The Equilibrium Constant Kc
For the reaction: \( aA + bB \rightleftharpoons cC + dD \)
\( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)
Only aqueous and gaseous species are included. Solids and pure liquids are omitted.
Interpreting K
- K >> 1 → products are favoured (equilibrium lies to the right)
- K << 1 → reactants are favoured (equilibrium lies to the left)
- K ≈ 1 → neither side is strongly favoured
Worked Example: Calculating Kc
H₂(g) + I₂(g) ⇌ 2HI(g) at 448°C
[H₂] = 0.46, [I₂] = 0.39, [HI] = 3.0 mol dm⁻³
\( K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3.0)^2}{(0.46)(0.39)} \)
\( K_c = \frac{9.0}{0.179} = 50.2 \)
K >> 1, so products are strongly favoured at this temperature.
Kp (HL Only)
\( K_p = \frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b} \)
where p = partial pressure. \( p_A = x_A \times P_{total} \)
Mole Fraction & Partial Pressure
\( x_A = \frac{n_A}{n_{total}} \) and \( p_A = x_A \times P_{total} \)
The sum of all mole fractions = 1. The sum of all partial pressures = total pressure.
Think About It
If you add more N₂ to the Haber process at constant volume, what happens to Kc?
Kc does not change - it only depends on temperature. The system will shift towards products until the concentrations again satisfy the Kc expression, but the value of Kc itself remains the same.
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