ICE Table Method for Equilibrium Calculations in Class 11

By Aniket Bhardwaj · 14 October 2026 · Updated 14 October 2026 · Class 11 Chemistry

Most Class 11 equilibrium numericals ask the same thing in different words: "Some amounts are given at the start. Find the amounts at equilibrium." The ICE table is the tidy way to do this. It turns a confusing paragraph into a short algebra problem. Learn the table once and you can solve almost every Kc problem in your chapter.

The ICE table for H₂ + I₂ ⇌ 2HI shows initial values 1.0, 1.0, and 0; changes −x, −x, and +2x; and equilibrium values 1.0 − x, 1.0 − x, and 2x. — Track each substance from its initial concentration through change to equilibrium.

What ICE stands for

ICE is a three-row table with one column for each substance in the reaction.

I = Initial. The concentration of each substance at the start, before any reaction.
C = Change. How much each substance changes. Use the balanced equation: the change is proportional to the coefficients. Reactants lose (−), products gain (+).
E = Equilibrium. Initial plus change. Only these values go into the Kc expression.

For aA + bB ⇌ cC + dD:
K_c = [C]^c[D]^d / ([A]^a[B]^b)
Change row: −ax, −bx, +cx, +dx (x is the unknown)

Use concentrations in mol/L. If the question gives moles and a volume, divide moles by the volume first. When the total moles of gas are the same on both sides (Δn = 0), the volume cancels out of K_c, but it is still safer to convert to concentrations.

The five steps

Write the balanced equation and the Kc expression.
Fill the I row with the starting concentrations (zero for anything not added).
Fill the C row using x and the coefficients.
Fill the E row. Put these into the Kc expression.
Solve for x, then calculate the equilibrium concentrations. Check by putting them back into Kc.

Worked examples

Example 1: a perfect square. H₂ + I₂ ⇌ 2HI, K_c = 50 (value given in the problem). Start with 1.0 mol/L each of H₂ and I₂ and no HI. Find the equilibrium concentrations.
I: [H₂] = 1.0, [I₂] = 1.0, [HI] = 0.
C: −x, −x, +2x.
E: 1.0 − x, 1.0 − x, 2x.
K_c = (2x)² / (1.0 − x)² = 50. The right side is a perfect square, so take the square root of both sides.
2x / (1.0 − x) = √50 = 7.071. So 2x = 7.071 − 7.071x, so 9.071x = 7.071, so x = 0.7795.
[H₂] = [I₂] = 1.0 − 0.7795 = 0.2205 M. [HI] = 2 × 0.7795 = 1.559 M.
Check: (1.559)² / (0.2205)² = 2.430 / 0.04862 = 50.0 ✓

Example 2: a quadratic equation. N₂O₄ ⇌ 2NO₂, K_c = 0.0046 (value given in the problem). Start with 0.10 M N₂O₄ only.
I: 0.10, 0. C: −x, +2x. E: 0.10 − x, 2x.
K_c = (2x)² / (0.10 − x) = 0.0046, so 4x² = 0.0046(0.10 − x) = 0.00046 − 0.0046x.
Rearrange: 4x² + 0.0046x − 0.00046 = 0.
x = [−0.0046 + √(0.0046² + 4 × 4 × 0.00046)] / (2 × 4) = [−0.0046 + √0.00738] / 8 = (−0.0046 + 0.08591) / 8 = 0.01016.
[NO₂] = 2x = 0.0203 M. [N₂O₄] = 0.10 − 0.01016 = 0.0898 M.
Check: (0.0203)² / 0.0898 = 0.0046 ✓ (the negative root of the quadratic is rejected because concentration cannot be negative).

Example 3: finding K_c from equilibrium data. 2.0 mol SO₂ and 1.0 mol O₂ are placed in a 1.0 L flask. At equilibrium 1.2 mol SO₃ is present. Find K_c for 2SO₂ + O₂ ⇌ 2SO₃.
I: 2.0, 1.0, 0. C: −2x, −x, +2x. The equilibrium SO₃ = 2x = 1.2, so x = 0.6.
E: SO₂ = 2.0 − 1.2 = 0.8; O₂ = 1.0 − 0.6 = 0.4; SO₃ = 1.2.
K_c = (1.2)² / [(0.8)² × 0.4] = 1.44 / 0.256 = 5.625 L/mol, about 5.6 L/mol.
Notice the unit: Δn = −1 here, so K_c has a unit. Many textbooks leave out units for K, so follow your teacher's style.

Example 4: when x is very small. A ⇌ B + C, K_c = 1.0 × 10⁻⁶. Start with 0.10 M A.
E: 0.10 − x, x, x. So x² / (0.10 − x) = 1.0 × 10⁻⁶.
Because K_c is tiny, x is far smaller than 0.10, so take 0.10 − x ≈ 0.10. Then x² = 1.0 × 10⁻⁷, and x = 3.16 × 10⁻⁴ M.
Check: x is 0.32% of 0.10, well under 5%, so the shortcut is acceptable. Solving the full quadratic gives x = 3.157 × 10⁻⁴ M, nearly the same.

When can you ignore x in the denominator?

If K_c is very small compared with the starting concentration, x is tiny and "initial − x" is almost the initial value. A common classroom rule is to check that x is less than 5% of the initial value. In Example 2, x is 10.2% of 0.10, so the shortcut would be a poor choice there; in Example 4 it works. Always verify the shortcut after you use it, and if it fails, solve the quadratic.

Using moles instead of concentration. If volume is given, convert to mol/L before putting values in Kc.
Forgetting the coefficients in the C row. For 2HI the change is 2x, and the Kc expression raises the concentration to the power 2.
Putting the I row values into Kc. Only the E row belongs there.
Including pure solids and liquids. They do not appear in the Kc expression.
Wrong sign of x. If only reactants are present at the start, the reaction must go forward, so reactants decrease. If only products are present, it goes backward.
Keeping both roots of the quadratic. Reject any root that makes a concentration negative.

Exam summary

Given	Find	Method
Initial amounts and K_c	Equilibrium amounts	ICE table, solve for x
Initial amounts and one equilibrium amount	K_c	Find x from the known amount, complete E row
K_c very small	Equilibrium amounts	Ignore x in the denominator, then verify
Perfect-square expression	Equilibrium amounts	Take the square root of both sides

Check your textbook for how your board writes units of K_c and what level of algebra is expected.

Work the ICE table by hand, as the exam expects. The calculator suite has no dedicated equilibrium tool, but you can use its scientific calculator to check the algebra and square roots.

Open the Chemistry Calculator Suite →