ICE Tables Explained: How to Solve Equilibrium Problems for x, Kc and Kp
Short answer: An ICE table lists the Initial amount, the Change and the Equilibrium amount of each species in a reaction. Each change is the stoichiometric coefficient times x, negative for reactants and positive for products. Put the equilibrium row into the K expression and solve for x. The free ICE table and equilibrium solver does this exactly, shows the 5% check and tells you which way the reaction shifts.
What an ICE table is
An ICE table is a three-row table used to find equilibrium amounts. I is the initial concentration or pressure. C is the change on the way to equilibrium. E is the equilibrium amount, which is I + C.
The unknown x is the extent of reaction in mol/L (or atm for pressures). A species with coefficient ν changes by νx. For 2HI the change is +2x. For 3H₂ it is −3x.
Solids (s) and pure liquids (l) such as solvent water are left out of K, because their activity is 1. They still appear in the equation but get no column in the calculation.
The formulas
For a balanced reaction aA + bB ⇌ cC + dD:
- Kc = [C]^c [D]^d / ([A]^a [B]^b), using concentrations in mol/L.
- Kp = Kc (RT)^Δn, where Δn = moles of gaseous products − moles of gaseous reactants, R = 0.08206 L·atm/(mol·K) and T is in kelvin.
- Q has the same form as K but uses the current amounts. Q < K: the reaction goes forward. Q > K: it goes backward. Q = K: it is at equilibrium.
- Small-x approximation: if K is small, (C₀ − x) ≈ C₀. The result is valid when x is less than 5% of C₀.
How to use the ICE table solver
- Open the ICE table and equilibrium solver. Keep Solve equilibrium (find x) selected.
- Type the reaction in Balanced reaction, for example
N2O4(g) ⇌ 2NO2(g). Add (g), (aq), (s) or (l) after each species. Put a space on each side of every +. - Choose Amounts are: Concentrations (mol/L), Moles in a volume, Partial pressures (atm) or Partial pressures (bar).
- Pick K type (Kc or Kp) and type the Value of K. If K and the amounts do not match, the solver converts K using the Temperature.
- Type the initial amount of each species in the table. Leave a product at 0 if none is present.
- Read x and the equilibrium amounts on the right. Below them are the full ICE table, the equation, the polynomial in x and the 5% check.
- To go the other way, press Find K from equilibrium data. Enter every equilibrium amount, or the initial amounts plus one equilibrium amount. The Kp ↔ Kc converter tab converts a K value on its own.
Use the example buttons (H₂ + I₂ ⇌ 2HI, N₂O₄ ⇌ 2NO₂, Weak acid (acetic acid Ka), PCl₅ decomposition (Kp), N₂ + 3H₂ ⇌ 2NH₃) to load the problems below.
Worked examples
1. H₂ + I₂ ⇌ 2HI, Kc = 54.3 at 430 °C
Start with 0.500 M H₂ and 0.500 M I₂. The E row is 0.500 − x, 0.500 − x and 2x.
54.3 = (2x)² / (0.500 − x)². Take the square root: 2x / (0.500 − x) = 7.369, so x = 0.393 M.
At equilibrium [HI] = 0.787 M and [H₂] = [I₂] = 0.107 M. The other quadratic root, 0.686 M, would make [H₂] negative, so it is rejected.
2. N₂O₄ ⇌ 2NO₂, Kc = 4.63 × 10⁻³ at 25 °C
Start with 0.0400 M N₂O₄. 4.63 × 10⁻³ = (2x)² / (0.0400 − x).
The shortcut gives x ≈ √(4.63 × 10⁻³ × 0.0400 / 4) = 6.80 × 10⁻³ M. That is 17% of 0.0400 M, so the shortcut fails.
The quadratic 4x² + 4.63 × 10⁻³ x − 1.852 × 10⁻⁴ = 0 gives x = 6.25 × 10⁻³ M: [NO₂] = 0.0125 M and [N₂O₄] = 0.0337 M.
3. 0.10 M acetic acid, Ka = 1.8 × 10⁻⁵
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq). Water is left out. x² / (0.10 − x) = 1.8 × 10⁻⁵.
Shortcut: x ≈ √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M, which is 1.3% of 0.10 M. The shortcut is valid. The exact x is 1.33 × 10⁻³ M and pH = 2.88.
4. PCl₅ ⇌ PCl₃ + Cl₂, Kp = 1.05 at 250 °C
Start with P(PCl₅) = 1.00 atm. x² / (1.00 − x) = 1.05, so x² + 1.05x − 1.05 = 0 and x = 0.626 atm.
