ICE Tables & Calculating K

You’ll be able to

Set up an ICE (Initial, Change, Equilibrium) table for a reversible reaction
Calculate K from initial amounts and a single equilibrium measurement
Use a known K to solve for equilibrium concentrations

The ICE table: bookkeeping for equilibrium

An ICE table organizes the three states of every species. Initial concentrations are what you mix in. The Change is written in terms of a single unknown x, scaled by each coefficient — reactants change by −(coefficient)·x, products by +(coefficient)·x. The Equilibrium row is just Initial + Change. One measurement lets you pin down x, and then every concentration is known.

Two directions, one tool

ICE tables answer both classic questions. If you know the equilibrium amounts, you use them to find x, fill the table, and compute K. If you know K and the starting amounts, you write everything in terms of x, substitute into the K expression, and solve the resulting equation for x — which gives you the equilibrium concentrations.

ICE relationship

Equilibrium concentration = Initial + Change (Change = ± coefficient × x)

Reactants are consumed (−), products are formed (+). The coefficient in front of x must match the balanced equation.

Worked example

A flask is filled with 1.00 M H₂ and 1.00 M I₂ (no HI). For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium [HI] is measured as 1.56 M. Calculate Kc.

1.Set up ICE (all in M). Initial: [H₂] = 1.00, [I₂] = 1.00, [HI] = 0.
2.Change: H₂ and I₂ each lose x; HI gains 2x (coefficient 2). Equilibrium: [H₂] = 1.00 − x, [I₂] = 1.00 − x, [HI] = 2x.
3.Use the measurement to find x: 2x = 1.56, so x = 0.78.
4.Fill the equilibrium row: [H₂] = 1.00 − 0.78 = 0.22 M, [I₂] = 0.22 M, [HI] = 1.56 M.
5.Write and evaluate Kc: Kc = [HI]² / ([H₂][I₂]) = (1.56)² / (0.22 × 0.22).
6.= 2.4336 / 0.0484 ≈ 50.

Answer: Kc ≈ 50. The large value confirms that product (HI) is strongly favored at equilibrium.

Checkpoint

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) starts with 1.00 M PCl₅ only. At equilibrium, 0.20 M Cl₂ has formed. What is Kc?

Worked example

For the reaction X(g) ⇌ Y(g), Kc = 4.0. A flask starts with 1.00 M X and no Y. Find the equilibrium concentration of Y.

1.ICE. Initial: [X] = 1.00, [Y] = 0. Change: [X] = −x, [Y] = +x. Equilibrium: [X] = 1.00 − x, [Y] = x.
2.Substitute into K: Kc = [Y] / [X] = x / (1.00 − x) = 4.0.
3.Solve: x = 4.0(1.00 − x) = 4.0 − 4.0x → 5.0x = 4.0 → x = 0.80.
4.Therefore [Y] = 0.80 M and [X] = 1.00 − 0.80 = 0.20 M.
5.Check: 0.80 / 0.20 = 4.0 ✓.

Answer: [Y] = 0.80 M (and [X] = 0.20 M). Because K > 1, more than half the X converts to Y.

Checkpoint

In the worked X ⇌ Y example above, why does the equilibrium [X] end up at only 0.20 M rather than closer to 1.00 M?

Tip

When K is very small and the change x is being subtracted from a much larger initial value, you may approximate "initial − x ≈ initial" to avoid the quadratic. The approximation is valid when x is under ~5% of the initial concentration — always check that at the end.

Watch out

Match the coefficient to x. If a product has coefficient 2, its change is +2x, and its equilibrium value is 2x — not x. Forgetting the coefficient here is one of the most common ICE-table errors, and it corrupts both x and the final K.

Answer the 2 checkpoints as you read.