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.
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 nothing changes.
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.
A mixture of only reactants has Q = 0, which is below any positive K, so the reaction must go forward. A mixture of only products has Q infinite, so it must go in reverse. These limiting cases are a quick sanity check on any Q calculation.
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.
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.
Why it's wrong: K only describes the equilibrium state; the current mixture may be far from it.
Check instead: Evaluate Q with the given concentrations and then compare.
Why it's wrong: A small Q means too little product, so product must be made.
Check instead: Ask which side needs to grow for Q to rise to K.
Why it's wrong: Every concentration changes on mixing, often by a factor of two.
Check instead: Recalculate each concentration for the combined volume first.
Why it's wrong: Q follows exactly the same exclusion rules as K.
Check instead: Drop every (s) and (l) term.
No practice questions are available for this topic yet. You can still practice the whole unit.
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.
This lesson is original Chem Help content. No external sources were adapted.