Equilibrium Shift in a Δn = 0 Gas Reaction
In a 2.00 L rigid flask held at 400 K, 1.00 mol of A(g) and 1.00 mol of B(g) are mixed with 0.50 mol of C(g). The only reaction is A(g) + B(g) ⇌ 2C(g), for which Kc = 4.00 at this temperature. A student evaluates the reaction quotient using moles directly, obtains Q = 0.25, and compares it with Kc to conclude that the mixture must shift right. He then plans to increase the yield by compressing the mixture into a 1.00 L flask, arguing that higher pressure always favours the side with fewer gas molecules, and he expects Kc itself to change when the volume changes. A second student objects that for this particular reaction the value of Q is numerically the same whether moles or concentrations are used, and that the pressure argument must be checked against Δn before it is applied.
1. Establish the direction of shift and calculate all equilibrium concentrations.
Read the worked answer
Convert to concentrations (V = 2.00 L): [A] = [B] = 0.500 M, [C] = 0.250 M.
Q = [C]²/([A][B]) = 0.250²/(0.500 × 0.500) = 0.250 < Kc = 4.00 → net reaction is forward.
Let x mol/L of A react: [A] = [B] = 0.500 − x, [C] = 0.250 + 2x.
(0.250 + 2x)²/(0.500 − x)² = 4.00 → (0.250 + 2x)/(0.500 − x) = 2.00 (take the positive root) 0.250 + 2x = 1.000 − 2x → x = 0.1875 M
Equilibrium: [A] = [B] = 0.3125 M, [C] = 0.625 M. Check: 0.625²/0.3125² = 4.00. Moles: 0.625, 0.625, 1.25. The student's arithmetic was accidentally right because Δn = 0 makes Q identical in moles or in molarity.
2. Predict what happens when the mixture is compressed to 1.00 L, and whether Kc changes.
Read the worked answer
Δn = 2 − 2 = 0, so pressure has no lever here. Halving the volume doubles every concentration:
Q = (2[C])²/((2[A])(2[B])) = [C]²/([A][B])
The factor 4 cancels, so Q is still 4.00 = Kc and there is no shift. Amounts remain 0.625/0.625/1.25 mol; concentrations become 0.625/0.625/1.25 M. Kc is unchanged because it depends only on temperature.
The 'fewer gas molecules' rule is inapplicable, not reversed: with equal counts on both sides it predicts no direction at all.
3. At the original equilibrium, 0.25 mol of C is injected at constant volume. Which way does the system respond?
Read the worked answer
New [C] = (1.25 + 0.25)/2.00 = 0.750 M, while [A] = [B] = 0.3125 M.
Q = 0.750²/0.3125² = 0.5625/0.09766 ≈ 5.76 > Kc = 4.00
Q exceeds Kc, so the reverse reaction dominates until Q falls back to 4.00: some C is consumed and A and B are regenerated. The system does not simply 'use up the added C' - it restores the ratio, so the final [C] lies between the old and the injected value.