TL;DR:
Write the balanced equation and the K expression first, with products on top and each concentration raised to its coefficient.
Build an ICE table (initial, change, equilibrium) with the change in terms of x using the coefficients.
Solve, substitute back to check that K comes out right, and report with sensible significant figures.
An equilibrium free-response question often gives you a reaction, starting amounts and a value of K, and asks for the concentrations at equilibrium. Students who know the chemistry still lose time and marks to the setup: a wrong exponent, a change written with the wrong sign, a K expression that includes a pure liquid.
A fixed routine fixes this. The routine is short, and you can practise it on any equilibrium problem.
Why the setup is the skill
The algebra in these problems is usually simple once the table is right. What differs between a clean answer and a confused page is whether you wrote the K expression from the balanced equation, tracked the change with the coefficients, and kept the units straight. A reader marking your answer can follow a clear table even if your arithmetic slips.
Check the current course and exam description on the College Board site for exactly what is expected, because requirements change between years.
The routine
Write the balanced equation: Coefficients matter, because they set both the exponents in K and the size of each change.
Write the K expression: Products over reactants, each concentration raised to its coefficient. Leave out pure solids and pure liquids.
Draw the ICE table: Rows for Initial, Change and Equilibrium, one column per species.
Express the changes with x: Use the coefficients: if a reactant has coefficient 2, its change is −2x. Products get +.
Substitute into K and solve: Look for a perfect square before you reach for the quadratic formula.
Check: Put your values back into the K expression, and check that no concentration is negative.
Worked example: an ICE table for a gas-phase equilibrium
The reaction is H₂(g) + I₂(g) ⇌ 2HI(g), with K = 50 in terms of concentrations at the temperature given. A 1.00 L vessel starts with 1.00 mol H₂ and 1.00 mol I₂ and no HI. Find the equilibrium concentrations.
K expression. K = [HI]² / ([H₂][I₂]). The coefficient 2 on HI becomes the exponent.
Row | [H₂] | [I₂] | [HI] |
|---|---|---|---|
Initial (M) | 1.00 | 1.00 | 0 |
Change (M) | −x | −x | +2x |
Equilibrium (M) | 1.00 − x | 1.00 − x | 2x |
Substitute. 50 = (2x)² / (1.00 − x)². The right side is a perfect square, so take the square root of both sides: √50 = 7.07 = 2x / (1.00 − x).
Multiply out:
7.07(1.00 − x) = 2x, so7.07 − 7.07x = 2x.Collect:
7.07 = 9.07x, sox = 0.78.Equilibrium:
[H₂] = [I₂] = 1.00 − 0.78 = 0.22 M, and[HI] = 2 × 0.78 = 1.56 M, which is 1.6 M to two significant figures.
Check. Put the values back: K = 1.56² / 0.22² = 2.43 / 0.0484 ≈ 50. It matches the given K. Also check conservation: each mole of HI uses half a mole each of H₂ and I₂, so the hydrogen atoms are 2 × 0.22 + 1.56 = 2.00 per litre, equal to the 2 × 1.00 we started with.
Reading the reaction quotient first
Sometimes a question gives you all the concentrations right now and asks which way the reaction will shift. Compute the reaction quotient Q with the same expression as K, using the current values, and compare. If Q is smaller than K, the reaction proceeds toward products. If Q is larger, it shifts toward reactants. If they are equal, it is already at equilibrium.
In the example, the starting Q is 0 / (1.00 × 1.00) = 0, which is below 50, so the system must form HI. That matches the table, where the change for HI is positive. Checking the direction before you fill in the table catches sign errors early.
Units and significant figures
Concentrations are in mol/L (M). K for this reaction has the same number of concentration terms on top and bottom, so it has no units. For other reactions the units of K change, so follow the course's convention. Since K was given as 50 (two significant figures), report concentrations to two: 0.22 M and 1.6 M.
Carry an extra digit in your working and round only the final answer. Rounding 7.07 too early to 7.1 would change x in the second decimal.
When x is small
If K is very small and the starting concentrations are large, x is tiny compared with them and you may approximate 1.00 − x ≈ 1.00. After solving, check that x is small compared with the starting value. If not, solve the full equation. Say on the page that you made the approximation, and why.
Writing it up so a reader can follow
Show each stage in the same order every time: the balanced equation, the K expression, the ICE table, the substituted equation, the solved value of x, the final concentrations, and a check. Label the final answers with units and the species. A reader looking for specific steps finds them quickly, and you find your own slips more easily too.
Mistakes to avoid
Writing the K expression upside down.
Leaving out the coefficient in the change row, such as +x instead of +2x.
Including a pure liquid or solid in K.
Forgetting that the K expression uses equilibrium values, not initial ones.
Giving a negative concentration without noticing it is impossible.
Your plan for this week
Write the routine on a card.
Do three equilibrium problems, one with a perfect square, one needing the quadratic formula and one with a small x.
For each, substitute back to check K.
Compare your tables with the worked solutions.
Frequently asked questions
When do I use Kp instead of Kc?
Use Kp when the problem gives partial pressures and Kc when it gives concentrations. The ICE table works the same way with pressures in place of concentrations.
Do I always need the quadratic formula?
No. Try for a perfect square or a small-x approximation first, and use the quadratic formula when neither applies.
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