Chemical Equilibrium and Kc

At equilibrium:

• Rate of forward reaction = rate of reverse reaction
• Concentrations of reactants and products remain constant
• Both reactions are still occurring (dynamic, not static)
• The system must be closed (no matter enters or leaves)

For the general reaction: $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$

$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$

where concentrations are equilibrium concentrations in mol dm⁻³.

Key Points

• $K_c$ has a fixed value at a given temperature
• Only temperature changes the value of $K_c$
• Changing concentration, pressure, or adding a catalyst does NOT change $K_c$
• Large $K_c$: equilibrium favours products
• Small $K_c$: equilibrium favours reactants
• Pure solids and liquids are not included in the expression (their activity = 1)

Example: ICE Table Method

Question: 1.0 mol of ethanoic acid reacts with 1.0 mol of ethanol in a 1.0 dm³ flask. At equilibrium, 0.67 mol of ester is present.

$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$

Since volume = 1.0 dm³, concentrations = moles.

$K_c = \frac{(0.67)(0.67)}{(0.33)(0.33)} = \frac{0.449}{0.109} = 4.1$

For gaseous equilibria, $K_p$ uses partial pressures instead of concentrations.

Partial Pressure

$p_A = x_A \times P_{\text{total}}$

where $x_A$ = mole fraction of A = $\frac{\text{moles of A}}{\text{total moles}}$

For: $a\text{A}(g) + b\text{B}(g) \rightleftharpoons c\text{C}(g) + d\text{D}(g)$

$K_p = \frac{p_C^c \times p_D^d}{p_A^a \times p_B^b}$

Example

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, total pressure 100 kPa.

At equilibrium: 0.20 mol N₂O₄ and 0.80 mol NO₂. Total = 1.00 mol.

$x(\text{N}_2\text{O}_4) = 0.20$, $p(\text{N}_2\text{O}_4) = 0.20 \times 100 = 20$ kPa

$x(\text{NO}_2) = 0.80$, $p(\text{NO}_2) = 0.80 \times 100 = 80$ kPa

$K_p = \frac{(80)^2}{20} = \frac{6400}{20} = 320 \text{ kPa}$

The units depend on the stoichiometry:

• If total powers of products = total powers of reactants → no units
• Otherwise, work out the net power and apply concentration or pressure units

For $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$: units = $(\text{mol dm}^{-3})^{(c+d)-(a+b)}$

• Only include gases and aqueous species in $K_c$ expressions (not solids/pure liquids)
• Use an ICE table for calculations
• $K_c$ only changes with temperature
• Always state units for $K_c$ and $K_p$
• For $K_p$ questions, calculate mole fractions first, then partial pressures

• $K_c = \frac{[\text{products}]}{[\text{reactants}]}$ at equilibrium (products over reactants, each raised to stoichiometric power)
• $K_p$ uses partial pressures for gas-phase equilibria
• Only temperature changes $K_c$; concentration, pressure, and catalysts do not
• Large $K_c$ → products favoured; small $K_c$ → reactants favoured
• Le Chatelier predicts the direction of shift; $K_c$ tells you the new position