The Equilibrium Constant Kc
For the reaction: \( aA + bB \rightleftharpoons cC + dD \)
The Equilibrium Constant Expression (Kc)
\( K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b} \)
For the generalised reaction \(aA + bB \rightleftharpoons cC + dD\). Only aqueous and gaseous species appear in \(K_c\) expressions; pure solids and pure liquid solvents are omitted.
Physical Interpretation of the Magnitude of K
- \(K \gg 1\) (\(K > 10^2\)): Products are strongly favoured at equilibrium (reaction almost complete).
- \(K \ll 1\) (\(K < 10^{-2}\)): Reactants are strongly favoured at equilibrium (barely proceeds).
- \(K \approx 1\): Reactants and products are present in roughly comparable amounts.
Worked Example: Calculating Kc
Calculating Kc from Equilibrium Concentrations
For the reaction: \(\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})\) at \(448^\circ\text{C}\)
Equilibrium concentrations: \([\text{H}_2] = 0.46\text{ M}\), \([\text{I}_2] = 0.39\text{ M}\), \([\text{HI}] = 3.0\text{ M}\).
Step 1: Write the \(K_c\) expression
\(K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\)
Step 2: Substitute and calculate
\(K_c = \dfrac{(3.0)^2}{(0.46)(0.39)} = \dfrac{9.0}{0.1794} = \mathbf{50.2}\)
Conclusion: Because \(K_c \gg 1\), products (\(\text{HI}\)) predominate at \(448^\circ\text{C}\).
Kp (HL Only)
The Partial Pressure Equilibrium Constant (Kp)
\( K_p = \dfrac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b} \)
Where \(p_i\) represents the equilibrium partial pressure of gaseous component \(i\) (in \(\text{kPa}\) or \(\text{bar}\)).
Mole Fraction and Dalton's Law of Partial Pressures
Mole fraction (\(x_A\)): \( x_A = \dfrac{n_A}{n_{\text{total}}} \) | The sum of all mole fractions: \( \sum x_i = 1 \).
Dalton\'s Law: \( p_A = x_A \times P_{\text{total}} \)
Adding Reactant Does Not Change Kc
Adding \(\text{N}_2\) momentarily makes \(Q < K_c\), causing the system to shift forward to produce more \(\text{NH}_3\). When the new equilibrium state is established, the ratio \(\dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\) returns to the exact same value of \(K_c\).
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