General Chemistry II · Chemical Equilibrium
ICE Tables
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In 30 seconds
An ICE table is an organized bookkeeping tool for equilibrium calculations. It lays out the Initial concentrations, the Change that occurs as the system moves to equilibrium, and the resulting Equilibrium concentrations — all tied together by the stoichiometry of the reaction. Substituting the equilibrium row into the K expression yields an equation to solve for the unknown change x.
Why this matters
ICE tables are the standard technique for virtually every equilibrium calculation in chemistry — gas-phase equilibria, weak acid/base ionization, solubility, and complex-ion equilibria all use this same framework. Fluency with the table, the quadratic, and the 5% rule is the single highest-yield quantitative skill in the equilibrium unit.
The college version
Core Concept
An ICE table is an organized bookkeeping tool for equilibrium calculations. It lays out the Initial concentrations, the Change that occurs as the system moves to equilibrium, and the resulting Equilibrium concentrations — all tied together by the stoichiometry of the reaction. Substituting the equilibrium row into the K expression yields an equation to solve for the unknown change x.
Key Ideas
- The "Change" row is written in terms of a single unknown x, scaled by stoichiometric coefficients.
- For a reaction shifting right, reactants change by −(coefficient)x and products by +(coefficient)x.
- Substituting the Equilibrium row into K gives either a simple square-root case, a quadratic, or a candidate for approximation.
- The 5% rule justifies dropping x in a sum/difference only if x is small relative to the initial concentration.
Equations and Variables
For aA + bB ⇌ cC + dD, the ICE table:
| A | B | C | D | |
|---|---|---|---|---|
| Initial | [A]₀ | [B]₀ | [C]₀ | [D]₀ |
| Change | −ax | −bx | +cx | +dx |
| Equilibrium | [A]₀−ax | [B]₀−bx | [C]₀+cx | [D]₀+dx |
Then solve:
K = ([C]0 + cx)c([D]0 + dx)d([A]0 - ax)a([B]0 - bx)b
5% rule: the approximation x ≪ [X]₀ is acceptable when x/[X]₀ × 100% < 5%.
How It Works
The method is mechanical. (1) Write the balanced equation and K expression. (2) Fill the Initial row with starting concentrations. (3) Express the Change row as ±coefficient·x. (4) Write the Equilibrium row. (5) Substitute into K and solve for x. (6) Back-substitute to get every equilibrium concentration.
When K is very small, the reaction barely proceeds, so x is tiny compared with the initial concentrations, and terms like [A]₀ − x can be approximated as [A]₀. This turns a cubic or quadratic into something easy. The 5% rule tells you whether that shortcut was valid; if x is more than 5% of the initial concentration, you must solve the full equation.
Worked Example
For N₂O₄(g) ⇌ 2 NO₂(g), K_c = 4.63 × 10⁻³ at 25 °C. A flask is charged with 0.500 M N₂O₄ (no NO₂). Find the equilibrium concentrations.
Step 1 — ICE table:
| N₂O₄ | NO₂ | |
|---|---|---|
| Initial | 0.500 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.500 − x | 2x |
Step 2 — Substitute into K:
4.63 × 10-3 = (2x)20.500 - x = 4x20.500 - x
Step 3 — Approximate (K is small, so x ≪ 0.500):
4.63 × 10-3 ≈ 4x20.500 ⇒ 4x2 = 2.315 × 10-3 ⇒ x = 0.0241
Step 4 — Check the 5% rule:
x[N2O4]0 × 100% = 0.02410.500 × 100% = 4.8% < 5%
The approximation is valid. (The exact quadratic gives x = 0.0235, within 2.6% of the approximate value.)
Step 5 — Equilibrium concentrations:
[N₂O₄] = 0.500 − 0.0241 = 0.476 M; [NO₂] = 2(0.0241) = 0.0482 M.
