Chemistry · General chemistry II · Worked example
Solve an equilibrium with an ICE table
At a certain temperature, Kc = 0.25 for N₂O₄(g) ⇌ 2NO₂(g). A flask starts with 1.00 M N₂O₄ and no NO₂. Find the equilibrium concentrations.
Set up the ICE table
N₂O₄ loses x, and by the coefficients NO₂ gains 2x.
| N₂O₄ (M) | NO₂ (M) | |
|---|---|---|
| Initial | 1.00 | 0 |
| Change | −x | +2x |
| Equilibrium | 1.00 − x | 2x |
Substitute into K
Kc = [NO₂]²/[N₂O₄], with the equilibrium row in place of each concentration.
Rearrange into a quadratic
Multiply both sides by 1.00 − x and collect every term on one side.
Solve and choose the root
The quadratic formula gives x = 0.2207 or x = −0.2832. Only the positive root keeps both concentrations positive. Carry the extra digit until the end.
Find the concentrations and check
[NO₂] = 2(0.2207) = 0.441 M and [N₂O₄] = 1.00 − 0.2207 = 0.779 M. Substituting them back gives 0.250, which matches K.
Result
[NO₂] = 0.441 M and [N₂O₄] = 0.779 M.
Your turn
At a different temperature, Kc = 49.0 for H₂(g) + I₂(g) ⇌ 2HI(g). Starting from 0.100 M each of H₂ and I₂, find the equilibrium concentrations.
Show the answer and explanation
[HI] = 0.156 M; [H₂] = [I₂] = 0.0222 M.
Both sides of (2x)²/(0.100 − x)² = 49.0 are perfect squares, so take square roots: 2x/(0.100 − x) = 7.00. Then 9.00x = 0.700 and x = 0.0778, so [HI] = 2x = 0.156 M and [H₂] = [I₂] = 0.100 − x = 0.0222 M.
Keep exploring
In Equilibrium & ICE tables, start from 2.00 M N₂O₄ instead. [NO₂] rises only to 0.647 M, not to twice 0.441 M, because K fixes [NO₂]²/[N₂O₄], not a simple ratio.
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