Subjects and lessons to explore.

Law

Chemistry: Atoms First 2e · Fundamental Equilibrium Concepts

Equilibrium Calculations

7 min read
5 key termsSourcesReport an error
Equilibrium data: Kc = 50.2 at 445 °C for H2 + I2 ⇌ 2HI is a widely used textbook value; Kc = 4.1 × 10-4 for N2 + O2 ⇌ 2NO at 2000 °C is a commonly taught textbook value (small variations exist among reference tables and temperatures). All worked examples use self-consistent arithmetic that reproduces the stated K on verification.
Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Equilibrium calculations answer two questions: what is K? and what are the equilibrium concentrations? Both are solved with the — Initial, Change, Equilibrium bookkeeping. Fill in what you know from the balanced equation, let x be the unknown change, write the equilibrium concentrations in terms of x, and substitute into K:

K = (product terms in x)(reactant terms in x)

Find-K problems evaluate the expression directly. Find-concentration problems solve for x — directly, with the small-x approximation (when K is small, x is often negligible compared with initial concentrations), or with the . The approximation is valid only if it passes the (x < about 5% of the initial concentration it is subtracted from); otherwise use the exact quadratic.

Why this matters

Equilibrium calculations turn K values into predictions: the yield of an industrial synthesis, the concentration of a drug's active form at physiological pH, or the pH of a buffer (Chapter 14 extends the method to Ka and Kb problems). Laboratory data from one experiment give K, which predicts any other starting mixture at the same temperature. On exams, ICE tables are the chapter's computational backbone; acid-base and solubility problems in later chapters are variations on the same method.

The college version

Core Concepts

Setting up the ICE table

The Change row follows stoichiometry exactly. If x moles per liter of a reactant are consumed, a product with coefficient 2 gains 2x, a reactant with coefficient 3 loses 3x, and so on. Signs matter: reactants decrease (-x, -3x), products increase (+x, +2x). The Equilibrium row is Initial plus Change, and only those expressions go into K.

Type 1: finding K from equilibrium data

When the problem supplies equilibrium concentrations, substitute them directly into K — no x needed.

Type 2: finding equilibrium concentrations from K

When K and initial concentrations are given, the Equilibrium row contains x, and you solve the resulting equation. If the equation is a perfect square (equal starting concentrations of reactants), take the square root of both sides. Otherwise, rearrange into the quadratic form ax2 + bx + c = 0 and apply

x = -b ± b2 - 4ac2a

keeping only the root that gives physically sensible (non-negative) concentrations.

The small-x approximation and the 5% rule

When K is small (roughly 10-3 or smaller), the reaction barely proceeds and x is tiny compared with initial concentrations, so [initial] - x ≈ [initial] turns a quadratic into a simple power. The shortcut must be verified: if x < 5% of the initial concentration it was subtracted from, it is acceptable; otherwise solve exactly.

How It Works / Step-by-Step Process

  1. Write the balanced equation with physical states; write the K expression (omit pure solids and liquids).
  2. Build the ICE table: Initial concentrations from the problem, Change row with x and coefficients, Equilibrium row = Initial + Change.
  3. Substitute the Equilibrium row into K.
  4. Choose the route: direct evaluation, square root, small-x approximation (then apply the 5% rule), or quadratic formula; discard non-physical roots.
  5. Verify by substituting final concentrations back into K.

Common Confusions

Common ConfusionCorrect Understanding
"Forgetting the coefficients in the Change row."A coefficient of 2 means the change is 2x, and the concentration is squared in K.
"The 5% rule is optional."It is the validity test for the approximation; if x ≥ 5% of the initial concentration, solve exactly.
"Either quadratic root works."Only roots giving non-negative concentrations are physical; check both.
"Solids and liquids go in the ICE expression."They are omitted from K.
"K and Q are the same in the ICE table."The Equilibrium row goes into K; current non-equilibrium values go into Q.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

An ICE table is like balancing a budget. You start with what you have (Initial), then the reaction "spends" some amount x of reactants and "earns" some amount of products, in the exact proportions the recipe says (Change). At the end (Equilibrium), you plug what's left into the rule that says how the balance must look — and solve for x.

Worked examples

At 445 °C, a flask with 0.100 M H2 and 0.100 M I2 reaches equilibrium with [HI] = 0.156 M. Find Kc for H2(g) + I2(g) ⇌ 2HI(g).

ICE table:

[H2][I2][HI]
Initial0.1000.1000
Change-x-x+2x
Equilibrium0.100 - x0.100 - x2x

Since [HI]eq = 0.156 M, 2x = 0.156, so x = 0.078 M. Then [H2] = [I2] = 0.100 - 0.078 = 0.022 M. Substitute into the K expression:

Kc = [HI]2[H2][I2] = (0.156)2(0.022)(0.022) = 0.02430.000484 = 50.3 ≈ 50.2

The computed constant matches the accepted value at 445 °C — a built-in consistency check.

