AP Chemistry · Unit 7 · Topic 7.6
Properties of the Equilibrium Constant
A balanced equation is part of an equilibrium constant's identity. Change how the equation is written, and you must change the constant that goes with it.
Reverse the equation, invert the ratio
If \(A\rightleftharpoons B\) has \(K_1=[B]/[A]\), then \(B\rightleftharpoons A\) has \([A]/[B]=1/K_1\). This does not describe a new experiment or a new temperature. It is a different mathematical description of the same equilibrium composition.
Scaling changes every exponent
Writing \(2A\rightleftharpoons2B\) gives \([B]^2/[A]^2=K_1^2\). Dividing all coefficients by two takes the square root of the constant. Do not multiply K by the coefficient multiplier; that would confuse an exponential concentration ratio with an extensive quantity such as reaction enthalpy.
Three transformations side by side
| Equation operation | Constant operation | Reason |
|---|---|---|
| Reverse | \(1/K\) | Swap numerator and denominator |
| Multiply coefficients by m | \(K^m\) | Multiply every exponent by m |
| Add equations | \(K_1K_2\) | Multiply ratios; intermediates cancel |
All constants must refer to the same temperature and compatible standard-state conventions.
A worked combination
Suppose \(A\rightleftharpoons B\) has K₁ = 4.0 and \(B\rightleftharpoons C\) has K₂ = 0.50. Adding the reactions cancels B and gives \(A\rightleftharpoons C\).
For \(2C\rightleftharpoons2A\), reverse this overall reaction and then double it: \(K=(1/2.0)^2=0.25\). Keeping the operations in separate steps prevents a common inversion error.
Check the result with actual amounts
A hypothetical equilibrium set [A] = 0.020 M, [B] = 0.080 M, and [C] = 0.040 M satisfies both supplied ratios. It also gives [C]/[A] = 2.0 and [A]²/[C]² = 0.25. Constructing one consistent set is a useful independent check on symbolic manipulations.
Distinguish notation from intervention
Doubling coefficients on paper is not the same as doubling every concentration in a beaker. The first changes the definition of K. The second changes Q and may disturb equilibrium, depending on stoichiometry. Similarly, reversing an equation does not physically swap all reactants and products in an existing vessel.
Investigate an apparent disagreement
If two reference tables quote different K values, compare their equations before blaming the measurements. One table may use the reverse reaction, half the coefficients, gas partial pressures rather than molar concentrations, or a different temperature. Record that context beside the number; K alone is an incomplete piece of chemical information.
Practice with answers
- If K = 16, what is the constant when all coefficients are halved? 4.
- Two reactions add to a target reaction and have constants 3 and 5. Find the target K. 15, not 8.
- Does multiplying a reaction by three multiply K by three? No; it cubes K.
Factual reference: OpenStax Chemistry 2e, 13.2. All numerical systems, explanations, and check questions are independently authored.
Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.