E. Chemical Systems and Equilibrium · Chemical Equilibrium

ICE Tables III: The Small-K Approximation

E2.4 A1.12 A1.13About 65 minutes

Learning goals

I can recognize when K is small enough that x can be ignored in (c − x).
I can use the approximation to solve equilibrium problems, including ones that would be cubic.
I can apply the 5 % rule to decide whether the approximation is valid.
I can switch to an exact method when the approximation fails.

Words to know

small-K approximation
When K is very small compared with the initial concentration, the change x is so small that (c − x) ≈ c.

5 % rule
The approximation is accepted only if x is less than 5 % of the initial concentration it was subtracted from.

percent change (percent ionization or dissociation)
x ÷ initial concentration × 100 %, for the substance whose concentration was approximated.

significant figures
The digits in a measured or calculated value that carry meaning; equilibrium answers are usually given to 2 or 3 significant figures, matching the data.

When x is almost nothing

Nitrogen and oxygen make up about 99 % of dry air, yet they barely react with each other at room temperature. Even in a hot car engine, only a tiny fraction turns into nitrogen monoxide (NO), the pollutant behind smog. When K is this small, x is so tiny that subtracting it from a starting concentration hardly changes anything. Chemists use that fact to turn hard algebra into one line of work. This trick is also the key to weak-acid problems in the unit Acid-Base and Solubility Equilibria.

The idea

If K is very small compared with the initial concentrations, very little reacts, so for each reactant:

c − x ≈ c

Only drop x where it is added to or subtracted from a number. Never drop an x that stands alone (such as the 2x for a product), because that is the answer you are looking for.

After solving, always check with the 5 % rule: calculate x ÷ c × 100 %. If it is less than 5 %, the approximation is valid. If not, go back and solve exactly (quadratic formula). A quick screen: when the initial concentration is several hundred times K or more, the approximation usually passes, but the 5 % check is what decides.

Using the approximation safely: assume, solve, then check with the 5 % rule.

Worked example

Example 1: Nitrogen monoxide in a hot engine

For N₂(g) + O₂(g) ⇌ 2 NO(g), suppose K = 1.0 × 10⁻⁵ at a certain high temperature. A cylinder starts with [N₂] = 0.0400 mol/L and [O₂] = 0.0100 mol/L. Find [NO] at equilibrium.

ICE: E row: 0.0400 − x, 0.0100 − x, 2x. K is far smaller than both initial concentrations.
Exact: (2x)² ÷ [(0.0400 − x)(0.0100 − x)] = 1.0 × 10⁻⁵.
Approximate: 0.0400 − x ≈ 0.0400 and 0.0100 − x ≈ 0.0100. So 4x² ÷ (0.0400 × 0.0100) = 1.0 × 10⁻⁵.
4x² = (1.0 × 10⁻⁵)(4.00 × 10⁻⁴) = 4.0 × 10⁻⁹, so x² = 1.0 × 10⁻⁹ and x = 3.16 × 10⁻⁵.
Check: x ÷ 0.0100 × 100 % = 0.32 % < 5 % ✓ (the smaller starting value is the stricter test).
[NO] = 2x = 6.3 × 10⁻⁵ mol/L.

[NO] ≈ 6.3 × 10⁻⁵ mol/L. Solving exactly gives x = 3.156 × 10⁻⁵, the same to three significant figures.

Worked example

Example 2: Turning a cubic into a cube root

For 2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g), suppose K = 1.6 × 10⁻⁵ at a certain temperature. A flask starts with [NOCl] = 0.500 mol/L. Find the equilibrium concentrations.

E row: 0.500 − 2x, 2x, x. K = (2x)²(x) ÷ (0.500 − 2x)². Exactly, this is a cubic equation.
Approximate 0.500 − 2x ≈ 0.500: 4x³ ÷ 0.250 = 1.6 × 10⁻⁵.
4x³ = 4.0 × 10⁻⁶, so x³ = 1.0 × 10⁻⁶ and x = 0.0100 (cube root).
Check: the NOCl changed by 2x = 0.0200; 0.0200 ÷ 0.500 × 100 % = 4.0 % < 5 % ✓.
[NO] = 0.0200 mol/L; [Cl₂] = 0.0100 mol/L; [NOCl] = 0.500 − 0.0200 = 0.480 mol/L.

[NO] = 0.020 mol/L, [Cl₂] = 0.010 mol/L, [NOCl] ≈ 0.48 mol/L. The approximation turned an equation we could not solve by hand into a cube root.

Worked example

Example 3: When the approximation fails

For N₂O₄(g) ⇌ 2 NO₂(g), K = 4.6 × 10⁻³ at 25 °C. Start: [N₂O₄] = 0.100 mol/L. Try the approximation, check it, and fix it if needed.

