A Sealed N2O4/NO2 Flask: Whose Pressure Bookkeeping Is Right?
A sealed 2.00 L glass flask at 320 K is charged with 0.0200 mol of pure N2O4 and nothing else, then left until the colour stops deepening. The total pressure at that point is measured as 0.400 bar. The gas phase contains only N2O4 and NO2, and the equilibrium N2O4(g) ⇌ 2NO2(g) is established. A first student argues that because the flask was sealed and no gas entered or left, the number of gas moles must still be 0.0200, so the measured pressure is simply inconsistent with the ideal gas law. A second student disagrees. The mixture is then compressed isothermally to 1.00 L with a greased piston and a new equilibrium is awaited. Finally a third experiment is proposed: repeat the original loading in the same flask, but before sealing admit 0.010 mol of argon, keeping volume and temperature at 2.00 L and 320 K. Assume ideal-gas behaviour, and assume no reaction other than the stated equilibrium.
1. From the measured equilibrium pressure, calculate the degree of dissociation α and Kp for N2O4(g) ⇌ 2NO2(g) at 320 K. Was the first student right that the gas-phase mole count must still be 0.0200 mol? Show the mole bookkeeping explicitly.
Read the worked answer
Moles actually present. Treat the equilibrium mixture as ideal: n_total = PV/RT = (4.00 × 10^4 Pa)(2.00 × 10^-3 m^3)/(8.314 × 320) = 80.0/2660.5 = 3.01 × 10^-2 mol. That is about 1.5 times the 0.0200 mol charged, so roughly 0.0100 mol of new gas particles has appeared. The first student is wrong: sealing conserves N and O atoms, not molecules, because one N2O4 gives two NO2.
Mole table. Let α be the fraction dissociated. n(N2O4) = 0.0200(1 − α); n(NO2) = 0.0400α; total = 0.0200(1 + α) = 0.0301 mol, hence α ≈ 0.50.
Partial pressures. x(N2O4) = (1 − α)/(1 + α) = 0.4965/1.5035 = 0.330; x(NO2) = 0.670. So p(N2O4) = 0.132 bar and p(NO2) = 0.268 bar.
Kp = p(NO2)^2/p(N2O4) = (0.268)^2/0.132 = 0.54 bar (i.e. 0.54 relative to p° = 1 bar).
Cross-check: Kp = 4α^2·p_total/(1 − α^2) = 4(0.2535)(0.400)/(0.7465) = 0.54 bar, in agreement.
The usual trap is to convert the equilibrium moles to mol L^-1 and substitute those numbers into the Kp expression. That produces a Kc, not a Kp, and mixing the two is the commonest error here.
2. The mixture is compressed isothermally to 1.00 L. Predict the direction of shift using the reaction quotient immediately after compression, then calculate the new α. Is the result consistent with Le Chatelier's principle?
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What happens the instant the piston stops. Halving the volume doubles both partial pressures: p(NO2) = 0.536 bar, p(N2O4) = 0.264 bar. Q = (0.536)^2/0.264 = 1.09, which exceeds Kp = 0.54, so the net change is N2O4-forming; α must fall.
Quantitative treatment. Define the reference pressure p0 = n(N2O4,initial)·RT/V = (0.0200 × 2660.5)/1.00 × 10^-3 = 5.32 × 10^4 Pa = 0.532 bar. Then p(NO2) = 2α(0.532) and p(N2O4) = (1 − α)(0.532), so
Kp = 4α^2(0.532)/(1 − α) = 0.54
which gives 2.13α^2 + 0.54α − 0.54 = 0, i.e. α = (−0.54 + √(0.29 + 4.60))/4.26 = (−0.54 + 2.22)/4.26 ≈ 0.39.
Interpretation. α falls from about 0.50 to about 0.39, so the amount of NO2 falls even though its concentration rises. This agrees with Le Chatelier: on compression the system moves toward the side with fewer gas molecules. Note what Le Chatelier does not give - it predicts direction only. The extent has to be computed, and the shift is partial, not a complete reversal.
3. Does adding 0.010 mol of argon before sealing shift the position of equilibrium? Answer for (a) the sealed 2.00 L flask at fixed volume and temperature, and (b) the same amount of argon added while the gas is held at a fixed total pressure by a freely moving piston.
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(a) Constant V and T. The argon raises the total pressure (from about 0.40 bar to roughly 0.53 bar if added to the equilibrated mixture) but does not change p(N2O4) or p(NO2): ideal gases do not interact, and each gas's partial pressure is set by its own mole number, volume and temperature. Kp contains only the partial pressures of species appearing in the balanced equation, and Q is unchanged, so there is no shift. A student who writes 'total pressure rose, so the equilibrium shifts to the side with fewer gas molecules' has confused total pressure with partial pressure.
(b) Constant total pressure. Here the piston must move out to accommodate the argon, so the reacting mixture is diluted: both partial pressures fall in the same proportion, say by factor f < 1. Then Q = (f·p(NO2))^2/(f·p(N2O4)) = f·Kp < Kp, so Q < K and the reaction proceeds forward. Dissociation of N2O4 is promoted.
General rule worth memorising. For Δn_gas > 0, an inert gas at constant pressure increases the degree of dissociation; at constant volume it has no effect whatsoever. The distinction is a standard discriminator in this type of question.