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Chemistry 2e · Fundamental Equilibrium Concepts

Equilibrium Calculations

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Check yourself
  7. Study tools
  8. Sources & references

In 30 seconds

Knowing K and the initial amounts, or knowing the equilibrium amounts, you can solve for the missing quantities. There are two classic problem types:

  1. Compute K from equilibrium concentrations — plug the measured equilibrium concentrations into the equilibrium expression.
  2. Compute equilibrium concentrations from K and initial amounts — define a change variable x, build an (Initial, Change, Equilibrium), substitute into the equilibrium expression, and solve.

The second type often produces a quadratic equation, but a useful approximation applies when K is very small: the change in reactant concentration is negligible compared with the initial amount, turning the quadratic into a square-root problem. This topic builds on the previous two.

Why this matters

Equilibrium calculations convert the abstract constant K into concrete numbers: the actual concentration of every species in a mixture. That matters whenever an outcome depends on a specific amount:

  • Pharmaceuticals: The fraction of a weak drug in its uncharged (absorbable) form depends on acid-base equilibria in the stomach and bloodstream — exactly these K-type calculations.
  • Water treatment: Solubility equilibria (Chapter 15) use the same ICE method to predict precipitation and dissolved contaminant levels.
  • Industry: Yield predictions for the Haber or contact process are equilibrium calculations; engineers set reactor pressure and feed composition from computed equilibrium concentrations.
  • Biology: Hemoglobin–oxygen binding curves are equilibrium calculations in disguise; the same mathematics (with Kd) describes receptor–drug binding.

The college version

Core Concepts

The ICE table method

An ICE table organizes three stages:

  • I = initial concentrations (what was mixed),
  • C = change as the reaction moves to equilibrium,
  • E = equilibrium concentrations (I + C).

For aA + bB ⇌ cC + dD starting with only reactants, if x mol/L of A react, the changes follow balanced stoichiometry: A decreases by ax, B by bx, products increase by cx and dx. Substitute the equilibrium row into the equilibrium expression. The variable x is always defined by the balanced equation, never guessed.

Type 1: computing K from equilibrium concentrations

When equilibrium concentrations are measured (by spectroscopy or titration, for example), substitute them into the equilibrium expression and evaluate. No x is needed. If different starting mixtures give the same K, the system truly reached equilibrium.

Type 2: computing equilibrium concentrations from K

When K and initial concentrations are known:

  1. Write the balanced equation and the equilibrium expression.
  2. Build the ICE table; express every equilibrium concentration in terms of x.
  3. Substitute into the equilibrium expression, giving an equation in x.
  4. Solve — using the if needed, keeping the root that gives non-negative concentrations.
  5. Back-substitute x to find every equilibrium concentration.

The small-K approximation

If K is very small (roughly < 10-3), x is tiny compared with the initial concentrations, so 0.10 - x pprox 0.10, avoiding the quadratic. Always check: if x is more than about 5% of the initial concentration it was subtracted from, the approximation is invalid and the quadratic must be solved.

How It Works / Step-by-Step Process

Worked example 1: computing K from equilibrium concentrations

Problem. A mixture of H2, I2, and HI at 445 °C reaches equilibrium with [H2] = 0.110 M, [I2] = 0.110 M, and [HI] = 0.780 M for H2(g) + I2(g) ⇌ 2HI(g). Find Kc.

Solution.

  1. Write the equilibrium expression:

Kc = [HI]2[H2][I2]

  1. Substitute the equilibrium values:

Kc = (0.780)2(0.110)(0.110) = 0.60840.0121 = 50.3

  1. Round: Kc ≈ 50 — large, consistent with the substantial product present; equilibrium favors HI.

Worked example 2: computing equilibrium concentrations from K

Problem. For the same reaction at 445 °C (Kc = 50.2), start with [H2] = [I2] = 0.500 M and no HI. Find the equilibrium concentrations.

Solution.

  1. ICE table with change x:
H2I2HI
Initial0.5000.5000
Change−x−x+2x
Equilibrium0.500 − x0.500 − x2x
  1. Substitute into the equilibrium expression:

50.2 = (2x)2(0.500 - x)2 = 4x2(0.500 - x)2

  1. Take the square root of both sides (both sides are positive):

50.2 = 2x0.500 - x   ⇒  7.085 = 2x0.500 - x

  1. Solve for x:

7.085(0.500 - x) = 2x   ⇒  3.543 - 7.085x = 2x   ⇒  x = 3.5439.085 = 0.390

  1. Back-substitute:

[HI] = 2x = 0.780 M,   [H2] = [I2] = 0.500 - 0.390 = 0.110 M

Check by plugging back: (0.780)2/(0.110)2 = 50.3 ≈ Kc. The result is consistent.

Worked example 3: the small-K approximation with the 5% rule

Problem. For N2O4(g) ⇌ 2NO2(g), Kc = 4.6 × 10-3 at 25 °C. If pure N2O4 is placed in a flask at 0.100 M, find the equilibrium concentration of NO2.

