IB Chemistry R2.3 R2.3.5

The Equilibrium Constant Kc

Writing Kc expressions, ICE tables, and comparing the reaction quotient Q to K.

Reactivity 2.3 HL Extension ⏱️ ~6 min revision
IB Understanding

The Equilibrium Law and Stoichiometry

The equilibrium law describes how the equilibrium constant \(K_{\text{c}}\) is determined from reaction stoichiometry and equilibrium concentrations. The reaction quotient \(Q\) determines the direction of spontaneous shift toward equilibrium.

The Kc Expression

For a general homogeneous reaction:

\[aA + bB \rightleftharpoons cC + dD\] \[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
Core Rules

Formulating Equilibrium Law Expressions

  • Products on top, reactants on bottom: Concentrations are multiplied, never added.
  • Coefficients become exponents: For \(2A + B \rightleftharpoons 3C\), \([A]^2\) and \([C]^3\).
  • Reaction quotient \(Q\): If \(Q < K\), reaction shifts right; if \(Q > K\), reaction shifts left; if \(Q = K\), system is at equilibrium.

What Does Kc Tell Us?

Value of KcPosition of EquilibriumInterpretation
\(K \gg 1\)Far to the rightProducts strongly favoured; reaction almost goes to completion
\(K \approx 1\)BalancedSignificant amounts of both reactants and products
\(K \ll 1\)Far to the leftReactants strongly favoured; reaction hardly proceeds

The Reaction Quotient Q

Q is calculated using the same expression as K, but with concentrations at any point in time (not necessarily at equilibrium).

Comparing Q and K

R2.3.5 The Equilibrium Constant Kc - IB HL | ChemEasy Q < K Reaction proceeds → RIGHT (more products formed) Q = K System is at EQUILIBRIUM (no net change) Q > K Reaction proceeds ← LEFT (more reactants formed)

Worked Example: ICE Table

Worked Example

ICE Table Equilibrium Calculation

Problem: 0.100 mol of ethyl ethanoate is added to 0.100 mol of water (total volume = 1 dm³). At equilibrium, 0.0654 mol of water remains. Calculate Kc.

\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\)

EsterH₂OAcidAlcohol
I0.1000.10000
C−0.0346−0.0346+0.0346+0.0346
E0.06540.06540.03460.0346

\(K_c = \frac{(0.0346)(0.0346)}{(0.0654)(0.0654)} = \frac{0.001197}{0.004277} = \textbf{0.280}\)

Examiner Trap

Approximations and Square Brackets in Kc Expressions

Always use square brackets [ ] denoting molar concentration. In ICE calculations where \(K < 10^{-3}\), you may assume \(x \ll [ ext{initial}]\), but always verify your approximation yields \(< 5\%\) error.

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