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ICE Tables & K — Advanced

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The small-x shortcut and why it works

When K is very small, only a tiny amount of reactant converts, so the change x is negligible next to the initial concentration. In an equilibrium term like (0.100 − x), if x turns out to be ~0.001, then 0.100 − x ≈ 0.100 to three figures. Dropping the −x collapses a quadratic (or worse) into a term you can solve by a single square root. The approximation is a convenience, not a law — you must justify it afterward.

The 5% rule — always check

The standard test: the approximation is acceptable if x is less than 5% of the initial concentration you subtracted it from. A quick pre-screen is the ratio [initial]/K: when it is roughly ≥ 500, x will come out under 5% and the shortcut is safe. If x lands above 5% — or worse, comes out larger than the initial concentration (a physical impossibility) — you must discard the approximation and solve the full equation exactly.

When the shortcut fails: the quadratic

For a moderate K, a meaningful fraction of reactant reacts, so −x is not negligible and you cannot drop it. Keeping the full (initial − x) term produces a quadratic ax² + bx + c = 0, solved with x = (−b ± √(b² − 4ac)) / 2a. Two roots come out; keep only the physically real one — the root that makes every equilibrium concentration positive. A negative concentration, or one larger than the initial amount, is rejected.

Quadratic formula (for the exact ICE solve)
ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a
Use when the small-x approximation fails the 5% test. Discard the root that gives a negative concentration or a value exceeding the initial amount.
Kp from Kc
Kp = Kc (RT)^Δn, Δn = (moles of gaseous product) − (moles of gaseous reactant)
R = 0.0821 L·atm·mol⁻¹·K⁻¹ and T is in kelvin. If Δn = 0 the factor is 1, so Kp = Kc. Only gas-phase species count toward Δn.
Worked example

For A(g) ⇌ B(g) + C(g), Kc = 1.0×10⁻⁵. A flask starts with 0.100 M A and no products. Find the equilibrium concentrations using the small-x approximation, and verify it is valid.

  1. 1.ICE. Initial: [A] = 0.100, [B] = [C] = 0. Change: [A] = −x, [B] = +x, [C] = +x. Equilibrium: [A] = 0.100 − x, [B] = x, [C] = x.
  2. 2.Substitute: Kc = [B][C]/[A] = x·x / (0.100 − x) = x² / (0.100 − x) = 1.0×10⁻⁵.
  3. 3.Pre-screen: [initial]/K = 0.100 / (1.0×10⁻⁵) = 1.0×10⁴, far above 500, so assume 0.100 − x ≈ 0.100.
  4. 4.Then x² / 0.100 = 1.0×10⁻⁵ → x² = 1.0×10⁻⁶ → x = √(1.0×10⁻⁶) = 1.0×10⁻³.
  5. 5.Validity check: x / initial = 1.0×10⁻³ / 0.100 = 0.010 = 1.0%, which is under 5% ✓.
  6. 6.Equilibrium: [B] = [C] = 1.0×10⁻³ M, [A] = 0.100 − 0.001 = 0.099 M.
Answer: [B] = [C] = 1.0×10⁻³ M and [A] ≈ 0.099 M. Because x is only 1% of the initial 0.100 M, dropping −x was justified.
Worked example

For A(g) ⇌ B(g) + C(g), Kc = 0.10. A flask starts with 0.20 M A only. Find x. (Notice K is not small here.)

  1. 1.ICE gives [A] = 0.20 − x, [B] = x, [C] = x, so Kc = x² / (0.20 − x) = 0.10.
  2. 2.Test the shortcut first: [initial]/K = 0.20 / 0.10 = 2, nowhere near 500 — the approximation will fail, so solve exactly.
  3. 3.Clear the denominator: x² = 0.10(0.20 − x) = 0.020 − 0.10x.
  4. 4.Rearrange to standard form: x² + 0.10x − 0.020 = 0, so a = 1, b = 0.10, c = −0.020.
  5. 5.Discriminant: b² − 4ac = (0.10)² − 4(1)(−0.020) = 0.010 + 0.080 = 0.090, and √0.090 = 0.30.
  6. 6.Roots: x = (−0.10 ± 0.30) / 2. The two candidates are x = +0.10 and x = −0.20. Reject the negative root.
  7. 7.So x = 0.10. Equilibrium: [A] = 0.20 − 0.10 = 0.10 M, [B] = [C] = 0.10 M. Check: (0.10)(0.10)/0.10 = 0.10 ✓.
Answer: x = 0.10 M, giving [A] = [B] = [C] = 0.10 M. Here the approximation would have predicted x ≈ 0.141 M — larger than the initial 0.20 M reactant left over would allow — flagging that the quadratic was mandatory.
Worked example

For N₂O₄(g) ⇌ 2NO₂(g), Kc = 0.10 at 298 K. Calculate Kp. Use R = 0.0821 L·atm·mol⁻¹·K⁻¹.

  1. 1.Find Δn from the gas moles: 2 mol gas product (NO₂) − 1 mol gas reactant (N₂O₄) = +1.
  2. 2.Apply Kp = Kc(RT)^Δn = 0.10 × (0.0821 × 298)¹.
  3. 3.Compute RT: 0.0821 × 298 = 24.5.
  4. 4.Kp = 0.10 × 24.5 = 2.45.
Answer: Kp ≈ 2.4. Because Δn = +1 (more gas moles on the product side), the (RT) factor makes Kp larger than Kc.
Checkpoint

For HA(aq) ⇌ H⁺(aq) + A⁻(aq), K = 4.0×10⁻⁷ and the initial [HA] = 0.10 M. Using the small-x approximation, what is [H⁺] at equilibrium?

Checkpoint

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which correctly relates Kp and Kc?

Checkpoint

For A(g) ⇌ B(g) + C(g), Kc = 0.20 and the initial [A] = 0.10 M. Is the small-x approximation appropriate here?

On the exam

On the free-response, the point is often awarded for the justification, not just the number. After a small-x solve, explicitly write the ratio (e.g. "x/[initial] = 1% < 5%, so the approximation is valid"). If it fails, say so and switch to the quadratic — graders reward recognizing the boundary.

Watch out

Never keep a quadratic root that yields a negative concentration or one exceeding the initial amount. Both roots are mathematically valid, but only one is chemically real. And when computing Δn for Kp, count gaseous species only — pure solids and liquids do not contribute.

Answer the 3 checkpoints as you read.

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