Solution.

  1. ICE table:
N2O4NO2
Initial0.1000
Change−x+2x
Equilibrium0.100 − x2x
  1. Equilibrium expression:

4.6 × 10-3 = (2x)20.100 - x = 4x20.100 - x

  1. Because K is small, try the approximation 0.100 - x ≈ 0.100:

4.6 × 10-3 ≈ 4x20.100   ⇒  x2 ≈ 4.6 × 10-3 × 0.1004 = 1.15 × 10-4

x ≈ 1.15 × 10-4 = 1.07 × 10-2

  1. Check the : x/0.100 = 0.107 — 10.7% > 5% — so the approximation is invalid. Solve the quadratic exactly:

4x2 + 4.6 × 10-3x - 4.6 × 10-4 = 0

Using the quadratic formula with a = 4, b = 4.6 × 10-3, c = -4.6 × 10-4:

x = -4.6 × 10-3 + (4.6 × 10-3)2 + 4(4)(4.6 × 10-4)2(4) = 1.02 × 10-2

  1. Equilibrium concentrations: [NO2] = 2.04 × 10-2 M, [N2O4] = 0.0898 M. The approximation would have overstated [NO2] by ~5% — exactly why the 5% check matters.

Common Confusions

Do Not ConfuseWithDifference
Change row signsInitial concentrationsChange always follows stoichiometry of the balanced equation; sign tells you whether a species is consumed (−) or produced (+).
Using initial concentrations in KUsing equilibrium concentrations in KOnly equilibrium concentrations belong in K; initials go in the I row and into Q.
x being the amount reactedx being a concentration of one specific speciesx is the amount reacted in mol/L; a species with coefficient 2 has concentration change 2x, not x.
Approximating whenever K < 1Approximating only when K is very smallThe approximation is a mathematical shortcut, not a rule of thumb about K < 1; always apply the 5% check.
Solving only for xReporting the equilibrium concentrationsThe final answer is the E row values, not x itself — a classic lost-points error.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

An ICE table is like a balance sheet for a chemical reaction. You write down what you started with (Initial), how much changed (Change), and what you end up with (Equilibrium). If you know how the reaction "likes" to sit (the number K), you can use the balance sheet to figure out exactly how much of each chemical is left when everything stops changing.

Key takeaways

  • ICE = Initial, Change, Equilibrium; the Change row follows balanced-equation stoichiometry.
  • Type 1 problems: plug measured equilibrium concentrations straight into K.
  • Type 2 problems: define x, build the ICE table, solve the resulting equation for x.
  • Small K (roughly < 10-3) allows the approximation "initial − x ≈  initial" — but verify with the 5% rule.
  • When a quadratic appears, discard roots that give negative concentrations.
  • If the reaction starts with only reactants, x is positive and appears with minus signs on the reactant side; if it starts with only products, run the reaction "in reverse" and x appears with minus signs on the product side.
  • Q vs K (previous topic) tells you which direction the system moves before you build the table — the Change signs must match the predicted direction.
  • Unit discipline: keep concentrations in mol/L; the equilibrium expression is unitless by convention (standard-state concentrations), so no units are carried into K.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

  1. What does ICE stand for, and what does the Change row encode?

    Show answer

    Initial, Change, Equilibrium. The Change row shows how each concentration changes as the reaction moves to equilibrium, in the stoichiometric ratio of the balanced equation.

  2. In Worked Example 2, why was the square root taken before solving for x?

    Show answer

    Both sides were perfect squares, so taking the square root converted the quadratic into a linear equation — faster than the quadratic formula.

  3. When is the small-K approximation valid, and how do you verify it?

    Show answer

    The approximation is valid when K is very small (roughly < 10-3); verify that x is less than about 5% of the initial concentration it was subtracted from.

  4. In Worked Example 3, why did the approximation fail, and what was the exact value of [NO2]?

    Show answer

    The approximation gave x = 1.07 × 10-2, which is 10.7% of 0.100 M — above the 5% threshold. The exact quadratic solution gives x = 1.02 × 10-2, so [NO2] = 2.04 × 10-2 M.

  5. If a problem starts with only product present and K is large, in which direction does the reaction proceed, and how does that affect the ICE Change row?

    Show answer

    With only products present, the reaction proceeds in reverse; the Change row has minus signs on the product side and plus signs on the reactant side. (This mirrors Le Châtelier logic: Q = ∞> K, so reverse direction is required.)

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

ICE table
Three-row bookkeeping: Initial, Change, Equilibrium
change variable (x)
The unknown amount (in mol/L) by which the reaction proceeds
quadratic formula
x = -b ± b2 - 4ac2a
small-K approximation
Assuming initial − x ≈  initial when K is tiny
5% rule
The check that x is less than ~5% of the concentration it was subtracted from

Sources & references

  1. openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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