AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-6)
Unit 7 · Topic 7.6
7.6 Properties of the Equilibrium Constant
When you change a balanced equation, its K changes in a predictable way: reversing the reaction gives 1/K, multiplying the coefficients by n raises K to the nth power, and adding reactions multiplies their K values. These rules let you find K for a reaction from the K values of other reactions.
Key terms
- reversing a reaction
- multiplying coefficients
- adding reactions
- overall K
Reversing a reaction
Reversing swaps products and reactants, which flips the fraction upside down. So the new K is the reciprocal of the old one: K(reverse) = 1/K(forward).
Example: for N₂O₄ ⇌ 2NO₂, K = [NO₂]²/[N₂O₄]. For 2NO₂ ⇌ N₂O₄, K = [N₂O₄]/[NO₂]², which is 1 divided by the first one.
Multiplying coefficients
Multiplying every coefficient by a factor n multiplies every exponent by n, so the new K is the old K raised to the nth power: K(new) = Kⁿ.
Doubling an equation squares K. Halving an equation takes the square root of K, because n = ½.
Adding reactions
When two reactions are added to make an overall reaction, the overall K is the product of the individual K values: K(overall) = K₁ × K₂.
Why: multiplying the two K expressions together makes any species that appears on both sides (an intermediate) cancel out, leaving exactly the expression for the overall reaction.
The same rules apply to Q
Q and K have the same mathematical form, so everything here works for Q too. If Q = 4.0 for a reaction, Q is 0.25 for the reverse reaction at the same moment.
Why the rules work
Take two steps: A ⇌ B with K₁ = [B]/[A], and B ⇌ C with K₂ = [C]/[B]. Adding them gives A ⇌ C. Multiplying the expressions gives K₁ × K₂ = ([B]/[A]) × ([C]/[B]) = [C]/[A], which is exactly K for A ⇌ C. The intermediate B cancels, just as it does in the equations.
A useful example: add the ionization of a weak acid, HA + H₂O ⇌ H₃O⁺ + A⁻ (Ka), to the reaction of its conjugate base with water, A⁻ + H₂O ⇌ HA + OH⁻ (Kb). HA and A⁻ cancel, leaving 2H₂O ⇌ H₃O⁺ + OH⁻, whose constant is Kw. So Ka × Kb = Kw (topic 8.3).
- Strategy: write the target equation first.
- For each given equation, decide whether to reverse it and what to multiply it by so its species match the target.
- Apply the matching change to each K (reciprocal, power).
- Add the equations, check that the result is exactly the target, then multiply the adjusted K values.
Don't mix these up with ΔH rules
| Change to the equation | What happens to K | What happens to ΔH |
|---|---|---|
| Reverse it | 1/K | −ΔH |
| Multiply coefficients by n | Kⁿ | n × ΔH |
| Add two equations | K₁ × K₂ | ΔH₁ + ΔH₂ |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Reverse and halve
At 25 °C, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has K = 6.0 × 10⁵. Find K for NH₃(g) ⇌ ½N₂(g) + 3/2 H₂(g).
Show the solutionHide the solution
- Step 1: The target is the original reaction reversed and then multiplied by ½.
- Step 2: Reverse: K = 1 ÷ (6.0 × 10⁵) = 1.67 × 10⁻⁶.
- Step 3: Multiply by ½: raise to the ½ power, which is a square root. √(1.67 × 10⁻⁶) = 1.3 × 10⁻³.
- Step 4: Equivalently, K(new) = 1 ÷ √(6.0 × 10⁵).
Answer: K ≈ 1.3 × 10⁻³
- Example 2Calculator allowed
Combining two reactions
Given (1) HF(aq) ⇌ H⁺(aq) + F⁻(aq), K₁ = 6.8 × 10⁻⁴ and (2) H₂C₂O₄(aq) ⇌ 2H⁺(aq) + C₂O₄²⁻(aq), K₂ = 3.8 × 10⁻⁶, find K for 2HF(aq) + C₂O₄²⁻(aq) ⇌ 2F⁻(aq) + H₂C₂O₄(aq).
Show the solutionHide the solution
- Step 1: HF appears as 2HF on the left of the target, so double reaction 1: 2HF ⇌ 2H⁺ + 2F⁻, K = (6.8 × 10⁻⁴)² = 4.6 × 10⁻⁷.
- Step 2: H₂C₂O₄ is a product in the target, so reverse reaction 2: 2H⁺ + C₂O₄²⁻ ⇌ H₂C₂O₄, K = 1 ÷ (3.8 × 10⁻⁶).
- Step 3: Add the two: 2H⁺ cancels from both sides, leaving the target.
- Step 4: K = (4.6 × 10⁻⁷) ÷ (3.8 × 10⁻⁶) = 0.12.
Answer: K ≈ 0.12
- Example 3Calculator allowed
Trap: halving is a square root, not division by 2
For N₂O₄(g) ⇌ 2NO₂(g), Kp = 0.15 at 25 °C. Find Kp for NO₂(g) ⇌ ½N₂O₄(g).
Show the solutionHide the solution
- Step 1: The target is the original reversed, then halved.
- Step 2: Reverse: 1 ÷ 0.15 = 6.7. (Not −0.15; negative K values never happen.)
- Step 3: Halve: take the square root, √6.7 = 2.6. (Not 6.7 ÷ 2 = 3.3; that's using a ΔH-style rule on K.)
Answer: Kp ≈ 2.6
Common mistakes
- Making K negative when reversing a reaction. Reversing takes the reciprocal; K is always positive.
- Multiplying K by n instead of raising it to the nth power.
- Adding K values when reactions are added. Add ΔH values, but multiply K values.
- Forgetting to apply a reversal to K after reversing the equation on paper.
On the exam
- Expect questions that give one or two reactions with K values and ask for K of a related reaction. Write each manipulated equation next to its new K so the logic is clear.
- A classic application is Ka × Kb = Kw for a conjugate pair (topic 8.3): adding the acid's and its conjugate base's reactions with water gives the water autoionization reaction.
Connected topics
Videos
Check yourself
4 questions on 7.6 Properties of the Equilibrium Constant. Pick an answer to see if you got it, and why.
For the reaction A(g) ⇌ 2B(g), K = 4.0 at a certain temperature. What is K for the reaction B(g) ⇌ ½A(g) at the same temperature?
Reaction 1 has K₁ = 2.0 × 10³ and reaction 2 has K₂ = 5.0 × 10⁻², both at the same temperature. Adding reaction 1 and reaction 2 gives reaction 3. What is K₃?
At a certain temperature, reaction 1, A₂(g) + B₂(g) ⇌ 2AB(g), has K₁ = 1.0 × 10⁻⁴, and reaction 2, 2AB(g) + B₂(g) ⇌ 2AB₂(g), has K₂ = 4.0 × 10⁶. What is K for ½A₂(g) + B₂(g) ⇌ AB₂(g) at this temperature?
Reaction 1, A + B ⇌ C, has equilibrium constant K₁, and reaction 2, C ⇌ D, has equilibrium constant K₂, at the same temperature. What is the equilibrium constant for D ⇌ A + B?
0 of 4 answered