Distinguish fixed-volume and fixed-pressure equilibrium disturbances

Try: total pressure is not the reaction quotient

The original ideal model starts with D(g) ⇌ 2M(g), one mole D and two moles M in 24 L. D contains two conserved M units. Temperature is fixed so RT/p° = 24 L/mol, with standard pressure p° = 1 bar. Initial partial pressures are 1 and 2 bar. The assigned dimensionless equilibrium constant is K = 4, matching Q = (pM/p°)²/(pD/p°).

Compress to 12 L before chemical response. Both partial pressures double, but the numerator is squared: Q becomes 8, not 4. The reverse reaction associates M into D until Q returns to 4. The ledger separates the immediate disturbance from this later equilibrium calculation.

Predict: two ways to add the same inert gas

Add three moles inert I at fixed volume. The amounts of D and M and their volume have not changed, so their partial pressures remain 1 and 2 bar. Total pressure rises from 3 to 6 bar, yet Q remains 4 and there is no shift. The rule “higher total pressure means fewer gas molecules” cannot be applied without knowing what caused the pressure change.

Now add the same three moles inert gas at fixed total pressure, 3 bar. The vessel expands immediately from 24 to 48 L. The reacting partial pressures halve; Q becomes 2. Predict the direction before inspecting the final state.

Experiment: keep the constraint throughout the reaction

At fixed total pressure the forward reaction forms more gas moles, so volume continues increasing as equilibration proceeds. The endpoint has signed dissociation extent x = 0.25 mol: D = 0.75 mol, M = 2.50 mol, inert I = 3 mol, and volume = 50 L. The final partial pressures are 0.36, 1.20 and 1.44 bar. Their sum is 3 bar and Q = 1.20²/0.36 = 4.

Holding volume at the immediate 48 L would solve a different experiment. Compare both inert modes across several added amounts. Then compare volume changes with adding D or M at fixed volume. Capture the immediate Q and final conserved-unit ledger for each explanation.

Understand: amounts and constraints enter the quotient

After the disturbance, let nD,0 and nM,0 be the starting amounts for chemical response, and nI the unchanged inert amount. At signed dissociation extent x, D has nD,0 − x moles and M has nM,0 + 2x. Hence 2nD + nM is conserved during re-equilibration, though adding D or M changes it relative to the original batch.

At fixed volume V, substitute pi/p° = niRT/(Vp°) into Q. Solve (RT/(Vp°))(nM,0 + 2x)² = K(nD,0 − x). At fixed total pressure P, use mole fractions: pi = Pni/(nD + nM + nI). The corresponding equation is (P/p°)(nM,0 + 2x)² = K(nD,0 − x)(nD,0 − x + nM,0 + 2x + nI). The solver brackets physically allowed amounts and performs a bounded root calculation; no reaction rate is assumed.

Apply: a stoichiometric counterexample

Complete the practice for X(g) ⇌ Y(g). At fixed pressure, inert addition dilutes both reacting partial pressures equally. Because the quotient contains one power of each, the factors cancel. A volume change does not shift every equilibrium. Explain the role of the equation’s gas stoichiometry rather than memorizing a direction.

Continue: what the states do not establish

The model is hypothetical and ideal, not measured gas chemistry. Temperature and K stay fixed; no response to temperature change can be computed without additional information. Mechanical/thermal control is assumed, but heat, work, kinetics, nonideal interactions and apparatus transients are not modeled. The three displayed stages are not timed animation frames.

The earlier composition constructor separates inventory from equilibrium. AP Chemistry CED topics 7.9–7.10 connect disturbances, Q and K. OpenStax Chemistry 2e 13.3 explains partial-pressure and volume effects. The inert-gas comparisons here are derived explicitly from the ideal-gas relations and specified constraints, rather than presented as empirical observations.