ICE-Table Calculations

Initial, Change, Equilibrium — a three-row grid that turns messy equilibrium problems into simple bookkeeping.

High schoolIntro Gen ChemUni Year 1
⏱️ About 22 min

You know the starting concentrations and the equilibrium constant K. You want the concentrations once the reaction settles. It sounds like it needs a leap of insight — but it's really just careful bookkeeping. Three rows, one unknown, and out comes the answer.

💡
The big idea: An ICE table lays out the Initial amounts, the Change as the reaction proceeds (written in terms of one unknown x, scaled by the coefficients), and the Equilibrium amounts (initial + change). Plug that bottom row into the K expression and solve for x.
🎯 By the end, you'll be able to
  • Set up an ICE table with the correct signs and coefficient ratios
  • Solve for equilibrium concentrations given initial amounts and K
  • Calculate K from initial amounts and one measured equilibrium concentration
  • Decide when to use the small-x approximation and when to solve the quadratic
📎 Helpful to know first

What ICE stands for

An ICE table has one column per species and three rows:

  • I — Initial: the concentrations you start with (products often 0).
  • C — Change: how much each concentration changes as the reaction moves to equilibrium, written with a single unknown x. Reactants that get used up are −; products that form are +; each is scaled by its coefficient.
  • E — Equilibrium: simply Initial + Change for each column.

Then you substitute the E row into the K expression and solve for x.

🔑 The change row follows the coefficients
For N₂ + 3H₂ ⇌ 2NH₃, if x is the amount of N₂ that reacts, the changes are −x for N₂, −3x for H₂, and +2x for NH₃. The coefficients become the multipliers on x; the signs show what's consumed versus formed.
\[ [\text{species}]_{\text{eq}} = [\text{species}]_{\text{initial}} + (\text{coefficient})\times(\pm x) \]
Every entry in the Equilibrium row is just the Initial value plus the coefficient-scaled change.

Case 1 — finding K from a measured concentration

The simplest ICE problems give you the initial amounts and one measured equilibrium concentration, and ask for K. You use the measurement to pin down x, fill in the rest of the E row, and compute K directly — no equation to solve.

📝 Worked example: H₂(g) + I₂(g) ⇌ 2HI(g). A flask starts with [H₂] = [I₂] = 0.100 mol/L and no HI. At equilibrium [HI] = 0.156 mol/L. Find K.
  1. Let x be the amount of H₂ that reacts. Change row: H₂ −x, I₂ −x, HI +2x.
  2. HI is formed as +2x and starts at 0, so at equilibrium [HI] = 2x = 0.156 → x = 0.078.
  3. Equilibrium row: [H₂] = [I₂] = 0.100 − 0.078 = 0.022 mol/L; [HI] = 0.156 mol/L.
  4. K = [HI]² / ([H₂][I₂]) = (0.156)² / (0.022 × 0.022) = 0.024336 / 0.000484.
  5. K ≈ 50.3.
✓ K ≈ 50 — strongly product-favoured at this temperature.

Case 2 — solving for x from K

When you're given K and asked for the equilibrium concentrations, x appears inside the K expression and you have to solve for it. That leaves you with either a quadratic to solve exactly, or — when x is tiny compared with the initial concentration — a shortcut approximation.

