (Write equilibrium expressions for the following reactions)

  • Reaction: $HF_{(aq)} + H_2O_{(l)} <br>ightleftharpoons H_3O^+<em>{(aq)} + F^-</em>{(aq)}$

  • Reaction: PbCl_2_{(s)}
    ightleftharpoons Pb^{2+}{(aq)} + 2Cl^{-}{(aq)}

Gas Equilibrium Expressions

• Gas Moles Change:

  • Equilibrium expressions can be formulated in terms of concentrations $K_c$ or partial pressures $K_p$ for gas systems.

• Example Reaction:

  • $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$

    • Expressions:

      • $K_c = rac{[PCl_3][Cl_2]}{[PCl_5]}$

      • $K_p = rac{P_{PCl_3} imes P_{Cl_2}}{P_{PCl_5}}$

• Calculation Relation:

  • $K_p = K_c R T^{ riangle n_g}$

    • Where:

      • $riangle n_g = n_{gas, products} - n_{gas, reactants}$

      • R - gas constant $= 0.0821 rac{L atm}{mol K}$

      • T - Temperature in Kelvin.

• Special Case:

  • If $riangle n_g = 0$ (equal moles on both sides), then:

    • $K_p = K_c$

Example Problems

Calculate $K_p$ from known $K_c$

• Reaction:

  • $CO(g) + 2 H_2(g) <br>ightleftharpoons CH_3OH(g)$,

  • given $K_c = 57.32$ at 200 °C.

• Calculation Steps:

  • Calculate $riangle n_g$:

    • $riangle n_g = 1 - 3 = -2$.

    • Use: $T = 200 + 273 = 473 K$

  • Apply $K_p$ formula:

    • $K_p = K_c R T^{ riangle n_g}$

    • $K_p = 57.32 imes 0.0821 imes 473^{-2} = 3.80 imes 10^{-2}$.

Understanding K Values

• K Significance:

  • Depending on the value of K:

    • When $K ext{~} 1$, products ≈ reactants at equilibrium.

    • When K ext{~} > 1, products favored (more products).

    • When K ext{~} < 1, reactants favored (more reactants).

  • Range of K:

    • $K$ values vary widely usually from $10^{100}$ to $10^{-100}$ and can be larger or smaller.

    • For 10^{-2} < K < 10^2, significant amounts of both products and reactants are present.

Reaction Quotient (Q)

• Q Comparison:

  • Same form as equilibrium expression.

• Comparison Before Equilibrium:

  • At equilibrium, $Q = K$.

  • If Q > K, there are too many products, the reaction shifts left (towards reactants).

  • If Q < K, there are not enough products, the reaction shifts right (towards products).

• Reaction Quotient Formula:

  • $Q = rac{[products]}{[reactants]}$.

  • Determine Q by using current concentrations or pressures.

• Example Checking for Equilibrium:

  • Given: $[H_3O^+] = [CH_3COO^-] = 0.30 M$ and $[CH_3COOH] = 2.0 M$, Calculate Q:

    • $Q_a = rac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]} = rac{0.30 imes 0.30}{2.0} = 0.045 = 4.5 imes 10^{-2}$.

Factors Affecting Equilibrium

Temperature Dependence

• Effect of Temperature:

  • Changing temperature affects equilibrium constants.

    • Temperature is given for information purposes and is not always needed for calculations.

• Checking Equilibrium:

  • Example: $[H_2] = [I_2] = 0.085 M$, $[HI] = 0.35 M$

  • Reaction: $H_2(g) + I_2(g) <br>ightleftharpoons 2HI(g)$, with $K_c = 54$ at 430 °C.

• Next Steps:

  • Determine if the reaction is at equilibrium, if not state which way it will shift.

Le Châtelier’s Principle

• Principle Definition:

  • If a system at equilibrium is disturbed, it shifts to negate the stress.

• Types of Disturbance:

  • Changing concentration of reactants/products.

  • Changing volume or pressure in a gaseous system.

Changing Concentration and its Effects

• Increasing or decreasing concentration leads to a shift:

  • Example: For the reaction $N_2(g) + 3 H_2(g) ightleftharpoons 2 NH_3(g)$:

    • Increase $N_2$: shifts right (towards products).

    • Increase $NH_3$: shifts left.

    • Decrease $H_2$: shifts left.

    • Decrease $NH_3$: shifts right.

Changing Temperature Effects

• Forward Reaction Types:

  • Exothermic Reaction:

    • riangle H < 0 (heat is released)

      • Increase in T shifts left (toward reactants).

      • Decrease in T shifts right (toward products).

  • Endothermic Reaction:

    • riangle H > 0 (heat is absorbed)

      • Increase in T shifts right (towards products).

      • Decrease in T shifts left (towards reactants).

Changing Pressure Effects

• Impact on Gaseous Systems:

  • Pressure changes shift equilibrium:

    • Decrease in volume (pressure increases): shifts towards less moles of gas.

    • Increase in volume (pressure decreases): shifts towards more moles of gas.

  • Note: The value of K remains unchanged with pressure changes.

Changing Pressure Examples

• When adding an inert gas (like helium):

  • Total pressure increases but partial pressures do not change thus it does not shift equilibrium.

Calculating K Values

• Knowing equilibrium concentrations (or partial pressures) is essential for finding K.

• ICE Table Methodology:

  • Requires initial concentrations, changes, and equilibrium concentrations.

  • Setup:

    • Initial: Set known concentrations (e.g. $N_2 = 0.515, H_2 = 0.282, NH_3 = 0$)

    • Change: Represent decreases/increases (e.g. -x, -3x, +2x)

    • Equilibrium: Solve for equilibrium concentrations.

Example Problem Using Kc

• Given concentrations at equilibrium:

  • $N_2 + 3H_2 ightleftharpoons 2NH_3$:

    • $K_c = rac{[NH_3]^2}{[N_2][H_2]^3}$,

    • Using determined equilibrium values: $N_2 = 0.500 M, H_2 = 0.237 M, NH_3 = 0.030 M$;

    • Calculate:
      $K_c = rac{(0.030)^2}{(0.500)(0.237)^3} = 0.135$.

Continuing Calculations for Equilibrium Concentrations

Example Reaction: Using $SO_3 + NO</h5><p>ightarrow NO_2 + SO_2$

• Given quantities allow setup of equilibrium expression using the previous format.

## End of Part 1

Chapter 16 - Part 2

Book Sections 16.6, 16.2

Le Châtelier’s Principle Revisited

• When stress is placed on a system in equilibrium, the system shifts to reduce the stress.

• Sources of Stress Include:

  • Change in concentrations of reactants or products.

  • Change in temperature.

  • Change in volume/pressure in a gaseous system.

Changing Concentration

• Modifying Concentrations:

  • Increase reactants: Shift equilibrium left.

  • Decrease reactants: Shift equilibrium right.