2.3.3—Equilibrium constant (K)

K range	Equilibrium tendency
K << 1	reactants strongly favoured
K < 1	reactants favoured
K = 1	comparable amounts
K > 1	products favoured
K >> 1	products strongly favoured

$Kreverse = 1 / Kforward$

K describes a ratio, not reaction speed: a very large K can still belong to a slow reaction. Reversing the equation gives 1/K, while multiplying every coefficient by a factor raises K to that factor. Interpret 'favoured' as equilibrium composition, not complete conversion.

Worked reading: if K = 0.0665 at 100 C, K < 1, so reactants are favoured and the forward reaction has a small extent. This describes equilibrium composition, not reaction speed. At the same temperature, reversing the equation gives K = 1/0.0665.

Representative question

[Maximum number: 1]

At $100^{\circ} \mathrm{C} \mathrm{K}_{\mathrm{c}}$ for this reaction is 0.0665 . Outline what this indicates about the extent of this reaction.

reaction hardly proceeds

OR

reverse reaction/formation of $\mathrm{NO}_{2}$ is favoured

OR

«concentration of» reactants greater than «concentration of» products «at equilibrium»

Marking guidance:

Accept equilibrium lies to the left.

Retrieve the route: define dynamic equilibrium, write K, interpret its magnitude, predict Le Châtelier shifts, compare Q with K, solve a RICE table, and connect K with ΔG.

Check closed-system and equal-rate language, exponents and direction, whether a change affects K, current versus equilibrium concentrations, stoichiometric x changes, and kelvin/unit consistency in ΔG calculations.

2.3.3—Equilibrium constant (K)

Interpreting K

Using K to Describe Extent

Assessment in practice

Extent of Chemical Change Summary

Question 1