At equilibrium P(PCl₅) = 0.374 atm and P(PCl₃) = P(Cl₂) = 0.626 atm. Δn = +1, so Kc = 1.05 / (0.08206 × 523.15) = 0.0245.
5. N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.105 at 472 °C
Start with 1.00 M N₂ and 3.00 M H₂. 0.105 = (2x)² / [(1.00 − x)(3.00 − 3x)³]. This is a quartic with no neat shortcut.
The solver finds x = 0.353 M: [NH₃] = 0.705 M, [N₂] = 0.647 M and [H₂] = 1.94 M. With Δn = −2, Kp = 0.105 / (0.08206 × 745.15)² = 2.81 × 10⁻⁵.
6. Find K from one equilibrium amount
1.00 mol H₂ and 1.00 mol I₂ are heated in a 2.00 L flask. At equilibrium there is 1.56 mol HI.
[HI] = 0.780 M = 2x, so x = 0.390 M. [H₂] = [I₂] = 0.500 − 0.390 = 0.110 M. Kc = 0.780² / 0.110² = 50.3.
Summary table
| Problem | K | x | Shortcut valid |
|---|---|---|---|
| 0.500 M H₂ + 0.500 M I₂ | Kc = 54.3 | 0.393 M | No (K large) |
| 0.0400 M N₂O₄ | Kc = 4.63 × 10⁻³ | 6.25 × 10⁻³ M | No (17%) |
| 0.10 M acetic acid | Ka = 1.8 × 10⁻⁵ | 1.33 × 10⁻³ M | Yes (1.3%) |
| 1.00 atm PCl₅ | Kp = 1.05 | 0.626 atm | No |
| 1.00 M N₂ + 3.00 M H₂ | Kc = 0.105 | 0.353 M | No (quartic) |
Common mistakes
- Writing 2x² instead of (2x)² for a species with coefficient 2.
- Putting solids, pure liquids or solvent water into K.
- Using moles instead of concentrations when Δn is not zero.
- Using the small-x shortcut without checking the 5% rule.
- Keeping the quadratic root that makes a concentration negative.
- Counting non-gases in Δn, or using T in °C in Kp = Kc(RT)^Δn.
- Assuming the reaction goes forward when products are present at the start. Work out Q first.
Exam tips
- AP Chemistry: the free-response questions expect the ICE table, the K expression with numbers and a stated 5% check. Show units on each concentration.
- A-level: Kc and Kp questions often give moles in a stated volume. Divide by the volume first, and remember Kp uses mole fractions times total pressure.
- JEE and NEET: look for perfect squares, as in H₂ + I₂ ⇌ 2HI, and for degree of dissociation α instead of x. For PCl₅ with α: Kp = α²P / (1 − α²).
- Ratio of initial concentration to K above about 400 usually means the shortcut is safe. Below that, solve the quadratic.
Questions and answers
What does ICE stand for in chemistry?
ICE stands for Initial, Change and Equilibrium. The three rows list the starting amount of each species, the change in terms of x and the equilibrium amount, which is the sum of the first two.
How do I know if the 5% rule is satisfied?
Solve with the small-x shortcut, then divide x by the initial amount it was subtracted from and multiply by 100. If the result is under 5%, the shortcut is accepted. For 0.10 M acetic acid it is 1.3%; for 0.0400 M N2O4 it is 17%, so you must solve the quadratic.
Why can x be negative in an ICE table?
A negative x means the reaction moves backward, from products to reactants. This happens when Q is greater than K, for example when you start with only HI and let it form H2 and I2.
Which root of the quadratic should I choose?
Choose the root that keeps every equilibrium amount at zero or above. For H2 + I2 with 0.500 M each and Kc = 54.3, the roots are 0.393 and 0.686; only 0.393 is physical because 0.686 would make [H2] negative.
Do I use concentrations or pressures in an ICE table?
Use concentrations in mol/L with Kc and partial pressures with Kp. If you are given moles, divide by the volume in litres. If you are given K in the other form, convert it with Kp = Kc(RT)^Δn.
Why is water left out of the K expression for a weak acid?
In dilute solution water is the solvent and a pure liquid. Its concentration hardly changes and its activity is taken as 1, so it does not appear in Ka.
How is Kp related to Kc?
Kp = Kc(RT)^Δn, where Δn is the change in moles of gas, R = 0.08206 L·atm/(mol·K) and T is in kelvin. When Δn = 0, Kp equals Kc.