✨ The 5% approximation
If K is very small, only a little reactant converts, so x is tiny. Then a term like (0.100 − x) is barely different from 0.100, and you can drop the x to avoid the quadratic. The rule of thumb: the approximation is acceptable if x is less than 5% of the initial concentration. Always check afterward; if it's over 5%, solve the full quadratic instead.
📝 Worked example: N₂O₄(g) ⇌ 2NO₂(g), K = 0.36 at this temperature. A flask starts with [N₂O₄] = 0.100 mol/L and no NO₂. Find the equilibrium concentrations. (Here x is NOT small — solve the quadratic.)
  1. ICE change row: N₂O₄ −x, NO₂ +2x. Equilibrium: [N₂O₄] = 0.100 − x, [NO₂] = 2x.
  2. K = [NO₂]² / [N₂O₄] = (2x)² / (0.100 − x) = 4x² / (0.100 − x) = 0.36.
  3. Clear the fraction: 4x² = 0.36(0.100 − x) = 0.036 − 0.36x, so 4x² + 0.36x − 0.036 = 0.
  4. Divide by 4: x² + 0.09x − 0.009 = 0. Quadratic formula: x = [−0.09 ± √(0.09² + 4·0.009)] / 2 = [−0.09 ± √0.0441] / 2 = (−0.09 ± 0.21) / 2.
  5. Take the positive root: x = 0.12 / 2 = 0.060. (The negative root gives a negative concentration — rejected.)
  6. Equilibrium: [N₂O₄] = 0.100 − 0.060 = 0.040 mol/L; [NO₂] = 2(0.060) = 0.120 mol/L. Check: (0.120)²/0.040 = 0.0144/0.040 = 0.36 ✓.
✓ [N₂O₄] = 0.040 mol/L and [NO₂] = 0.120 mol/L. Note x = 0.060 is 60% of 0.100, so the small-x shortcut would have been wildly wrong here.
✏️ Practice: A ⇌ B + C, K = 4.0×10⁻⁵, starting with [A] = 0.20 mol/L and no B or C. K is tiny, so use the approximation [A] ≈ 0.20. Solve for x, the equilibrium value of [B]. (K ≈ x² / 0.20.) Give x in mol/L.
mol/L
Solution
  1. ICE: [A] = 0.20 − x, [B] = x, [C] = x. So K = x²/(0.20 − x).
  2. K is tiny, so assume x ≪ 0.20 and use K ≈ x²/0.20.
  3. x² = K × 0.20 = (4.0×10⁻⁵)(0.20) = 8.0×10⁻⁶.
  4. x = √(8.0×10⁻⁶) = 2.83×10⁻³ mol/L (0.00283).
  5. Check the approximation: x/0.20 = 0.00283/0.20 = 1.4%, which is under 5% — the shortcut was valid.
✏️ Practice: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). A flask starts with [PCl₅] = 0.100 mol/L and no products. At equilibrium [Cl₂] = 0.040 mol/L. Calculate K. (Round to 2 significant figures.)
Solution
  1. Change row: PCl₅ −x, PCl₃ +x, Cl₂ +x. Since [Cl₂] = x = 0.040, then x = 0.040.
  2. Equilibrium: [PCl₅] = 0.100 − 0.040 = 0.060; [PCl₃] = 0.040; [Cl₂] = 0.040 mol/L.
  3. K = [PCl₃][Cl₂] / [PCl₅] = (0.040 × 0.040) / 0.060 = 0.0016 / 0.060.
  4. K = 0.027 (2 sig figs). K < 1, so reactants are favoured.

Check your understanding

1. In an ICE table, what does the middle (Change) row represent?
The Change row records how much each species is consumed (−) or formed (+) as the reaction proceeds, scaled by the coefficients and written using one unknown x.
2. For 2A ⇌ B, if x is the amount of B formed, what is the change in [A]?
The coefficient of A is 2, and A is consumed, so its change is −2x while B's change is +x.
3. When is the small-x approximation (dropping x in a term like 0.10 − x) acceptable?
The approximation holds when x is small — under about 5% of the initial concentration. Always check afterward; if x exceeds 5%, solve the quadratic.
🔎 Related lessons
Dynamic Equilibrium & K The Reaction Quotient Q Le Châtelier's Principle pH, pOH & Strong Acids/Bases
✅ Key takeaways
  • ICE = Initial, Change, Equilibrium — one column per species, three rows.
  • The Change row uses the coefficients as multipliers on x, with − for reactants and + for products.
  • Substitute the Equilibrium row into the K expression and solve for x.
  • Given a measured equilibrium concentration, you can find K directly (no equation to solve).
  • Use the small-x approximation only when x is under ~5% of the initial concentration; otherwise solve the quadratic.
➡️ ICE tables are the workhorse for every equilibrium calculation ahead — weak-acid and weak-base problems, buffers, and solubility all run on exactly this method. Next up, acids and bases put ICE tables to work on the pH scale.
Want to test yourself on this? Try the Chemistry practice test →