What you'll be able to do: Calculate Q for any mixture and use the comparison with K to predict which way the reaction will proceed.
Give a chemist a flask of reactants and products in arbitrary amounts and they can tell you which way it will go, without a rate law and without waiting. The tool is Q.
These are recommended, not required. You can start this lesson at any time.
Q is built from the identical products-over-reactants expression, with the same exponents and the same exclusions for pure solids and liquids. The only difference is that Q uses current concentrations while K uses equilibrium ones.
Compute Q for the mixture immediately after a change but before substantial reaction. If 10.0 mL of 0.200 M A is mixed with 30.0 mL of a solution containing no A, its initial mixed [A] is (0.200)(0.0100)/0.0400 = 0.0500 M, assuming additive volumes. Stock concentration is not the concentration to put into Q. Use the same expression, phases and pressure/concentration convention as K, at the same temperature.
Q = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ) at any instant
If Q is smaller than K the product terms are too small, so the reaction must make more product and proceeds to the right. If Q is larger than K the reaction proceeds to the left.
If Q equals K there is no net composition change, although both directions continue.
Q moves continuously towards K as the reaction proceeds. Making product raises the numerator and lowers the denominator, so Q always increases when the reaction runs forward. The system therefore travels in whichever direction closes the gap.
For a forward-defined reaction, Q < K means forward conversion lowers Gibbs energy; Q > K means reverse conversion does. At Q = K there is no net conversion, but forward and reverse molecular events continue. The rule predicts direction, not how far the system moves or how quickly. To find final concentrations, follow it with a stoichiometric material balance such as an ICE table.
For an ordinary homogeneous A ⇌ B mixture, no initial B gives Q = 0; no A with B present makes the quotient diverge. Those cases support forward and reverse directions respectively, if the pathway is accessible. They are not universal rules for every phase combination: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), the pure solids are excluded, and Q depends on CO₂ pressure. If a required phase is absent, consider whether it can form before assuming the stated heterogeneous equilibrium applies.
Avoid evaluating undefined 0/0 expressions by calculator. Identify which species are initially present and use stoichiometry to decide an allowed initial direction. An overwhelmingly favorable direction may still be kinetically slow.
For a dissolving salt the same quotient is written from the ion concentrations and compared with Ksp. If Q exceeds Ksp the solution is supersaturated and a precipitate forms; below Ksp everything stays dissolved. This is the standard way to answer will a precipitate form.
For AgCl(s) ⇌ Ag⁺ + Cl⁻, Qsp = [Ag⁺][Cl⁻] in the dilute model. Equal volumes of 2.00 × 10⁻⁴ M AgNO₃ and NaCl give 1.00 × 10⁻⁴ M of each ion immediately after mixing, so Qsp = 1.00 × 10⁻⁸. With Ksp = 1.80 × 10⁻¹⁰, Qsp > Ksp and precipitation is favored. Conversely Qsp < Ksp means no precipitation from those ions; if solid is already present, more can dissolve. At Qsp = Ksp the solution is saturated. A supersaturated solution may persist until nucleation occurs, so equilibrium direction does not guarantee an instantaneous visible cloud.
An instantaneous concentration change changes the reaction quotient Q, while K stays fixed if temperature stays fixed. For A ⇌ B, adding B raises Q=[B]/[A] above K and the system responds by favoring the reverse reaction.
The final composition is new, but the equilibrium ratio returns to the same K.
The products-over-reactants expression evaluated with the concentrations present at a given moment.
The reaction proceeds forward, making more product.
The reaction proceeds in reverse, making more reactant.
The system is at equilibrium with no net change.
Comparing Q with Ksp tells you whether a solid will appear.
Q = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
For H₂(g) + I₂(g) ⇌ 2 HI(g), K = 50.0. A flask holds [H₂] = 0.20 M, [I₂] = 0.20 M, [HI] = 0.40 M. Which way does the reaction proceed?
Try it first: Write the quotient with the numbers you were given before comparing anything.
0 of 4 steps revealed.
For N₂O₄(g) ⇌ 2 NO₂(g), Kp = 0.15 atm. A vessel contains P(N₂O₄) = 0.10 atm and P(NO₂) = 0.50 atm. Predict the direction.
Try it first: Note that Kp is used with pressures, so Q must be built from pressures too.
0 of 3 steps revealed.
Equal volumes of 0.0020 M Pb(NO₃)₂ and 0.0020 M NaCl are mixed. Ksp for PbCl₂ is 1.7 × 10⁻⁵. Does a precipitate form?
Try it first: Remember that mixing equal volumes halves every concentration.
0 of 4 steps revealed.
A ⇌ B has K=2.0. At equilibrium [A]=1.0 M and [B]=2.0 M. Immediately after adding B to 3.0 M, which direction is favored?
Try it first: Calculate Q from the new, not old, concentrations.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Using K in place of Q for a non-equilibrium mixture. | K only describes the equilibrium state; the current mixture may be far from it. | Evaluate Q with the given concentrations and then compare. |
| Getting the direction backwards when Q < K. | A small Q means too little product, so product must be made. | Ask which side needs to grow for Q to rise to K. |
| Forgetting dilution when two solutions are mixed. | Every concentration changes on mixing, often by a factor of two. | Recalculate each concentration for the combined volume first. |
| Including a solid when writing Q. | Q follows exactly the same exclusion rules as K. | Drop every (s) and (l) term. |
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After the lesson · Question bank
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The reaction quotient Q has exactly the same algebraic form as K, but is evaluated with whatever concentrations or pressures the mixture happens to have right now. Comparing Q with K predicts the direction of change: Q < K means there is too little product, so the reaction runs forward; Q > K means too much product, so it runs in reverse; Q = K means the system is already at equilibrium. Because Q and K are computed the same way, any error in the expression affects both, so writing the expression carefully once solves both problems. Q is also the tool used to decide whether a precipitate forms in a solubility problem.
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