Three rules
The K expression multiplies and divides concentrations raised to powers, so K changes along with the equation:
- reverse the equation: numerator and denominator swap, so the new K = 1 ÷ K;
- multiply the coefficients by n: every exponent is multiplied by n, so the new K = Kⁿ (multiplying by ½ means taking the square root);
- add two equations: the expressions multiply, so the new K = K₁ × K₂.
These match the rules for ΔH in Hess’s law (U6.9) one for one, each “one operation up”: ΔH changes sign while K takes its reciprocal; ΔH is multiplied by n while K is raised to the power n; ΔH values add while K values multiply.
Rewrite an equation, or add two equations, and watch K change:
The second equation is the reverse of HCN ionizing, so its K = 1/Ka(HCN). The sum has K = Ka(HNO₂) ÷ Ka(HCN) ≫ 1: the stronger acid, HNO₂, gives its proton to CN⁻.
The first is acetic acid ionizing (Ka); the second is acetate reacting with water (Kb). Adding them cancels CH₃COOH and CH₃COO⁻, leaving the ionization of water, so Ka × Kb = Kw.
Adding the two steps cancels HCO₃⁻, so K = Ka1 × Ka2.
When equations are added, their K values are multiplied (compare: ΔH values are added).
Example 1: Reverse, then multiply by ½
At 400 °C, H₂(g) + I₂(g) ⇌ 2HI(g) has K = 50 (OpenStax). Find K at the same temperature for HI(g) ⇌ ½H₂(g) + ½I₂(g).
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Compare with the target. In the original equation HI is on the right with coefficient 2; in the target it is on the left with coefficient 1. So the target is the original equation reversed and then multiplied by ½.
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Reverse: take the reciprocal of K. For 2HI(g) ⇌ H₂(g) + I₂(g):
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Multiply by ½: raise K to the ½ power. For HI(g) ⇌ ½H₂(g) + ½I₂(g):
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Check with the expressions. Write the expressions and compare them:
The target’s expression is exactly the square root of the reciprocal of the original, matching steps 2 and 3.
Example 2: Adding two equations
At 25 °C, Ka for HF is 6.4 × 10⁻⁴ and Ka for CH₃COOH is 1.8 × 10⁻⁵ (OpenStax). Find K for HF(aq) + CH₃COO⁻(aq) ⇌ F⁻(aq) + CH₃COOH(aq).
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First equation: the ionization of HF, used as written. HF is on the left of the target and F⁻ on the right:
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Second equation: the ionization of acetic acid, reversed. In the target CH₃COO⁻ is on the left and CH₃COOH on the right, the opposite of acetic acid’s ionization, so take the reciprocal of its K:
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Add the equations. H⁺ appears once on each side and cancels, leaving exactly the target: HF + CH₃COO⁻ ⇌ F⁻ + CH₃COOH.
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Adding equations multiplies the K values.
In one step: K = Ka(HF) ÷ Ka(CH₃COOH) = 6.4 × 10⁻⁴ ÷ 1.8 × 10⁻⁵ = 36.
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Meaning. K > 1, so products are favored: the stronger acid, HF, gives its proton to the stronger base, CH₃COO⁻, forming the weaker acid, CH₃COOH, and the weaker base, F⁻.
- 2NH₃ ⇌ N₂ + 3H₂
Reversed: 1 ÷ 0.50.
- ½N₂ + 3/2H₂ ⇌ NH₃
Multiplied by ½: √0.50.
- 2N₂ + 6H₂ ⇌ 4NH₃
Multiplied by 2: 0.50².
- NH₃ ⇌ ½N₂ + 3/2H₂
Reversed and multiplied by ½: 1 ÷ √0.50.
- 4NH₃ ⇌ 2N₂ + 6H₂
Reversed and multiplied by 2: 1 ÷ 0.25.
- 3H₂ + N₂ ⇌ 2NH₃
Only the order of the reactants changed, so K is the same.
- N₂ + 3H₂ ⇌ 2NH₃, run in a larger container
K depends only on temperature.
- 2NH₃ ⇌ N₂ + 3H₂, with every coefficient then multiplied by 2
2.0² = 4.0.
Common mistakes
Mistake 1: “Multiplying the coefficients by 2 multiplies K by 2.” The coefficients are exponents in the expression, so multiplying them by 2 squares K. Multiplying by a number is the rule for ΔH.
Mistake 2: “Reversing the equation makes K negative.” K is always positive. Reversing the equation makes K its reciprocal.
