Equilibrium Constant from an ICE Table

Introduction

Many equilibrium problems give you the starting mixture and just one piece of information about the final state: the amount of a product formed, the fraction of a reactant left, or a colour intensity that corresponds to one concentration. From that single measurement you must reconstruct the whole equilibrium composition and then the equilibrium constant. The ICE table (Initial, Change, Equilibrium) is the bookkeeping tool that makes this reliable. It combines mass balance and coefficient ratios, which you met earlier, with the equilibrium law.

Core explanation

One variable controls every change. For a reaction such as H₂(g) + I₂(g) ⇌ 2HI(g), if x mol of H₂ reacts, then x mol of I₂ must also react and 2x mol of HI must form. The changes are not independent: they all follow from one reaction extent multiplied by the stoichiometric coefficients. Reactants get a minus sign, products a plus sign.

The three rows.

H₂ I₂ HI --- --- --- --- Initial / mol 1.00 1.00 0 Change / mol −x −x +2x Equilibrium / mol 1.00 − x 1.00 − x 2x

If the problem states that 1.56 mol of HI is present at equilibrium, then 2x = 1.56 and x = 0.78. The equilibrium amounts are therefore 0.22 mol H₂, 0.22 mol I₂ and 1.56 mol HI.

Amounts versus concentrations. Kc uses concentrations, so each equilibrium amount must be divided by the volume V of the vessel:

Kc = [HI]² / ([H₂][I₂]) = (1.56/V)² / ((0.22/V)(0.22/V))

Here the V² terms cancel because the number of moles is the same on both sides (Δn = 0), and Kc = 1.56² / 0.22² = 2.434 / 0.0484 ≈ 50. When Δn is not zero, the volume does not cancel and forgetting to divide by V gives a wrong answer with the wrong units.

Units of Kc. Each concentration carries mol dm⁻³, so Kc has units of (mol dm⁻³)^Δn. For N₂O₄(g) ⇌ 2NO₂(g), Δn = +1 and Kc has units of mol dm⁻³; for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = −2 and the units are dm⁶ mol⁻². Writing the units is a useful check that the expression is the right way up.

Where the ICE table can be built. It is usually easiest to fill the table in moles, because mass balance holds for amounts regardless of volume, and to convert to concentrations only in the final row. If the volume is constant you may instead work in concentrations throughout; both routes give the same answer provided you are consistent.

Formulae

For aA + bB ⇌ cC + dD: Kc = [C]^c [D]^d / ([A]^a [B]^b), with concentrations at equilibrium. Change in amount of species i = νᵢx, where νᵢ is negative for reactants. Units of Kc = (mol dm⁻³)^Δn, with Δn = (c + d) − (a + b).

Step-by-step reasoning

1. Write the balanced equation and the Kc expression. 2. Enter the initial amounts (use zero for anything absent). 3. Write every change in terms of one unknown x, using the coefficients. 4. Use the given equilibrium datum to find x. 5. Complete the equilibrium row and check that no amount is negative. 6. Divide amounts by the volume, substitute into Kc and state its units.

Visual explanation

Picture the ICE table as a ledger with one column per species. The "Change" row is a single stamp, x, pressed through a stencil cut in the ratio 1 : 1 : 2; reactant columns receive a withdrawal and the product column receives a deposit. Reading down each column always gives Initial + Change = Equilibrium.

Real-world analogy

A café starts the morning with fixed stocks of bread and cheese. Each sandwich uses one slice of cheese and two slices of bread. If you know only how many sandwiches were sold, you can work out the remaining stock of both ingredients, because every sale changes the stocks in a fixed ratio. The ICE table applies the same logic to molecules.

Real-world example

Industrial chemists studying the synthesis of hydrogen iodide or ammonia measure the composition of an equilibrium mixture, often by analysing only one component, such as the ammonia content by absorption. The ICE table lets them reconstruct the full composition and compare Kc values at different temperatures, which guides the choice of operating conditions.

Why?

Why can one measured quantity determine all the others? Because atoms are conserved and the reaction proceeds only in the ratio set by the balanced equation. A single reaction has one degree of freedom, its extent, so fixing one amount fixes x and therefore every other amount.

Common misconception

"You can put initial concentrations into the Kc expression." Kc is defined only with equilibrium concentrations. Substituting starting values gives a reaction quotient for the initial mixture, which is a different quantity and usually not equal to Kc.

Worked example

Question: 0.100 mol of N₂O₄ is sealed in a 1.00 dm³ flask at 298 K. At equilibrium the flask contains 0.0200 mol of NO₂. Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g).

Reasoning: Initial: N₂O₄ 0.100, NO₂ 0. Change: −x, +2x. Given 2x = 0.0200, so x = 0.0100. Equilibrium: N₂O₄ 0.0900 mol, NO₂ 0.0200 mol. With V = 1.00 dm³, [N₂O₄] = 0.0900 mol dm⁻³ and [NO₂] = 0.0200 mol dm⁻³. Kc = (0.0200)² / 0.0900 = 4.00 × 10⁻⁴ / 0.0900 = 4.44 × 10⁻³.

Answer: Kc = 4.44 × 10⁻³ mol dm⁻³.

Quick check

1. In the reaction 2SO₂ + O₂ ⇌ 2SO₃, if 0.30 mol of SO₃ forms from pure reactants, how much O₂ has reacted? Answer: 0.15 mol of O₂, because the extent x is 0.15 and O₂ has a coefficient of 1 while SO₃ has 2.

Exam focus

Examiners reward a clearly labelled ICE table with units, a correct link between the given datum and x, and a final Kc with units. Common lost marks: forgetting to divide by volume when Δn ≠ 0, squaring the wrong term, and quoting too many significant figures.

Advanced insight

The thermodynamic equilibrium constant is dimensionless because each concentration is divided by a standard concentration c° = 1 mol dm⁻³, and strictly it uses activities rather than concentrations. For dilute gases and solutions the numerical values agree closely, so the concentration Kc with units is a practical approximation that fails mainly in concentrated ionic solutions and non-ideal gases.

Summary

An ICE table records initial amounts, changes tied to one reaction extent through the coefficients, and equilibrium amounts. One measured equilibrium quantity fixes the extent and hence the whole composition. Convert amounts to concentrations using the volume before substituting into Kc, whose units are (mol dm⁻³)^Δn; the volume cancels only when Δn = 0.

Practice questions

1. 0.200 mol of PCl₅ is placed in a 2.00 dm³ vessel and 40.0% dissociates: PCl₅ ⇌ PCl₃ + Cl₂. Calculate Kc. Answer: x = 0.0800 mol; equilibrium concentrations are PCl₅ 0.0600, PCl₃ 0.0400 and Cl₂ 0.0400 mol dm⁻³, so Kc = (0.0400 × 0.0400) / 0.0600 = 0.0267 mol dm⁻³. 2. Explain why the vessel volume matters for the PCl₅ equilibrium but not for H₂ + I₂ ⇌ 2HI. Answer: PCl₅ dissociation has Δn = +1, so one factor of 1/V is left in Kc; for the hydrogen iodide equilibrium Δn = 0 and the volume factors cancel. 3. For H₂ + I₂ ⇌ 2HI with Kc = 49 at a certain temperature, 0.50 mol of each reactant is placed in a 1.00 dm³ flask. Find the equilibrium concentration of HI. Answer: (2x)² / (0.50 − x)² = 49, so 2x / (0.50 − x) = 7, giving x = 0.389 and [HI] = 0.78 mol dm⁻³. 4. State the units of Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Answer: Δn = 2 − 4 = −2, so the units are (mol dm⁻³)⁻², which is dm⁶ mol⁻².