The larger K, the more product
Products are in the numerator of K and reactants in the denominator, so:
- K ≫ 1: products are strongly favored at equilibrium, and the reaction goes nearly to completion (Chinese textbooks often take K > 10⁵ as “essentially complete”);
- K ≪ 1: reactants are strongly favored, and very little product forms (below about 10⁻⁵ the reaction barely proceeds);
- K near 1: both reactants and products are present in significant amounts.
These are what AP calls “product-favored” and “reactant-favored.” Drag log K to see the equilibrium composition of A ⇌ B, then click some real reactions to see where their K values fall:
K says nothing about speed
K is set by the composition at equilibrium, not by how long it takes to get there. From the free energy of formation of water in OpenStax Appendix G, K for 2H₂(g) + O₂(g) ⇌ 2H₂O(l) at 25 °C is about 10⁸³, yet hydrogen and oxygen mixed at room temperature barely react: the activation energy is high, so the reaction is extremely slow (U5.5). Once ignited, the reaction is fast and goes essentially to completion.
Conversely, a reaction with a very small K can reach equilibrium quickly; there is just very little product when it does.
Small K allows an approximation
When K is very small, hardly any reactant is used up, so “initial concentration − x” in an ICE table can be approximated by the initial concentration. The x you get must still be checked: if x is no more than 5% of the initial concentration, the approximation is acceptable (U7.7 covers this in detail).
Example 1: Which acid ionizes more?
At 25 °C, Ka for HF is 6.4 × 10⁻⁴ and Ka for HCN is 4.9 × 10⁻¹⁰ (OpenStax). For 0.10 M solutions of each, which has more ions, and by how much?
-
Compare the K values. Both ionizations have the same form (HA ⇌ H⁺ + A⁻), so the larger Ka means more product (ions) at equilibrium. HF has the much larger Ka, so the HF solution has more ions. Next, estimate how many more.
-
Set up an ICE table. Let x be the amount of HA that ionizes (concentrations in M):
HA H⁺ A⁻ I 0.10 0 0 C −x +x +x E 0.10 − x x x So Ka = x² ÷ (0.10 − x). When Ka is small, x ≪ 0.10 and the denominator is about 0.10:
-
Estimate for HF.
x is 8% of 0.10, a little over 5%, so the approximation is rough; solving the quadratic gives about 7.7 × 10⁻³ M, which is close enough for comparing orders of magnitude.
-
Estimate for HCN.
-
Divide to compare.
The HF solution has about 1000 times as many ions as the HCN solution. Both acids are reactant-favored (Ka < 1), but to very different degrees.
Example 2: Estimating amounts from K
At 2000 °C, N₂(g) + O₂(g) ⇌ 2NO(g) has K = 4.1 × 10⁻⁴. Starting with [N₂] = [O₂] = 1.0 M, about how much NO is present at equilibrium?
-
Set up an ICE table. There is no NO at the start, so the reaction runs forward (concentrations in M):
N₂ O₂ 2NO I 1.0 1.0 0 C −x −x +2x E 1.0 − x 1.0 − x 2x -
Substitute into K.
-
Approximate. K is small, so very little reacts: x ≪ 1.0 and 1.0 − x ≈ 1.0:
-
Solve. Taking the square root gives 2x directly, which is [NO]:
-
Test the approximation. x ≈ 0.010, which is 1% of the starting concentration, well under 5%, so the approximation holds. Only about 1% of the N₂ becomes NO.
- K = 3.2 × 10⁸
K ≫ 1.
- K = 1.6 × 10⁻¹⁰
K ≪ 1.
- K = 0.640
Close to 1.
- K = 1.0 × 10⁻¹⁴
The ionization of water, for example.
- K = 2.5
Close to 1.
- The reverse of a reaction with K = 1.0 × 10⁻⁶
The reverse reaction has K = 1.0 × 10⁶.
- K = 6.4 × 10⁻⁴
Much less than 1.
- K = 9.4 × 10⁵
K ≫ 1.
Common mistakes
Mistake 1: “The larger K is, the faster the reaction.” K describes only the composition at equilibrium. How fast a reaction goes depends on factors such as activation energy and temperature.
Mistake 2: “When K = 1, every concentration is equal at equilibrium.” That is true only for a simple form such as A ⇌ B. With different coefficients, K = 1 doesn’t mean the concentrations are equal.
