Worked example 1
For H₂(g) + I₂(g) ⇌ 2 HI(g), K = 50.0. Starting with 1.00 M H₂ and 1.00 M I₂, find the equilibrium concentrations.
Try it first: Write the change row using the coefficients 1, 1 and 2.
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What you'll be able to do: Set up an ICE table for any equilibrium and solve for the unknown concentrations, using the small-x approximation when it is valid.
Once you can predict direction, the natural next question is how far. An ICE table is the bookkeeping that turns a starting mixture and a value of K into actual equilibrium concentrations.
These are recommended, not required. You can start this lesson at any time.
Initial holds the concentrations before any reaction. Change holds the amounts consumed and formed, written in terms of x. Equilibrium is simply the sum of the two rows above. Everything you need for the K expression lives in the bottom row.
For N₂O₄ ⇌ 2 NO₂ starting with 0.500 M N₂O₄ and no NO₂:
| Row | [N₂O₄] / M | [NO₂] / M |
|---|---|---|
| Initial | 0.500 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.500 − x | 2x |
The variable x is the concentration decrease of N₂O₄, not automatically the concentration of each product. Substitute only the last row into Kc = [NO₂]²/[N₂O₄], giving Kc = 4x²/(0.500 − x). All entries in a concentration table must use the same final vessel volume; do a rapid stoichiometric reaction or dilution first if the problem requires it.
The changes follow the coefficients exactly. For N₂ + 3 H₂ ⇌ 2 NH₃ the changes are -x, -3x and +2x. Reactants that are consumed get a minus sign and products formed get a plus sign, and the multipliers are never optional.
If the mixture starts with products present, calculate Q first. When Q > K the reaction runs in reverse and the signs in the change row flip. Assigning the direction before writing the change row prevents a negative concentration later.
When an initial concentration is much larger than the expected change, replacing C − x with C may simplify the equilibrium equation. Whether this is valid depends on the reaction stoichiometry, K, and starting mixture; no single initial-concentration-to-K ratio works for every ICE table. Use the approximation to estimate x, then check the percent change in each concentration whose C ± change term you simplified.
Never replace a product term 2x by zero merely because the product was initially absent. A small-x approximation applies to a sum or difference such as 0.500 − x, not to the x needed to create products. If a simplified term is C − 3x, test 3x/C, not x/C. A small K by itself is not a universal validation criterion because stoichiometric powers and starting products affect the solution.
After solving with an approximation, compare the magnitude of the concentration change with the nonzero initial concentration it was subtracted from or added to. If the change is 5% or less for each approximated concentration, the shortcut is acceptable; otherwise solve the full equation and keep only roots that give physically possible concentrations.
For A ⇌ 2 B with C = 0.100 M and Kc = 0.100, the equation is 0.100 = 4x²/(0.100 − x). Ignoring x in the denominator predicts x = 0.0500 M, a 50% change, so the shortcut fails. Keep all terms: 4x² + 0.100x − 0.0100 = 0. The roots are about +0.0390 and −0.0640 M. The negative root would give [B] < 0 and is rejected. The valid concentrations are [A] ≈ 0.0610 M and [B] ≈ 0.0781 M; substitution returns Kc ≈ 0.100. Testing physical concentrations and back-substituting are stronger checks than accepting whichever root the calculator shows first.
percent change = 100 × |concentration change| / initial concentration
An ICE table does not always start with products increasing. Calculate Q from initial concentrations when both sides are present: Q<K means forward change, Q>K means reverse change. Then use one variable x with stoichiometric coefficients and reject roots that imply negative concentrations.
When initial products are present, Q chooses a convenient sign convention but does not change equilibrium mathematics. A signed x is also possible if every entry remains consistent; the final test is that all concentrations are nonnegative and the material balance and K relation hold. Define x in words before building the row, especially when comparing a forward-oriented table with a reverse-oriented one.
A three-row layout of Initial, Change and Equilibrium concentrations used to solve equilibrium problems.
Every change is a coefficient multiple of x, negative for consumption and positive for formation.
Check the full omitted change, such as 3x, against each nonzero initial concentration that was approximated; ≤5% is the usual classroom criterion.
A quadratic root that gives a negative concentration is discarded.
Compare Q with K before assigning signs in the change row.
x = (-b +/- sqrt(b² - 4ac)) / (2a)
100 × |omitted concentration change| / initial concentration ≤ 5%
For H₂(g) + I₂(g) ⇌ 2 HI(g), K = 50.0. Starting with 1.00 M H₂ and 1.00 M I₂, find the equilibrium concentrations.
Try it first: Write the change row using the coefficients 1, 1 and 2.
0 of 5 steps revealed.
For N₂O₄(g) ⇌ 2 NO₂(g), Kc = 4.6 × 10⁻³. Starting with 0.50 M N₂O₄, find [NO₂] at equilibrium.
Try it first: Set up the equilibrium expression, find a candidate x, and then check whether the concentration approximation is within 5%.
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For A(g) ⇌ B(g), K = 2.0, starting with 1.0 M A and 0.0 M B. Solve exactly.
Try it first: Would the small-x approximation be valid here?
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For A ⇌ B, K=4.0. Initially [A]=2.0 M and [B]=4.0 M. Find equilibrium concentrations.
Try it first: Qinitial = 4.0/2.0 = 2.0, below K.
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Leaving the coefficient off the change row, for example writing +x instead of +2x. | The change must be in the stoichiometric ratio. | Copy each coefficient into the change row before substituting. |
| Using the change row instead of the equilibrium row in the K expression. | K is defined by equilibrium concentrations only. | Always substitute the bottom row. |
| Applying the small-x approximation when K is large. | A large K means x is comparable to the initial concentration. | Run the 5% check, or solve the quadratic from the start. |
| Keeping a quadratic root that gives a negative concentration. | Concentrations cannot be negative. | Discard that root and keep the physically possible one. |
| Reporting x when the question asks for a product concentration of 2x. | x is the extent of reaction, not necessarily the answer. | Read the equilibrium row for the species asked about. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
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An ICE table tracks Initial, Change and Equilibrium concentrations for every species. Changes follow the balanced equation’s stoichiometric ratios. Substituting the equilibrium row into K gives an equation in x. A small-x approximation may simplify that equation, but its validity depends on the reaction and starting mixture; estimate x and check the percent change in each concentration being approximated. If any change exceeds 5%, solve the full equation and keep only roots that give physically possible concentrations.
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