📘 IB Understanding
The equilibrium law describes how K can be determined from the stoichiometry of a reaction. The reaction quotient Q uses the same expression but with non-equilibrium concentrations to predict the direction of change.
The Kc Expression
For a general homogeneous reaction:
💡 Key Rules
- Products on top, reactants on bottom
- Coefficients become powers
- Solids and pure liquids are omitted from the expression
- Use square brackets [ ] for concentration (mol dm⁻³)
- K has no units in the IB syllabus
What Does Kc Tell Us?
| Value of Kc | Position of Equilibrium | Interpretation |
|---|---|---|
| \(K \gg 1\) | Far to the right | Products strongly favoured; reaction almost goes to completion |
| \(K \approx 1\) | Balanced | Significant amounts of both reactants and products |
| \(K \ll 1\) | Far to the left | Reactants 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
Worked Example: ICE Table
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.
\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\)
| Ester | H₂O | Acid | Alcohol | |
|---|---|---|---|---|
| I | 0.100 | 0.100 | 0 | 0 |
| C | −0.0346 | −0.0346 | +0.0346 | +0.0346 |
| E | 0.0654 | 0.0654 | 0.0346 | 0.0346 |
\(K_c = \frac{(0.0346)(0.0346)}{(0.0654)(0.0654)} = \frac{0.001197}{0.004277} = \textbf{0.280}\)
📋 Exam Tips
- Always use square brackets [ ] in K expressions — round brackets will lose marks
- If K < 10⁻³, you can assume [reactant]initial ≈ [reactant]eqm to simplify (state this assumption!)
- K is only affected by temperature — catalysts and concentration changes do not alter K