ICE Tables: Solving for Equilibrium Step by Step

What the three rows of an ICE table record

ICE stands for Initial, Change and Equilibrium. Write the balanced equation across the top and give each species its own column. The Initial row holds the concentrations (or partial pressures) before the reaction shifts. The Change row holds how much each one goes up or down. The Equilibrium row is the sum of the first two.

The Change row is where the chemistry lives. Every entry is a multiple of a single unknown, x, and the multiples are the coefficients of the balanced equation. For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) running forward, the changes are −2x, −x and +2x. Reactants lose and products gain when the reaction runs forward; the signs flip when it runs in reverse.

Use molar concentrations for Kc and partial pressures for Kp. Pure solids and pure liquids do not appear in the equilibrium expression, so leave them out of the table as well. Convert any moles to concentrations by dividing by the container volume before you fill in the Initial row.

Choosing the direction of change with Q

Before you assign signs, decide which way the reaction must go. Compute the reaction quotient Q from the initial values, using the same form as the K expression. If Q is less than K, there are too few products and the reaction runs forward. If Q is greater than K, it runs in reverse. If Q equals K, the mixture is already at equilibrium and nothing changes.

When a product starts at zero, Q is zero and the reaction must run forward, so you can skip the calculation. When every species is present, compute Q. For example, take the sulfur trioxide reaction with Kc = 37.0 (found in the next section) and a mixture with [SO₂] = 0.100 M, [O₂] = 0.050 M and [SO₃] = 0.500 M. Q = (0.500)² / ((0.100)² × 0.050) = 500. Q is greater than K, so SO₃ decomposes until the ratio falls to 37.0.

Worked example: K from one measured concentration

Sometimes the unknown is K itself. Suppose a vessel starts with [SO₂] = 0.400 M and [O₂] = 0.200 M and no SO₃, and at equilibrium [SO₃] is measured as 0.250 M. These values were constructed for this guide to show the method; they are not literature data.

SO₃ went from 0 to 0.250 M, so 2x = 0.250 and x = 0.125 M. Every other change follows from the coefficients: SO₂ falls by 2x = 0.250 M to 0.150 M, and O₂ falls by x = 0.125 M to 0.075 M. Then Kc = [SO₃]² / ([SO₂]²[O₂]) = (0.250)² / ((0.150)² × 0.075) = 37.0. One measured concentration plus the stoichiometry was enough to fix all three.

Worked example: when the equation is a perfect square

For H₂(g) + I₂(g) ⇌ 2 HI(g), use Kc = 50.5, the value OpenStax Chemistry 2e section 13.4 uses for this reaction under the conditions of its example. Start with [H₂] = [I₂] = 0.500 M and no HI. Q is zero, so the reaction runs forward. The table gives equilibrium values of 0.500 − x for each reactant and 2x for HI.

The expression is 50.5 = (2x)² / (0.500 − x)². Both sides are perfect squares, so take the square root instead of expanding: 7.106 = 2x / (0.500 − x). Rearranging gives x = 0.500 × 7.106 / (2 + 7.106) = 0.390 M. At equilibrium [H₂] = [I₂] = 0.110 M and [HI] = 0.780 M. Check: (0.780)² / (0.110)² = 50.3, which matches 50.5 within the rounding of the concentrations.

The square-root shortcut works only when the expression is a perfect square, which needs equal starting amounts of the two reactants here. With unequal amounts you must solve a quadratic.

Worked example: solving the full quadratic

For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), OpenStax section 13.4 uses Kc = 0.0211. Start with [PCl₅] = 0.400 M and [Cl₂] = 0.100 M but no PCl₃. Q is zero, so the reaction runs forward. Equilibrium values are 0.400 − x, x and 0.100 + x.

Substitute: 0.0211 = x(0.100 + x) / (0.400 − x). Multiply out and collect terms to get x² + 0.1211x − 0.00844 = 0. The quadratic formula gives x = 0.0495 or x = −0.171. The negative root would make [PCl₃] negative, so reject it. At equilibrium [PCl₅] = 0.351 M, [PCl₃] = 0.0495 M and [Cl₂] = 0.150 M. Substituting back gives 0.0211, so the answer checks.

Notice that the Cl₂ already present held the reaction back. Starting from 0.400 M PCl₅ alone, the same K gives x = 0.0819 M, so the added product reduced the extent of decomposition, as Le Chatelier's principle predicts.

When x is small: the approximation and the 5 percent check

When K is small and the starting concentrations are not, the reaction barely moves, and 0.0400 − x is almost exactly 0.0400. Dropping x from those terms turns a quadratic into a one-step calculation. The approximation is accepted when x is less than 5 percent of the concentration it was subtracted from.

Worked example. For N₂(g) + O₂(g) ⇌ 2 NO(g), OpenStax section 13.4 gives Kc = 4.1 × 10⁻⁴ at 2000 °C. Start with [N₂] = 0.0400 M and [O₂] = 0.0100 M. Then 4.1 × 10⁻⁴ = (2x)² / ((0.0400 − x)(0.0100 − x)). Approximate: 4x² = 4.1 × 10⁻⁴ × 0.0400 × 0.0100, so x = 2.0 × 10⁻⁴ M and [NO] = 2x = 4.0 × 10⁻⁴ M. Check against the smaller starting value: 2.0 × 10⁻⁴ / 0.0100 = 2 percent, under 5 percent. The exact quadratic gives the same [NO] to two significant figures.

The approximation fails when K is not small compared with the concentrations. For PCl₅ alone at 0.400 M with Kc = 0.0211, assuming x is small gives x = 0.092 M, which is 23 percent of 0.400. The exact answer is 0.0819 M, so the shortcut would be about 12 percent too high. Always run the check; never assume.

Partial pressures and the reverse reaction

An ICE table works the same way with partial pressures and Kp. For a reaction with no change in the number of gas moles, Kp equals Kc. That is true of the hydrogen iodide equilibrium, which has two moles of gas on each side.

Worked example. Pure HI is placed in a vessel at 2.00 atm and allowed to decompose: 2 HI(g) ⇌ H₂(g) + I₂(g). This is the reverse of the reaction above, so its constant is 1/50.5 = 0.0198. The changes are −2x for HI and +x for each product. Then 0.0198 = x² / (2.00 − 2x)². Taking the square root, 0.1407 = x / (2.00 − 2x), which gives x = 0.220 atm. At equilibrium P(H₂) = P(I₂) = 0.220 atm and P(HI) = 1.56 atm.

ICE-table slips and how to catch them

A final check catches nearly every error: substitute the equilibrium values back into the K expression and confirm you recover K to the precision of the data.

• Putting moles in the table when the container is not 1 L. Divide by volume first.
• Using 1 for every coefficient in the Change row, for example −x instead of −2x for SO₂.
• Forgetting to square or cube a term in the K expression to match its coefficient.
• Assuming the reaction runs forward when products are present at the start. Compute Q.
• Keeping a root that makes any concentration negative or larger than is physically possible.
• Dropping x without checking that it is under 5 percent of the starting value.
• Including a pure solid or liquid in the table or in the expression.
• Using K for the forward reaction when the problem is written in reverse. Reversing a reaction inverts K.