7.5 Magnitude of K

Magnitude of K

Click across 7 real AP reactions pinned on a log-K number line from 10⁻³⁰ to 10³³. A particulate flask rebalances red-to-green so a glance shows how far each reaction proceeds

Log-K Axis7 ReactionsExtent of ReactionParticulate View
Topic 7.5

Magnitude of the Equilibrium Constant

Explain the relationship between very large or very small values of K and the relative concentrations of chemical species at equilibrium.

Some equilibrium reactions have very large K values and proceed essentially to completion; others have very small K values and barely proceed at all.

Because K is a ratio of products to reactants:

  • K ≫ 1 — the numerator dominates. At equilibrium the mixture is mostly products. The reaction is "product-favored". For K > 10¹⁰ or so it is effectively complete.
  • K ≈ 1 — comparable amounts of both.
  • K ≪ 1 — the denominator dominates. Mostly reactants remain; the reaction is "reactant-favored".

Practical consequences:

  • Strong acids have Ka values so large that ionization is treated as complete (8.2). Weak acids have Ka around 10⁻⁵, so the "x is small" approximation works (7.7).
  • Ksp values around 10⁻¹⁰ mean a salt is called insoluble — though a tiny amount of ion really is in solution (7.11).

What K does not tell you:

  • Nothing about rate. A reaction with K = 10²⁰ can take centuries (see 9.4).
  • Nothing about the amounts you started with. K fixes a ratio, not absolute quantities.

Connection to thermodynamics: ΔG° = −RT ln K (topic 9.5). K > 1 corresponds to ΔG° < 0, and K < 1 to ΔG° > 0 — the same statement in two vocabularies.

Key points

  • K ≫ 1 → products favored. K ≪ 1 → reactants favored.
  • K says nothing about how fast equilibrium is reached.
  • K depends only on temperature; nothing else changes it.
  • K > 1 ⇔ ΔG° < 0 (topic 9.5).

Common mistakes

  • Equating a large K with a fast reaction. Independent properties.
  • Thinking adding reactant changes K. It changes Q, and the system responds to restore the same K.
  • Thinking a catalyst changes K. It does not.
  • Assuming K ≫ 1 means literally no reactant remains. A vanishingly small amount always does.

Worked example

Three reactions have K = 4.2 × 10⁻¹², K = 1.8, and K = 3.5 × 10¹⁵. For each, describe the composition at equilibrium and comment on whether the reaction is useful for producing the product.

K = 4.2 × 10⁻¹² (reactant-favored). The denominator vastly exceeds the numerator, so at equilibrium the mixture is essentially all reactants with only trace product. This reaction is not useful for making product directly — you would need to drive it by removing product continuously (Le Châtelier) or by coupling it to a favorable reaction (9.7).

K = 1.8 (comparable). Products and reactants are present in similar amounts. This is the situation where an ICE table is genuinely necessary and the "x is small" approximation will fail. Useful for making product, but the yield is limited and reaction conditions matter.

K = 3.5 × 10¹⁵ (product-favored). Essentially complete; only a negligible trace of reactant remains at equilibrium. Excellent for producing the product from a thermodynamic standpoint — though the reaction could still be far too slow to be practical, since K says nothing about rate.

Full notes for topic 7.5 →