AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-3)
Unit 7 · Topic 7.3
7.3 Reaction Quotient and Equilibrium Constant
The reaction quotient Q is a ratio of product amounts to reactant amounts, each raised to the power of its coefficient. At equilibrium Q equals the equilibrium constant K. You can write it with molar concentrations (Qc, Kc) or, for gases, partial pressures (Qp, Kp), and you leave out pure solids and liquids.
Key terms
- reaction quotient (Q)
- equilibrium constant (K)
- law of mass action
- Kc
- Kp
- partial pressure
Writing Q and K
For a balanced equation aA + bB ⇌ cC + dD, the reaction quotient in terms of concentration is Qc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ). Square brackets mean molar concentration (mol/L). Products go on top, reactants on the bottom, and each coefficient becomes an exponent. This pattern is called the law of mass action.
Q can be calculated at any moment. As a reaction proceeds, Q changes and moves toward a fixed value. When the system is at equilibrium, Q = K, the equilibrium constant. So K is just Q at equilibrium.
For a given balanced equation, K depends only on temperature. Changing the starting amounts changes Q, not K. On the AP exam, K is written without units.
Kc versus Kp
For reactions involving gases, you can also use partial pressures (usually in atm): Kp = (P_C)ᶜ(P_D)ᵈ / ((P_A)ᵃ(P_B)ᵇ). For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kp = (P_NH₃)² / ((P_N₂)(P_H₂)³).
Kc and Kp for the same reaction usually have different numbers. You won't need to convert between them, but you must notice which one a question gives and use the matching kind of data (concentrations for Kc, pressures for Kp).
Leave out solids and pure liquids
The concentration of a pure solid or pure liquid doesn't change no matter how much of it there is, so it isn't included in Q or K. Water as the solvent in an aqueous reaction is left out for the same reason.
Gases and dissolved (aq) species are always included.
| Reaction | Expression |
|---|---|
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | Kp = P_CO₂ |
| AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) | K = [Ag⁺][Cl⁻] |
| CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) | K = [CH₃COO⁻][H₃O⁺] / [CH₃COOH] |
| Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s) | K = [Zn²⁺] / [Cu²⁺] |
Writing an expression step by step
- Start with a balanced equation that shows every phase.
- Cross out pure solids and pure liquids, including water when it's the solvent.
- Put the remaining products on top and reactants on the bottom, multiplied together.
- Turn each coefficient into an exponent.
- Remember the expression belongs to the equation as written. Doubling or reversing the equation changes K (topic 7.6).
What's out of scope
You won't be asked to convert Kc to Kp, or to do equilibrium calculations for a gas dissolving in a liquid (such as CO₂ dissolving in water).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Writing Kc and Kp
Write Kc and Kp for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), and write K for Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq).
Show the solutionHide the solution
- Step 1: Products over reactants, coefficients as exponents. NH₃ has coefficient 2; H₂ has 3.
- Step 2: Kc = [NH₃]² / ([N₂][H₂]³).
- Step 3: Kp = (P_NH₃)² / ((P_N₂)(P_H₂)³).
- Step 4: All species in the second reaction are aqueous, so all are included: K = [FeSCN²⁺] / ([Fe³⁺][SCN⁻]). There are no gases, so only Kc makes sense.
Answer: Kc = [NH₃]²/([N₂][H₂]³); Kp = (P_NH₃)²/((P_N₂)(P_H₂)³); K = [FeSCN²⁺]/([Fe³⁺][SCN⁻])
- Example 2Calculator allowed
Calculating Q at a moment
At one moment in a container, [N₂] = 0.20 M, [H₂] = 0.30 M and [NH₃] = 0.10 M. Calculate Qc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Show the solutionHide the solution
- Step 1: Qc = [NH₃]² / ([N₂][H₂]³).
- Step 2: Numerator: (0.10)² = 0.010. Denominator: (0.20)(0.30)³ = (0.20)(0.027) = 0.0054.
- Step 3: Qc = 0.010 ÷ 0.0054 = 1.9.
- Step 4: Q alone doesn't tell you the direction. Compare it with K at that temperature (topic 7.7).
Answer: Qc ≈ 1.9
- Example 3
Trap: solids, liquids and exponents
A student writes K = [CaO][CO₂] / [CaCO₃] for CaCO₃(s) ⇌ CaO(s) + CO₂(g). Correct the expression.
Show the solutionHide the solution
- Step 1: CaCO₃ and CaO are pure solids. Their concentrations don't change with amount, so they are left out.
- Step 2: Only the gas remains: Kc = [CO₂] or Kp = P_CO₂.
- Step 3: This means the CO₂ pressure at equilibrium is fixed at a given temperature, no matter how much solid is present (as long as both solids are there).
Answer: Kc = [CO₂] (or Kp = P_CO₂)
Common mistakes
- Including pure solids, pure liquids or solvent water in Q or K.
- Adding concentrations instead of multiplying them, or forgetting coefficients as exponents.
- Putting reactants on top. It's always products over reactants.
- Using concentrations when the question gives Kp, or pressures when it gives Kc.
On the exam
- Writing an equilibrium expression is a frequent first part of a longer free-response question. Getting the exponents and the omitted phases right is worth a point by itself.
- Read whether the question says Kc or Kp, and whether the data are concentrations or partial pressures.
Connected topics
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Check yourself
4 questions on 7.3 Reaction Quotient and Equilibrium Constant. Pick an answer to see if you got it, and why.
Which of the following is the equilibrium expression, Kc, for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)?
Which of the following is the expression for Kp for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g)?
Which of the following is the expression for Kc for the reaction 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)?
For 2NO(g) + O₂(g) ⇌ 2NO₂(g) at a certain temperature, an equilibrium mixture has P(NO) = 0.40 atm, P(O₂) = 0.20 atm and P(NO₂) = 0.80 atm. What is Kp?
0 of 4 answered