AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-5)
Unit 7 · Topic 7.5
7.5 Magnitude of the Equilibrium Constant
The size of K tells you what the equilibrium mixture looks like. A very large K means the mixture is almost all products, so the reaction essentially goes to completion; a very small K means it is almost all reactants, so the reaction barely happens.
Key terms
- large K
- small K
- product-favored
- reactant-favored
- goes to completion
Reading the size of K
K is products over reactants. If K is huge, the products term must be much bigger than the reactants term at equilibrium, so products dominate. If K is tiny, reactants dominate. If K is near 1, you'll find significant amounts of both.
| Value of K | Equilibrium mixture | Example |
|---|---|---|
| Very large (like 10¹⁰) | Almost all products; essentially complete | H₃O⁺ + OH⁻ → 2H₂O, K = 1.0 × 10¹⁴ at 25 °C |
| Near 1 | Noticeable amounts of reactants and products | 2NO₂ ⇌ N₂O₄, Kp ≈ 7 at 25 °C |
| Very small (like 10⁻¹⁰) | Almost all reactants; barely proceeds | AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = 1.8 × 10⁻¹⁰ |
Going to completion
When K is very large, you can treat the reaction like a regular stoichiometry problem: assume the limiting reactant is used up. A tiny amount of it is actually left, but too little to matter. This is why neutralizing a strong acid with a strong base, or precipitating AgCl from Ag⁺ and Cl⁻, is treated as complete.
When K is very small, you can assume almost nothing reacts. Nitrogen and oxygen make up most of the air, but at 25 °C the K for N₂ + O₂ ⇌ 2NO is around 10⁻³⁰, so the air doesn't turn into nitrogen monoxide.
Using K's size to simplify math
The size of K tells you which shortcut is safe. A small K means the change x is small compared with the starting amounts, so you can often drop x when subtracting it (the small-x approximation in topic 7.7). A large K means you can assume the reaction goes to completion first, then figure out the tiny amount of reactant left.
What K does not tell you
K says nothing about speed. A reaction with an enormous K can be so slow that nothing seems to happen (topic 9.4). K also doesn't change with starting amounts; only temperature changes it.
If you reverse the reaction, K flips to 1/K. So 'product-favored' always refers to the reaction as it is written.
K = 1 doesn't mean equal amounts
A K of exactly 1 doesn't guarantee equal concentrations of reactants and products, because concentrations are raised to powers and the values depend on the starting amounts. For A ⇌ B, K = 1 does mean [A] = [B] at equilibrium. But for A ⇌ 2B with K = 1, [B]²/[A] = 1, which could be [B] = 0.50 M and [A] = 0.25 M. Use K = 1 as a sign that neither side is strongly favored, not as a promise of equal amounts.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Interpreting three K values
At 25 °C, reaction X has K = 2.0 × 10¹², reaction Y has K = 0.85, and reaction Z has K = 3.0 × 10⁻⁹. Describe the equilibrium mixture for each.
Show the solutionHide the solution
- Step 1: X: K is much greater than 1, so products are favored overwhelmingly. The reaction essentially goes to completion.
- Step 2: Y: K is close to 1, so the equilibrium mixture has comparable amounts of reactants and products (slightly favoring reactants).
- Step 3: Z: K is much less than 1, so the mixture is almost entirely reactants; very little product forms.
Answer: X: almost all products; Y: significant amounts of both; Z: almost all reactants.
- Example 2Calculator allowed
How complete is 'complete'?
For Ag⁺(aq) + Cl⁻(aq) ⇌ AgCl(s), K = 5.6 × 10⁹ at 25 °C. Equal amounts of Ag⁺ and Cl⁻ are mixed so that each would be 0.010 M if no reaction occurred. Estimate the [Ag⁺] left at equilibrium and the percent remaining.
Show the solutionHide the solution
- Step 1: K is very large, so assume nearly all Ag⁺ and Cl⁻ precipitate. Since they were equal, a tiny equal amount of each stays dissolved: [Ag⁺] = [Cl⁻] = y.
- Step 2: K = 1 / ([Ag⁺][Cl⁻]) = 1/y², so y² = 1 / (5.6 × 10⁹) = 1.8 × 10⁻¹⁰.
- Step 3: y = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M.
- Step 4: Percent remaining = (1.3 × 10⁻⁵ ÷ 0.010) × 100% ≈ 0.13%. More than 99.8% has precipitated, so treating it as complete is reasonable.
Answer: [Ag⁺] ≈ 1.3 × 10⁻⁵ M, about 0.13% of the original
- Example 3
Trap: large K doesn't mean fast
For 2H₂(g) + O₂(g) ⇌ 2H₂O(l), K is on the order of 10⁸³ at 25 °C. Yet a balloon of hydrogen and oxygen can sit at room temperature without reacting. A student concludes the K value must be wrong. Explain.
Show the solutionHide the solution
- Step 1: K describes where the reaction ends up (the equilibrium mixture), not how fast it gets there.
- Step 2: The huge K means that once equilibrium is reached, essentially only water would be present.
- Step 3: The mixture doesn't react at room temperature because the activation energy is high, so the rate is extremely slow. A spark supplies enough energy to start it, and then it reacts explosively.
Answer: The K is fine; the reaction is product-favored but extremely slow without a spark because of its high activation energy.
Common mistakes
- Thinking a large K means a fast reaction. K is about extent, not rate.
- Saying a reaction with a small K produces no products at all. Some product always forms; it's just a tiny amount.
- Forgetting that K refers to the equation as written: if K is large for a reaction, it's small for the reverse.
On the exam
- Questions often ask you to justify why a reaction can be treated as going to completion, or why an approximation is valid. Cite the size of K directly.
- Be ready to rank mixtures or reactions by how far they proceed using given K values.
Connected topics
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Check yourself
5 questions on 7.5 Magnitude of the Equilibrium Constant. Pick an answer to see if you got it, and why.
For the reaction N₂(g) + O₂(g) ⇌ 2NO(g), K is about 10⁻³⁰ at 25 °C. Which of the following best describes air at 25 °C once this reaction has reached equilibrium?
| Reaction | K at 25 °C |
|---|---|
| W | 2.5 × 10⁻¹² |
| X | 0.85 |
| Y | 3.1 × 10⁴ |
| Z | 6.0 × 10¹⁵ |
Hypothetical data. Each reaction has the form R(aq) ⇌ P(aq), with one reactant and one product, so K = [P]/[R].
For which reaction would the equilibrium mixture contain similar amounts of R and P?
For reaction Z, a solution starts with [R] = 0.10 M and no P. Which of the following is closest to [R] at equilibrium?
A student makes a solution of R for reaction Z, and after an hour finds that almost no P has formed. Which of the following is the best explanation?
Which of the following correctly gives K for the reverse of reaction Y, P(aq) ⇌ R(aq), and describes its equilibrium mixture?
0 of 5 answered