Mistake 3: “A reaction with a very small K doesn’t happen at all.” A small K only means there is very little product at equilibrium, not none. Kw for water is tiny, yet pure water still contains 1.0 × 10⁻⁷ M H⁺.
Quick check
Which of these reactions has the largest fraction of product at equilibrium?
The larger K, the more product at equilibrium.
A reaction has K = 2.5 × 10⁻¹² at 25 °C. Which statement is correct?
K ≪ 1, so reactants are favored. K says nothing about speed (C), and the amount of product isn’t zero (D).
A reaction has K ≫ 1, but when the reactants are mixed at room temperature no change is seen for hours. What is the best explanation?
K reflects only the equilibrium composition; how fast a reaction goes depends on factors such as activation energy.
A reaction has K = 1.0 × 10⁶. At the same temperature, its reverse reaction has:
The reverse reaction’s K is the reciprocal (U7.6). Its “reactants” are the original products, so the equilibrium position is the same.
Compare 0.10 M HNO₂ (Ka = 4.6 × 10⁻⁴) with 0.10 M HClO (Ka = 2.9 × 10⁻⁸):
At the same concentration and with the same form, the acid with the larger Ka ionizes more.
For A(g) ⇌ B(g), K = 1.0. At equilibrium:
K = [B] ÷ [A] = 1.0. For a reaction with different coefficients, though, K = 1 doesn’t mean equal concentrations.
AP-style free response
The table lists K for three reactions (OpenStax):
| Reaction | Temperature | K |
|---|---|---|
| I: N₂(g) + O₂(g) ⇌ 2NO(g) | 2000 °C | 4.1 × 10⁻⁴ |
| II: H₂(g) + I₂(g) ⇌ 2HI(g) | 400 °C | 50 |
| III: 2NO₂(g) ⇌ N₂O₄(g) | 25 °C | 1.6 × 10² |
(a) Which reaction is reactant-favored at equilibrium? Justify your answer.
(b) For reaction I, the initial concentrations are [N₂] = [O₂] = 0.50 M. Estimate [NO] at equilibrium, and check your approximation.
(c) Give K for the reverse of reaction II, 2HI(g) ⇌ H₂(g) + I₂(g), and state which side is favored at equilibrium for this reverse reaction.
(d) A student says, “Reaction III has the largest K, so it is the fastest of the three.” Identify the error in this claim.
(e) At 25 °C, K for 2H₂(g) + O₂(g) ⇌ 2H₂O(l) is about 10⁸³, yet a mixture of H₂ and O₂ can be stored at room temperature for a long time. Explain.
Model answer and scoring
(a) Reaction I. K = 4.1 × 10⁻⁴ ≪ 1, so there is very little product at equilibrium.
(b) (2x)² ÷ (0.50 − x)² = 4.1 × 10⁻⁴; approximating 0.50 − x ≈ 0.50, 2x ≈ 0.50 × √(4.1 × 10⁻⁴) ≈ 0.010, so [NO] ≈ 0.010 M. x ≈ 0.0050, which is 1% of 0.50, so the approximation is reasonable.
(c) K = 1/50 = 0.020. The reactant of the reverse reaction, HI, is favored (the same thing as products being favored in the original reaction).
(d) K reflects only the composition at equilibrium, not the reaction rate. Besides, the three reactions are at different temperatures, so the K values can’t show which is faster anyway.
(e) The large K means that at equilibrium the mixture would be almost all water, but the reaction has a high activation energy and is extremely slow at room temperature, so no reaction is seen.
Scoring: (a) K ≪ 1; (b) approximate calculation and the 5% check; (c) the reciprocal and its meaning; (d) K is unrelated to rate; (e) thermodynamics versus kinetics.
Summary
- K ≫ 1 means products are favored, K ≪ 1 means reactants are favored, and K near 1 means both are present in significant amounts.
- K reflects only the composition at equilibrium, not how fast the reaction goes.
- When K is very small, the approximation “initial concentration − x ≈ initial concentration” often works, but check that x is no more than 5%.
AP tip: Use K to answer “which side is favored at equilibrium” and activation energy and rate to answer “how fast.” When a question asks why a reaction with a large K shows no change, the answer is always kinetics (it is slow), not thermodynamics.