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Chemistry · Unit 7: Equilibrium · 14 min read · Updated 2026-05-11

Calculating the equilibrium constant K — AP Chemistry

AP Chemistry · Unit 7: Equilibrium · 14 min read

1. Calculating K from Known Equilibrium Amounts ★★☆☆☆ ⏱ 4 min

The simplest K calculation uses directly given equilibrium concentrations or partial pressures. By the law of mass action, for a general balanced reaction $aA + bB \rightleftharpoons cC + dD$, the equilibrium constant expression follows reaction stoichiometry.

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \quad \text{or} \quad K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}

Exam tip: Always convert moles to molar concentration before plugging into $K_c$; MCQ traps often use distractors that match results from raw mole values.

2. Calculating K Using ICE Tables ★★★☆☆ ⏱ 5 min

Most AP exam problems do not give all equilibrium values directly. Instead, you use an ICE (Initial-Change-Equilibrium) table to map stoichiometric changes and solve for unknown equilibrium values. The ICE table has three core rows:

  • **Initial**: Concentrations/pressures before the reaction begins to reach equilibrium
  • **Change**: Change in concentration as the system approaches equilibrium, following stoichiometry: negative for reactants consumed, positive for products formed, with $x$ as the unknown change
  • **Equilibrium**: Final equilibrium value, calculated as *Initial + Change*

Once you use the known equilibrium value to solve for $x$, you can find all other equilibrium values and calculate $K$.

Exam tip: Always match the coefficient of $x$ in the change row to the stoichiometric coefficient of the species. If 2 moles of reactant are consumed, the change is $-2x$, not $-x$.

3. Relating Kc and Kp, and Manipulating K for Modified Reactions ★★★☆☆ ⏱ 4 min

Two common exam problems require converting between $K_c$ and $K_p$, and finding the new $K$ when a reaction is reversed, scaled, or combined. The relationship between $K_c$ and $K_p$ comes from the ideal gas law:

K_p = K_c(RT)^{\Delta n}

Where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$, $R = 0.0821\ L·atm/(mol·K)$, and $T$ is absolute temperature in Kelvin. Three core rules apply for modifying K:

  1. Reversing a reaction: $K_{new} = \frac{1}{K_{original}}$
  2. Multiplying all coefficients by a factor $n$: $K_{new} = (K_{original})^n$
  3. Adding two reactions to get a total reaction: $K_{total} = K_1 \times K_2$

Exam tip: $\Delta n$ only counts gaseous species; do not include aqueous solutes, pure solids, or pure liquids when calculating $\Delta n$ for the Kc-Kp conversion.

4. Calculating K from Standard Gibbs Free Energy Change ★★★★☆ ⏱ 3 min

This is a common cross-unit topic connecting Unit 7 (Equilibrium) and Unit 9 (Thermodynamics). The relationship between standard Gibbs free energy change $\Delta G^\circ$ and $K$ is:

\Delta G^\circ = -RT \ln K

Rearranging to solve for $K$ gives:

\ln K = -\frac{\Delta G^\circ}{RT}

Key note: $\Delta G^\circ$ must be in units of J/mol to match $R = 8.314\ J/(mol·K)$. If $\Delta G^\circ < 0$, $\ln K$ is positive so $K > 1$ (products favored at equilibrium); if $\Delta G^\circ > 0$, $K < 1$ (reactants favored).

Exam tip: Always convert $\Delta G^\circ$ from kJ/mol to J/mol before plugging into the formula. Using kJ directly will give a K that is orders of magnitude off, costing points on FRQs.

Common Pitfalls

Why: Students often plug all species from the balanced equation into the expression, forgetting that activity of pure condensed phases is 1

Why: Problems often give moles and volume, and students skip the division step because it seems trivial

Why: Students mix up the order of subtraction, or forget only gaseous species count for Δn

Why: Students confuse scaling reaction coefficients with scaling the equilibrium constant, mixing addition with exponentiation

Why: ΔG° is almost always reported in kJ/mol, but R uses Joules in the standard formula

Why: Students default to x as the change for all species regardless of stoichiometry

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