At 2000 °C, Kc = 4.1 × 10-4 for N2(g) + O2(g) ⇌ 2NO(g). A vessel starts with 0.25 M N2 and 0.25 M O2. Find the equilibrium [NO].

ICE table:

[N2][O2][NO]
Initial0.250.250
Change-x-x+2x
Equilibrium0.25 - x0.25 - x2x

Substitute into Kc:

Kc = [NO]2[N2][O2] = (2x)2(0.25 - x)(0.25 - x) = 4.1 × 10-4

Since Kc is tiny, approximate 0.25 - x ≈ 0.25:

4x2(0.25)2 = 4.1 × 10-4

4x2 = (4.1 × 10-4)(0.0625) = 2.56 × 10-5,   x2 = 6.4 × 10-6,   x = 2.5 × 10-3 M

So [NO] = 2x = 5.0 × 10-3 M and [N2] = [O2] = 0.25 - 0.0025 = 0.2475 M.

5% rule check: x/[N2]0 = 1.0% < 5%, so the approximation is valid. This is why NO forms only in trace amounts even in a hot engine — a small K means reactants dominate.

At 445 °C, Kc = 50.2 for H2(g) + I2(g) ⇌ 2HI(g). A flask starts with 0.100 M H2 and 0.050 M I2, no HI. Find the equilibrium concentrations.

ICE table:

[H2][I2][HI]
Initial0.1000.0500
Change-x-x+2x
Equilibrium0.100 - x0.050 - x2x

Substitute into Kc:

50.2 = (2x)2(0.100 - x)(0.050 - x)

The initial concentrations are unequal, so no square-root shortcut. Expand:

4x2 = 50.2(0.00500 - 0.150x + x2)

0 = 46.2x2 - 7.53x + 0.251

Apply the quadratic formula with a = 46.2, b = -7.53, c = 0.251:

x = 7.53 ± (7.53)2 - 4(46.2)(0.251)2(46.2) = 7.53 ± 56.7 - 46.492.4 = 7.53 ± 3.2192.4

The roots are x = 0.116 M and x = 0.0467 M. The first exceeds the initial [I2] = 0.050 M and would give a negative concentration, so keep x = 0.0467 M:

[HI] = 2x = 0.0935 M,   [H2] = 0.100 - 0.0467 = 0.0533 M,   [I2] = 0.050 - 0.0467 = 0.0033 M

Verify: (0.0935)2/((0.0533)(0.0033)) = 49.7 ≈ 50.2 (rounding). The iodine is nearly exhausted — it was the limiting reactant, exactly as a large K predicts.

Key takeaways

  • ICE = Initial, Change, Equilibrium; the Change row follows the balanced stoichiometry with signs.
  • Find-K problems: substitute equilibrium concentrations directly — no x solving needed.
  • Equal starting reactant concentrations → take the square root of both sides.
  • Small-K shortcut: approximate [A]0 - x ≈ [A]0, then verify with the 5% rule.
  • Quadratic formula: keep only the root giving positive concentrations.
  • Always verify: substituted concentrations must reproduce the given K.

Check yourself

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

  1. Write the ICE table (symbolically) for N2 + 3H2 ⇌ 2NH3 starting from 0.50 M N2 and 1.50 M H2.

    Show answer

    Initial: 0.50, 1.50, 0; Change: −x, −3x, +2x; Equilibrium: 0.50 − x, 1.50 − 3x, 2x.

  2. When is the small-x approximation allowed, and how do you verify it?

    Show answer

    When K is small enough that x ≪  the initial concentrations; verify that x is less than 5% of the initial concentration it is subtracted from.

  3. In Example 3, why was the root x = 0.116 M rejected?

    Show answer

    x = 0.116 M exceeds the initial [I₂] = 0.050 M, which would make the equilibrium [I₂] negative — physically impossible.

  4. After solving any equilibrium problem, what final check confirms the answer?

    Show answer

    Substitute the equilibrium concentrations back into the K expression; it should reproduce the given K (within rounding).

Keep learning

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

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

ICE table
Initial–Change–Equilibrium accounting of concentrations
change (x)
The unknown amount consumed/produced, scaled by coefficients
small-x approximation
Treating [A]0 - x as approximately [A]0 when K is tiny
5% rule
Approximation valid only if x < 5% of the initial concentration
quadratic formula
x = (-b ± b2 - 4ac)/(2a)

Sources & references

  1. openstax.org — Chemistry Atoms First 2e

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

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.