Construct and challenge an equilibrium quotient

One matching number can hide a wrong expression

Suppose every activity in a gas mixture is one. The products of the activities, their reciprocals and their squares are all one. Checking your equation only in this state cannot identify the right powers.

Start with the unequal-activity mixture. Build the quotient for N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Then switch to the all-one state, deliberately enter wrong powers, and inspect the difference between numerical agreement and structural correctness.

Build from the equation and phases

An activity is a dimensionless measure relative to a stated reference. Here an ideal gas uses its partial pressure divided by 1 bar; an ideal dissolved ion uses its concentration divided by 1 mol/L. These idealizations do not describe all real solutions or gases. We do not convert a concentration constant to a pressure constant.

Each product coefficient becomes a positive power; each reactant coefficient becomes a negative power. Thus the gas quotient is a(NH₃)² / (a(N₂) × a(H₂)³). At the assigned activities 0.10, 0.20 and 0.50 respectively, the numerator is 0.01 and the denominator is 0.025, giving Q=0.4. Reversing the reaction reverses every exponent and gives Q=2.5 at this same state.

The diagram places both your powers and the simplified reference on the same fixed −3 to +3 scale. Repair the powers individually; do not select a complete expression from a list.

A pure solid is different from an absent species

For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), the separate pure solid phase has activity one under this model. Its factor can be simplified away, leaving Q=a(Ca²⁺)a(F⁻)². At 0.02 and 0.04, Q=0.000032. Keeping 1 to a nonzero power is mathematically equivalent, though less clear than omitting it.

The two solid packets hold solution composition and temperature fixed while changing how much solid remains. The solid amount is not its activity. This does not say that finite solid inventory never matters: if a phase is exhausted, the conditional saturation and material constraints in the finite-solid investigation still apply.

A variable gas activity of zero has a different role. A zero numerator with a positive denominator gives Q=0. A positive numerator over a denominator tending to zero has a divergent positive limit. If numerator and denominator both vanish, the expression is undefined without specifying how the state is approached. No boundary case is silently replaced with a finite number.

Use Q without overstating the evidence

Q describes a composition and convention. Only at established equilibrium does it equal the corresponding K at that temperature. These hypothetical packets provide neither measured equilibrium nor K, so they do not justify a shift direction or reaction rate. Compare the historical measurement investigation for a separate measured-to-derived evidence chain.

Capture an expression that numerically agrees only in the all-one state and a corrected expression in the unequal state. Explain the failure of that single-state check. Then solve the unfamiliar three-species solution question in practice.

Source boundary

AP Chemistry Course and Exam Description, topic 7.3, printed 130–131 specifies quotient representations and pure-phase omission. Original packets and learner tasks implement these actions. Kc/Kp conversion and dissolved/gas partition calculations excluded by that topic are not added as requirements. Broader expression systems and independent experimental assessment remain open.

Transfer: pure liquid versus a dissolved liquid component

A separate pure liquid phase also has activity approximately one under the usual reference-state convention. For the idealized vaporization X(l) ⇌ X(g), with pure liquid present, Q = p(X)/p°. The liquid factor is one, not zero. At p(X) = 0.20 bar and p° = 1 bar, predict Q before checking: it is 0.20. Doubling the amount of already-present pure liquid does not directly double this quotient.

Now change the physical description: X is a component of a liquid mixture, not a separate pure phase. Its activity need not be one. For the same formal transfer, Q = a(X gas)/a(X liquid mixture), and composition or a suitable activity model is needed. Treating every substance labeled “liquid” as an omitted constant would be wrong. Likewise, if the pure phase is exhausted, a saturation-equality argument cannot assume it remains available.

Write the two expressions side by side and identify the extra evidence needed for the mixture case. This is a prose construction task, not a new branch of the solid/gas widget. Reference checked 30 October 2026: OpenStax equilibrium constants.