How to Solve ICE Table Equilibrium Problems
Setting up the table, solving for x, and checking whether the shortcut was actually valid.
An ICE table organizes an equilibrium calculation into three rows: Initial concentrations, the Change each species undergoes, and the resulting Equilibrium concentrations. Here's the full setup, worked through one complete example, including the shortcut that avoids the quadratic formula and the check that confirms whether that shortcut was actually allowed.
Setting Up the Table
- Initial row: the starting concentration of each species (0 for anything not present yet).
- Change row: write each species' change in terms of one unknown, x, scaled by its coefficient in the balanced equation. Reactants decrease (−x), products increase (+x).
- Equilibrium row: add the Initial and Change rows for each species.
- Substitute the Equilibrium row into the K expression and solve for x.
The Simplifying Approximation, and How to Check It
When K is small and the initial concentration is much larger than K, x usually turns out tiny compared to that initial concentration. That lets you approximate (initial − x) ≈ initial, which turns a quadratic equation into a simple square root. You can't just assume this worked, though, always check it with the 5% rule: divide your calculated x by the initial concentration you subtracted it from. Under 5%, the approximation was valid. 5% or more, and you have to go back and solve the full quadratic instead.
Full Worked Example
For the gas-phase equilibrium A(g) ⇌ B(g) + C(g), Kc = 1.00 × 10−4. A 0.100 M sample of A is allowed to reach equilibrium with no B or C present initially. Find the equilibrium concentration of each species.
| A | B | C | |
|---|---|---|---|
| Initial | 0.100 | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | 0.100 − x | x | x |
Substitute into Kc: Kc = [B][C]/[A] = x²/(0.100 − x) = 1.00 × 10−4.
Apply the approximation: since Kc is small relative to 0.100, assume 0.100 − x ≈ 0.100: x²/0.100 = 1.00 × 10−4, so x² = 1.00 × 10−5, giving x = 3.16 × 10−3.
Check the 5% rule: x ÷ 0.100 = (3.16 × 10−3) ÷ 0.100 = 0.0316 = 3.16%, under 5%. The approximation was valid.
Final equilibrium concentrations: [A] = 0.100 − 0.00316 ≈ 0.0968 M; [B] = [C] = 3.16 × 10−3 M.
Common Mistakes
- Forgetting stoichiometric coefficients in the Change row. A reaction like A ⇌ 2B needs +2x for B, not +x, matching the balanced equation.
- Skipping the 5% check. An approximation that isn't verified is just an assumption, always confirm it before trusting the answer.
- Forgetting to go back and use the quadratic formula when the 5% check fails. The approximate x isn't just "close enough" once it's over 5%, it's genuinely the wrong answer.
- Including pure solids or liquids in the ICE table's concentration rows. Only gases and aqueous species get concentration entries; solids and liquids don't appear in the K expression at all.
Related Resources
- Unit 7 Review: Equilibrium
- Unit 8 Review: Acids and Bases
- AP Chemistry Limiting Reactant & Percent Yield
- AP Chemistry Significant Figures Rules
- AP Chemistry Heating Curve
- AP Chemistry Study Guide
Frequently Asked Questions
What does ICE stand for in an ICE table?
Initial, Change, Equilibrium, the three rows of the table. Initial lists starting concentrations, Change lists how much each species changes (in terms of an unknown x, scaled by stoichiometric coefficients), and Equilibrium adds the first two rows to give each species' concentration once the reaction has settled.
When can you assume x is negligible compared to the initial concentration?
When K is small and the initial concentration is much larger than K, generally at least a few hundred times larger. That lets you approximate (initial − x) as just initial, avoiding the quadratic formula. You still have to check the assumption held, using the 5% rule.
What is the 5% rule?
After solving for x using the simplifying approximation, divide x by the initial concentration it was subtracted from. If that ratio is under 5%, the approximation was valid and your answer stands. If it's 5% or more, the approximation wasn't valid, and you have to redo the problem with the full quadratic formula instead.
Do you always need the quadratic formula for an ICE table problem?
No. You only need it when the simplifying approximation fails the 5% check, or when the problem's algebra doesn't reduce to a simple square root in the first place (for example, when the stoichiometry isn't a clean 1:1 ratio setup). Many AP-level ICE table problems are deliberately built so the approximation holds.
Does an ICE table work the same way for both gas-phase and acid-base equilibria?
Yes, the setup is identical either way, initial concentrations, a change in terms of x, and equilibrium concentrations. What changes is which equilibrium constant expression you plug the equilibrium row into, Kc or Kp for a general reaction, Ka or Kb for a weak acid or base.
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