Check: (0.0482)²/0.476 = 4.9 × 10⁻³ ≈ K. ✓
How it works
The method is mechanical. (1) Write the balanced equation and K expression. (2) Fill the Initial row with starting concentrations. (3) Express the Change row as ±coefficient·x. (4) Write the Equilibrium row. (5) Substitute into K and solve for x. (6) Back-substitute to get every equilibrium concentration.
When K is very small, the reaction barely proceeds, so x is tiny compared with the initial concentrations, and terms like [A]₀ − x can be approximated as [A]₀. This turns a cubic or quadratic into something easy. The 5% rule tells you whether that shortcut was valid; if x is more than 5% of the initial concentration, you must solve the full equation.
Common confusions
- Sign errors in the Change row. Products gain (+) and reactants lose (−) for a forward shift; reverse the signs for a backward shift.
- Forgetting to scale x by coefficients. NO₂ changes by +2x, not +x, in the N₂O₄ example.
- Using the approximation without the 5% check. Always verify.
- Forgetting to back-substitute x to report actual equilibrium concentrations.
Quick review
- What do the three rows of an ICE table represent?
- How is the Change row related to stoichiometry?
- When is the "x is negligible" approximation reasonable?
- State the 5% rule.
- For the N₂O₄ example, why is [NO₂] = 2x and not x?
Eli explains
The same idea, in plain words
Explain it like I’m 10
An ICE table is just a tidy ledger for a reaction's accounting. Start with what you have (Initial), record what shifts (Change, all in terms of one unknown x), and total it up (Equilibrium). Then plug the totals into K and solve. When K is tiny, the shift x is so small that "0.500 − x" is basically still 0.500, so you can ignore the minus-x and breeze through — but you must prove x really is negligible with the 5% rule, or you'll build your answer on a shaky assumption.
Worked example
Worked Example
For N₂O₄(g) ⇌ 2 NO₂(g), K_c = 4.63 × 10⁻³ at 25 °C. A flask is charged with 0.500 M N₂O₄ (no NO₂). Find the equilibrium concentrations.
Step 1 — ICE table:
| N₂O₄ | NO₂ | |
|---|---|---|
| Initial | 0.500 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.500 − x | 2x |
Step 2 — Substitute into K:
4.63 × 10-3 = (2x)20.500 - x = 4x20.500 - x
Step 3 — Approximate (K is small, so x ≪ 0.500):
4.63 × 10-3 ≈ 4x20.500 ⇒ 4x2 = 2.315 × 10-3 ⇒ x = 0.0241
Step 4 — Check the 5% rule:
x[N2O4]0 × 100% = 0.02410.500 × 100% = 4.8% < 5%
The approximation is valid. (The exact quadratic gives x = 0.0235, within 2.6% of the approximate value.)
Step 5 — Equilibrium concentrations:
[N₂O₄] = 0.500 − 0.0241 = 0.476 M; [NO₂] = 2(0.0241) = 0.0482 M.
Check: (0.0482)²/0.476 = 4.9 × 10⁻³ ≈ K. ✓
Key takeaways
- ### High-Yield Facts
- Change row = ±(coefficient)x; products gain, reactants lose (for forward shift).
- K small ⇒ x small ⇒ approximation (x ≪ [X]₀) may apply.
- 5% rule: approximation valid only if x/[X]₀ × 100% < 5%.
- If the 5% test fails, solve the full quadratic.
- Always back-substitute and check against K.
Study tools & related lessonsYou’ll learn to · Related
You’ll learn to
- Set up an ICE (Initial, Change, Equilibrium) table for any equilibrium problem.
- Solve the resulting equation exactly (quadratic) or with a justified approximation.
- Apply and verify the 5% rule when using the approximation.
- Compute equilibrium concentrations and, from them, K or a single unknown concentration.
Sources & references
- OpenStax. *Chemistry 2e*. Ch. 13, "Equilibrium Calculations." https://openstax.org/books/chemistry-2e/pages/13-4-equilibrium-calculations
- IUPAC Compendium of Chemical Terminology ("Gold Book"), "equilibrium constant." https://goldbook.iupac.org/
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
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