Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.
The concentrations or partial pressures of species at equilibrium can be predicted from the balanced reaction, the initial concentrations, and the appropriate K. The tool is the ICE table.
Initial — Change — Equilibrium. Fill the Initial row from the problem, express the Change row in terms of x scaled by the coefficients, add the two rows to get Equilibrium, then substitute into the K expression and solve for x.
The "x is small" approximation. When K is very small (roughly K < 10⁻⁴) the reaction barely proceeds, so x is negligible compared to the initial concentration and terms like (0.100 − x) can be approximated as 0.100. This turns an unwieldy polynomial into a simple root extraction.
The 5% rule: the approximation is valid if x is less than 5% of the initial concentration you approximated. Always check it. If it fails, you must use the quadratic formula.
When the approximation fails — typically for K near or above 10⁻³, or for very dilute solutions — set up the quadratic and use x = (−b ± √(b² − 4ac))/(2a). Discard the root that gives a negative concentration.
A useful shortcut for large K. If K is very large, it is easier to assume the reaction goes to completion first, then let it come back a small amount. Build the ICE table from the "all products" side.
Sanity checks before you write down an answer: no concentration may be negative, and substituting your equilibrium values back into the K expression must reproduce K.
For H₂(g) + I₂(g) ⇌ 2 HI(g), Kc = 50.5 at 448 °C. A 1.00 L flask is charged with 0.500 mol H₂ and 0.500 mol I₂. Find all equilibrium concentrations.
ICE table (M):
| H₂ | I₂ | 2 HI | |
|---|---|---|---|
| I | 0.500 | 0.500 | 0 |
| C | −x | −x | +2x |
| E | 0.500−x | 0.500−x | 2x |
Substitute: K = 50.5 is far too large for the "x is small" approximation, so solve exactly.
50.5 = (2x)² / [(0.500 − x)(0.500 − x)] = (2x)² / (0.500 − x)²
Both sides are perfect squares — take the square root:
√50.5 = 2x / (0.500 − x)
7.106 = 2x / (0.500 − x)
7.106(0.500 − x) = 2x
3.553 − 7.106x = 2x
3.553 = 9.106x → x = 0.390
Equilibrium concentrations:
[H₂] = [I₂] = 0.500 − 0.390 = 0.110 M
[HI] = 2(0.390) = 0.780 M
Check: (0.780)² / (0.110)² = 0.6084 / 0.0121 = 50.3 ≈ 50.5 ✓