Mistake 3: “When two equations are added, their K values add.” The expressions multiply, so the K values multiply.
Quick check
A reaction has K = 4.0 × 10⁻³. At the same temperature, what is K for the reverse reaction with all its coefficients halved?
Reversing gives 1 ÷ (4.0 × 10⁻³) = 2.5 × 10²; halving the coefficients takes the square root: √(2.5 × 10²) = 16. B forgets to halve; D is √K.
If every coefficient in an equilibrium equation is multiplied by 3, the new K is the original K:
Every exponent is multiplied by 3, so K becomes K³.
Reaction ① has K₁ = 2.0 and reaction ② has K₂ = 5.0 × 10⁻³. What is K for the reaction ① + ②?
For added equations the K values multiply: 2.0 × 5.0 × 10⁻³ = 1.0 × 10⁻².
At 25 °C, H₂CO₃ has Ka₁ = 4.3 × 10⁻⁷ and Ka₂ = 4.7 × 10⁻¹¹ (OpenStax). K for CO₃²⁻(aq) + 2H⁺(aq) ⇌ H₂CO₃(aq) is about:
Adding the two ionization steps gives H₂CO₃ ⇌ 2H⁺ + CO₃²⁻ with K = Ka₁ × Ka₂ = 2.0 × 10⁻¹⁷; the target is the reverse, so K = 1 ÷ (2.0 × 10⁻¹⁷) = 4.9 × 10¹⁶.
At 25 °C, Ka for HCN is 4.9 × 10⁻¹⁰. What is K for CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq), that is, Kb for CN⁻? (Kw = 1.0 × 10⁻¹⁴)
Adding the ionization of HCN and the reaction of CN⁻ with water gives the ionization of water, so Ka × Kb = Kw and Kb = 1.0 × 10⁻¹⁴ ÷ 4.9 × 10⁻¹⁰ = 2.0 × 10⁻⁵.
Two reactions are added. Which statement about ΔH and K is correct?
ΔH values add by Hess’s law; the K expressions multiply.
AP-style free response
At 25 °C, NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) has Kb = 1.8 × 10⁻⁵, H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) has Kw = 1.0 × 10⁻¹⁴, and HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq) has Ka = 4.6 × 10⁻⁴ (OpenStax).
(a) Explain how to combine the equations above to get NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq), and find its K (Ka for NH₄⁺).
(b) Give K for NH₃(aq) + H⁺(aq) ⇌ NH₄⁺(aq).
(c) Find K for NH₃(aq) + HNO₂(aq) ⇌ NH₄⁺(aq) + NO₂⁻(aq), and state which side is favored at equilibrium.
(d) If every coefficient in the equation in (c) is multiplied by 2, what is K?
(e) If ΔH were known for the two steps used in (c), how would you get ΔH for (c)? Compare with the way you found K.
Model answer and scoring
(a) Take the ionization of water and add the reverse of the reaction of NH₃ with water: H₂O ⇌ H⁺ + OH⁻ (Kw) plus NH₄⁺ + OH⁻ ⇌ NH₃ + H₂O (1/Kb) gives NH₄⁺ ⇌ NH₃ + H⁺. Ka = Kw ÷ Kb = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.
(b) K = 1 ÷ (5.6 × 10⁻¹⁰) = 1.8 × 10⁹.
(c) Add the ionization of HNO₂ and the reaction in (b); H⁺ cancels: K = 4.6 × 10⁻⁴ × 1.8 × 10⁹ = 8.3 × 10⁵. K ≫ 1, so products are favored.
(d) K = (8.3 × 10⁵)² = 6.9 × 10¹¹.
(e) Add the two ΔH values (changing the sign for any step written in reverse). For K the values multiply (taking the reciprocal for a reversed step); multiplying coefficients by n multiplies ΔH by n but raises K to the power n.
Scoring: (a) correct combination giving Kw/Kb; (b) the reciprocal; (c) multiplying K values and the judgment; (d) squaring; (e) contrasting adding ΔH with multiplying K.
Summary
- Reverse: K → 1/K; multiply coefficients by n: K → Kⁿ; add two equations: K = K₁ × K₂.
- Compare ΔH: change sign, multiply by n, add.
- Common combinations: Ka × Kb = Kw; for a polyprotic weak acid, the overall K = Ka₁ × Ka₂.
AP tip: When combining K values, write next to each known equation what you did to it (reversed, multiplied by what) and its new K, then multiply. Check the cancellations using the equations before doing the arithmetic.