The equilibrium-constant expression
For aA + bB ⇌ cC + dD at equilibrium at a given temperature:
Products go in the numerator and reactants in the denominator, each raised to its coefficient, and every concentration is the equilibrium concentration (in M). This is the expression for the equilibrium constant平衡常数 (2.11.4). AP uses two forms:
- Kc uses molar concentrations and applies to gases and dissolved species;
- Kp uses partial pressures of gases (in atm) and includes only gases. For example, for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kp = (PSO₃)² ÷ [(PSO₂)²(PO₂)].
Pure solids and pure liquids (including water as the solvent) are left out, because their “concentrations” don’t change during the reaction and are absorbed into K. K depends only on temperature and is written without units.
Click each substance to decide whether it goes in the numerator, the denominator, or not at all:
Click each substance to cycle through “left out → numerator → denominator,” then check.
The reaction quotient Q
Putting the concentrations (or partial pressures) at any moment into the same expression gives the reaction quotient反应商, Q. Compare Q with K:
- Q < K: there is relatively too little product; the forward rate exceeds the reverse rate, and the reaction proceeds forward;
- Q = K: the system is at equilibrium;
- Q > K: there is relatively too much product, and the reaction proceeds in reverse.
As the reaction proceeds, Q keeps changing until it equals K. Change the concentrations and compare Q with K:
Example 1: Writing Kc and Kp
Write Kc for each equilibrium, and Kp where possible.
(1) C(s) + CO₂(g) ⇌ 2CO(g) (2) Ag₂SO₄(s) ⇌ 2Ag⁺(aq) + SO₄²⁻(aq)
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The rule. Products go in the numerator and reactants in the denominator, each raised to its coefficient; pure solids and pure liquids are left out.
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Kc for (1). C(s) is a pure solid and is left out; the coefficient 2 of CO becomes an exponent:
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Kp for (1). CO and CO₂ are both gases, so partial pressures replace concentrations:
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Kc for (2). Ag₂SO₄(s) is the only reactant and is a solid, so it is left out and the denominator is 1:
This is the solubility product Ksp of Ag₂SO₄ (U7.11).
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No Kp for (2). The expression contains only dissolved ions and no gases, so there is no Kp.
Example 2: Comparing Q with K
At 400 °C, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has K = 0.50 (OpenStax). At some moment a container holds [N₂] = 1.0 M, [H₂] = 1.0 M, and [NH₃] = 0.80 M. Which way does the reaction proceed?
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Write the Q expression. Q has exactly the same form as K, but uses the concentrations at this moment, which need not be at equilibrium:
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Substitute the current concentrations.
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Compare Q with K.
The numerator (products) is relatively too large: there is relatively too much NH₃.
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Conclusion. The reaction proceeds in reverse: NH₃ decreases while N₂ and H₂ increase, so Q falls until Q = K. Meanwhile the reverse rate is greater than the forward rate.
- Q = 2.0 × 10⁻³, K = 4.1 × 10⁻⁴
Q > K.
- Q = 0.10, K = 0.640
Q < K.
- Q = 50, K = 50
Q = K.
- Only reactants are present (Q = 0)
Q = 0 < K.
- Only products are present (Q is infinite)
Q > K.
- Q = 1.0 × 10⁻⁸, K = 1.0 × 10⁻¹⁰
Q is 100 times K.
- Q = 3.0 × 10⁻⁶, K = 1.8 × 10⁻⁵
Q < K.
- Q = 1.6 × 10², K = 1.6 × 10²
They are equal.
Common mistakes
Mistake 1: “Every substance goes in the K expression.” Pure solids and pure liquids (including water as the solvent) are left out. For MgO(s) + SO₃(g) ⇌ MgSO₄(s), for example, Kc = 1 ÷ [SO₃].
Mistake 2: “Coefficients multiply the concentrations.” Coefficients are exponents: 2NH₃ becomes [NH₃]², not 2[NH₃].
Mistake 3: “Q is the same thing as K.” K uses only equilibrium concentrations and is fixed for a reaction at a given temperature; Q uses the concentrations at any moment, changes as the reaction proceeds, and equals K only at equilibrium.
Quick check
What is the Kc expression for 2H₂S(g) + 3O₂(g) ⇌ 2SO₂(g) + 2H₂O(g)?
Products over reactants, each raised to its coefficient. Water is a gas here and is included, so D is wrong.
What is the K expression for Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s)?
Zn(s) and Cu(s) are pure solids and are left out.
What is the Kp expression for Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)?
Solids are left out; water is a gas here and is included; the exponents are the coefficients.
At 800 °C, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has K = 0.640. At some moment [CO] = [H₂O] = 0.10 M and [CO₂] = [H₂] = 0.20 M. What happens to [CO] next?
Q = (0.20 × 0.20) ÷ (0.10 × 0.10) = 4.0 > 0.640, so the reaction proceeds in reverse and CO increases.
When is Q = 0?
With zero product concentration the numerator is zero, so Q = 0 < K and the reaction can only proceed forward.
At some moment Q < K. Which statement about the forward and reverse rates is correct?
When Q < K the reaction proceeds forward, so the forward reaction is faster (U7.2).
AP-style free response
At 800 °C, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has Kc = 0.640 (OpenStax). A sealed 2.00 L vessel at 800 °C is charged with 0.40 mol of CO, 0.40 mol of H₂O, 0.20 mol of CO₂, and 0.20 mol of H₂.
(a) Write the expression for Kc for this reaction.
(b) Calculate the initial reaction quotient, Q.
(c) Predict which way the reaction proceeds, and justify your answer.
(d) Before equilibrium is reached, how does [CO₂] change? At the start, which is larger, the forward rate or the reverse rate?
(e) A student calculates Q using moles instead of concentrations and gets the same result as in (b). Explain why this works for this reaction.
Model answer and scoring
(a) Kc = [CO₂][H₂] ÷ ([CO][H₂O]).
(b) Concentrations: CO and H₂O are 0.20 M, CO₂ and H₂ are 0.10 M; Q = (0.10 × 0.10) ÷ (0.20 × 0.20) = 0.25.
(c) Forward, because Q = 0.25 < Kc = 0.640: there is relatively too little product.
(d) [CO₂] increases; at the start the forward rate is greater than the reverse rate.
(e) The numerator and denominator each multiply two concentrations, each equal to n ÷ V, so V appears twice on top and twice on the bottom and cancels (both sides have the same number of gas moles).
Scoring: (a) the expression; (b) converting to concentrations before calculating Q; (c) comparing Q with K; (d) direction and rates; (e) the reason the volume cancels.
Summary
- The K expression is the product of product concentrations over the product of reactant concentrations, each raised to its coefficient; pure solids and liquids are left out.
- Kc uses concentrations and Kp uses partial pressures of gases; a reaction with no gases has only Kc.
- Q uses the concentrations at any moment: Q < K means forward, Q > K means reverse, and Q = K means equilibrium.
AP tip: To predict direction, write the expression for Q, substitute the values, and then give the full reasoning “Q ? K, so …”. When you are given moles and a volume, convert to concentrations first, unless both sides have the same number of gas moles.