Solving Equilibrium Concentrations with ICE Tables
An ICE table (Initial, Change, Equilibrium) is a systematic way to find the equilibrium concentrations of a reaction from the initial concentrations and the equilibrium constant, by letting x be the amount that reacts and substituting into the K expression. In brief: for H2 + I2 ⇌ 2HI with Kc = 54.3 and 0.500 M of both H2 and I2 at the start, the equilibrium concentrations are [H2] = [I2] = 0.107 M and [HI] = 0.787 M.
Knowing K, we want to find out how much of each substance there is at equilibrium. The ICE table organizes the algebra and makes it easy to follow the stoichiometry. This article describes the method step by step, solves a problem with a perfect square, a problem with a quadratic equation and a problem with the small-x approximation, and explains how to check the answer.
What are the steps of the ICE method?
- Write the balanced equation and the K expression.
- Make a table with a column for each species and three rows: Initial, Change and Equilibrium.
- Fill in the initial concentrations. If necessary, use Q to find which way the reaction will go.
- Fill in the change row using x and the coefficients: reactants −ax (or +ax), products +cx (or −cx).
- Add the change to the initial value to get the equilibrium row.
- Substitute the equilibrium row into the K expression and solve for x.
- Calculate the equilibrium concentrations and check that they are positive and satisfy K.
Example 1: a perfect-square equation
For H2(g) + I2(g) ⇌ 2HI(g), Kc = 54.3 at 430 °C. The initial concentrations are [H2] = [I2] = 0.500 M and [HI] = 0. Find the equilibrium concentrations.
| [H2] | [I2] | [HI] | |
|---|---|---|---|
| Initial (M) | 0.500 | 0.500 | 0 |
| Change (M) | −x | −x | +2x |
| Equilibrium (M) | 0.500 − x | 0.500 − x | 2x |
- Kc = (2x)2 / (0.500 − x)2 = 54.3.
- Take the square root of both sides: 2x / (0.500 − x) = 7.369.
- 2x = 7.369 (0.500 − x) = 3.685 − 7.369x, so 9.369x = 3.685 and x = 0.393.
- [H2] = [I2] = 0.500 − 0.393 = 0.107 M and [HI] = 2 × 0.393 = 0.787 M.
- Check: (0.787)2 / (0.107)2 = 0.619 / 0.01145 = 54.1, which agrees with 54.3 within rounding.
About 79 % of the hydrogen has been converted to HI.
Example 2: a quadratic equation
For N2O4(g) ⇌ 2NO2(g), Kc = 0.0060 at 298 K. Start with [N2O4] = 0.100 M and no NO2.
| [N2O4] | [NO2] | |
|---|---|---|
| Initial (M) | 0.100 | 0 |
| Change (M) | −x | +2x |
| Equilibrium (M) | 0.100 − x | 2x |
- Kc = (2x)2 / (0.100 − x) = 4x2 / (0.100 − x) = 0.0060.
- Cross-multiply: 4x2 = 0.0060(0.100 − x) = 0.00060 − 0.0060x.
- Rearrange to standard form: 4x2 + 0.0060x − 0.00060 = 0.
- Quadratic formula: x = [−b + √(b2 − 4ac)] / 2a, with a = 4, b = 0.0060 and c = −0.00060.
- b2 − 4ac = 0.000036 + 0.0096 = 0.009636, whose square root is 0.09816.
- x = (−0.0060 + 0.09816) / 8 = 0.01152 (the negative root is rejected, since it would give a negative concentration).
[NO2] = 2x = 0.0230 M and [N2O4] = 0.100 − 0.0115 = 0.0885 M. Check: (0.0230)2 / 0.0885 = 0.000529 / 0.0885 = 0.0060. About 11.5 % of the N2O4 has dissociated.
Example 3: the small-x approximation
For N2(g) + O2(g) ⇌ 2NO(g), Kc = 4.0 × 10−4 at a high temperature. The initial concentrations are [N2] = 0.80 M and [O2] = 0.20 M.
- Kc = (2x)2 / ((0.80 − x)(0.20 − x)) = 4.0 × 10−4.
- Since K is small, x will be small compared with 0.20. Assume 0.80 − x ≈ 0.80 and 0.20 − x ≈ 0.20.
- 4x2 / (0.80 × 0.20) = 4x2 / 0.16 = 4.0 × 10−4, so x2 = 1.6 × 10−5 and x = 4.0 × 10−3.
- Check the assumption: 0.0040 / 0.20 = 2 %, below 5 %, so it is valid.
- [NO] = 2x = 0.0080 M, [N2] = 0.796 M and [O2] = 0.196 M.
Solving the exact equation gives x = 0.00395, which differs from the approximate value by about 1 %.
When can you use the approximation?
| Situation | Method |
|---|---|
| Both sides are perfect squares | Take the square root |
| K is very small (about 10−3 or less) relative to initial concentrations | Neglect x in the subtracted terms; check 5 % rule |
| K is moderate and x is not small | Quadratic formula |
| K is very large | Assume the reaction goes to completion, then let the reverse shift be x |
The 5 % rule says that the approximation is acceptable if x is less than 5 % of the initial concentration it was dropped from.
How do you check an answer?
- All concentrations must be positive.
- Substituting them in the K expression must give K.
- The direction of the shift must agree with the comparison of Q and K.
- The stoichiometry must be consistent: the change in each species is proportional to its coefficient.
Where does the method come from?
The ICE table is a teaching device used in general chemistry courses, and not a historical concept in itself; it is a way of organizing the stoichiometry and algebra implied by the law of mass action. Its foundations, the equilibrium constant expression, were set by Guldberg and Waage in 1864. Today, equilibrium calculations for complex mixtures are carried out by computer programs that minimize the Gibbs energy of the system.
Which mistakes are common?
- Forgetting the coefficients in the change row: the change is 2x for a coefficient of 2.
- Putting the wrong sign: reactants decrease and products increase if the reaction goes forward.
- Squaring only part of the term: (2x)2 = 4x2.
- Keeping a negative root: negative concentrations are impossible.
- Using the approximation without checking: verify the 5 % rule.