ICE tables: solving for equilibrium concentrations

Master ICE tables (Initial, Change, Equilibrium): the systematic method for computing equilibrium concentrations, with a fully worked example.

Quick answer

An ICE table organizes equilibrium calculations into three rows — Initial concentrations, Change (in terms of x), and Equilibrium concentrations — which are then substituted into the K expression and solved for x.

Example: For H₂ + I₂ ⇌ 2HI (Kc = 54.3) starting at 0.150 M H₂ and 0.150 M I₂: solving 4x²/(0.150−x)² = 54.3 gives x = 0.118, so [HI] = 0.236 M and [H₂] = [I₂] = 0.0320 M at equilibrium.

What you will learn

  • Set up an ICE table with correct signs from the direction of shift.
  • Write the equilibrium-constant expression in terms of x.
  • Solve for x and compute all equilibrium concentrations.
  • Verify the answer by checking that Q equals K.

The Initial–Change–Equilibrium framework

When you know the starting concentrations and K but not the final ones, an ICE table keeps the bookkeeping straight. The three rows are:

  • I — Initial: the starting concentration of each species
  • C — Change: how each concentration shifts, written in terms of x (the extent of reaction)
  • E — Equilibrium: initial plus change, the expressions you plug into K

The signs in the change row follow the direction of shift: species that are consumed get −x (times their coefficient); species that form get +x. When in doubt, compute Q first — Q < K means shift right, Q > K means shift left.

Setting up the table: a full example

Consider H₂(g) + I₂(g) ⇌ 2HI(g) with Kc = 54.3 at 430 °C. A flask starts with [H₂] = 0.150 M, [I₂] = 0.150 M, and [HI] = 0. Since Q = 0 < K, the reaction shifts right: H₂ and I₂ are consumed, HI forms.

| | H₂ | I₂ | 2HI |

|---|---|---|---|

| Initial (M) | 0.150 | 0.150 | 0 |

| Change (M) | −x | −x | +2x |

| Equilibrium (M) | 0.150 − x | 0.150 − x | 2x |

Substituting the equilibrium row into the K expression:

Kc=[HI]2[H2][I2]=(2x)2(0.150−x)(0.150−x)=54.3K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.150 - x)(0.150 - x)} = 54.3

Solving for x and verifying

Taking the square root of both sides simplifies the algebra:

2x0.150−x=54.3=7.369\frac{2x}{0.150 - x} = \sqrt{54.3} = 7.369

2x=7.369(0.150−x)=1.105−7.369x2x = 7.369(0.150 - x) = 1.105 - 7.369x

9.369x=1.105⇒x=0.1189.369x = 1.105 \quad\Rightarrow\quad x = 0.118

Now read the equilibrium concentrations from the table's last row:

  • [HI] = 2x = 2(0.118) = 0.236 M
  • [H₂] = [I₂] = 0.150 − 0.118 = 0.0320 M

Always sanity-check: concentrations must be positive (they are), and plugging back should recover K: (0.236)²/(0.0320)² = 0.0557/0.00102 ≈ 54.4, matching 54.3 within rounding. The check closes the loop.

The small-x approximation

When K is very small, x is tiny compared to the initial concentrations, and you can approximate (initial − x) ≈ initial. This turns a quadratic into a one-line calculation. The rule: after solving, check that x is less than 5% of the initial concentration. If it passes, the approximation stands; if not, solve the full quadratic.

Example: for a reaction with K = 1.0 × 10⁻⁵ starting at 0.20 M reactant, x ≈ √(K × 0.20) = √(2.0 × 10⁻⁶) = 1.4 × 10⁻³ M — just 0.7% of 0.20 M, so the approximation is fine.

Variables and units

SymbolNameUnitMeaning
IIInitial concentrationMInitial concentration of a species before any shift
xxChange variableMExtent of reaction; the unknown solved from the K expression
EEEquilibrium concentrationMConcentration once equilibrium is reached (initial ± change)

Common mistakes

✗ The 'change' row always subtracts x from reactants.

The signs depend on the direction of shift. If Q > K, the reaction moves left: reactants gain (+x) and products lose (−x). Always check Q versus K (or think physically) before assigning signs.

✗ x is the equilibrium concentration.

x is the change in concentration — the extent of reaction. Equilibrium concentrations are initial ± x (times coefficients). Reporting x itself as the answer is a classic mistake.

✗ You can ignore the coefficients in the change row and just use x everywhere.

Coefficients scale the change: for 2HI forming, the change is +2x, not +x. The coefficient appears in the ICE table and again as an exponent in the K expression — both are required.

Real-world applications

  • Industrial chemistry: predicting equilibrium yields before building a reactor.

  • Environmental science: computing how pollutants partition at equilibrium.

  • Biochemistry: working out the final pH of buffer mixtures.

  • Classroom problem-solving: the standard method for every 'find the equilibrium concentrations' question.

Frequently asked questions

How do I know whether x is added or subtracted?

Use Q first. If Q < K, the reaction shifts right: reactants decrease (−x) and products increase (+x). If Q > K, flip the signs. Guessing wrong gives a negative concentration at the end — a clear signal to reverse the signs.

Do stoichiometric coefficients appear twice — in the change row and as exponents?

Yes. In the equilibrium expression, a change of +2x for 2HI becomes (2x)² — the coefficient appears both in the ICE row and as an exponent. Forgetting either one is the most common ICE error.

What if the equation has coefficients like 2 or 3?

Multiply every change entry by the same x (with its coefficient): −x, −x, +2x. There is only ever one x per reaction — it represents the single extent of reaction. Never use x and y for the same reaction.

When is the small-x approximation allowed?

Drop the tiny term: if x is less than 5% of the initial concentration, the approximation is accepted. If it fails the 5% check, solve the full quadratic — the approximation was not justified.