E row: 0.100 − x, 2x. Approximate 0.100 − x ≈ 0.100: 4x² ÷ 0.100 = 4.6 × 10⁻³.
x² = 1.15 × 10⁻⁴, so x = 0.0107.
Check: 0.0107 ÷ 0.100 × 100 % = 10.7 % > 5 % ✗. The approximation is not valid.
Solve exactly: 4x² + 0.0046x − 0.00046 = 0 gives x = 0.0102 (from lesson 5).
[NO₂] = 2x = 0.0203 mol/L.

[NO₂] = 0.0203 mol/L. The approximate answer (0.0214) was about 5 % too high. K = 4.6 × 10⁻³ is small, but not small enough compared with 0.100 mol/L.

What the 5 % rule measures. In Example 1, x is a sliver of the starting concentration (drawn a little wider than scale so you can see it). In Example 3, x is a tenth of it, too large to ignore.

Just for fun

The small-K approximation is the chemist's version of saying "close enough", except chemists then check, write down a percentage and say "close enough" again with confidence.

Hands-on activity

How small is small enough?

You need: a calculator or a free spreadsheet app.

For the reaction A ⇌ B + C starting with [A] = c = 1.00 mol/L, the equation is x² ÷ (c − x) = K.

For K = 0.1, 0.01, 0.0025 and 0.001 (so c ÷ K = 10, 100, 400 and 1000), find x approximately (x = √(Kc)) and exactly (quadratic formula).
For each, calculate the percent change x ÷ c × 100 % and the error in the approximate x.
At what ratio c ÷ K does the approximation start to pass the 5 % rule?

What to look for:

c ÷ K = 10: approx 0.316, exact 0.270 (approx 17 % too high; fails)
c ÷ K = 100: approx 0.100, exact 0.0951 (10 %; fails)
c ÷ K = 400: approx 0.0500, exact 0.0488 (5 %; borderline)
c ÷ K = 1000: approx 0.0316, exact 0.0311 (3 %; passes)

For this kind of reaction, the approximation passes once c is about 400 or more times K.

Your turn

In which expression may x be dropped if K is tiny: (0.250 − x) or (2x)²?

For A ⇌ B + C, K = 1.0 × 10⁻⁶, [A]ᵢ = 0.10 mol/L. Find [B] at equilibrium and check the 5 % rule.

For 2 H₂O(g) ⇌ 2 H₂(g) + O₂(g), suppose K = 2.0 × 10⁻¹³ at a high temperature, with [H₂O]ᵢ = 0.100 mol/L. Find [O₂].

For N₂O₄ ⇌ 2 NO₂ (K = 4.6 × 10⁻³), would the approximation work for [N₂O₄]ᵢ = 1.00 mol/L?

For N₂ + O₂ ⇌ 2 NO, K = 1.0 × 10⁻⁵. Start: [N₂] = [O₂] = 0.0200 mol/L. Find [NO] both with the approximation and by the perfect-square method. Compare.

Why is the 5 % check done on the reactant with the smaller initial concentration (when the coefficients are equal)?

More practice: warm-up

What does the 5 % rule compare?

x = 0.0020 and c = 0.50 mol/L. Is the approximation valid?

x = 0.012 and c = 0.10 mol/L. Is the approximation valid?

Solve x² ÷ 0.40 = 2.5 × 10⁻⁵.

If the approximation fails, what should you do?

More practice: on your own

For COCl₂(g) ⇌ CO(g) + Cl₂(g), suppose K = 2.2 × 10⁻¹⁰ at 100 °C. A flask starts with [COCl₂] = 0.050 mol/L. Find [CO] at equilibrium.

For A ⇌ B + C with K = 4.0 × 10⁻⁴ and [A]ᵢ = 0.010 mol/L, try the approximation and check it. If it fails, solve exactly.

For 2 NOCl ⇌ 2 NO + Cl₂ with K = 1.6 × 10⁻⁵, start with [NOCl] = 1.00 mol/L. Find [Cl₂].

For N₂ + O₂ ⇌ 2 NO, suppose K = 4.0 × 10⁻³¹ at 25 °C. In air, [N₂] ≈ 0.032 mol/L and [O₂] ≈ 0.0086 mol/L. Estimate the equilibrium [NO].

Explain why the approximation is the same idea as rounding: 1000.003 ≈ 1000.

Challenge

A reaction with a large K can be handled by a similar trick. For A + B ⇌ C, K = 1.0 × 10⁸, start with [A] = [B] = 0.100 mol/L. Assume the reaction goes to completion first, then let a little C come back. Find [A] at equilibrium.

For A ⇌ 2 B with K = 1.0 × 10⁻⁴, find the smallest [A]ᵢ for which the approximation passes the 5 % rule.

For H₂(g) + I₂(g) ⇌ 2 HI(g) the K is about 50. Explain why the small-K approximation is never appropriate for starting mixtures of H₂ and I₂, and what you should do instead.

Check your understanding

Quiz

The small-K approximation lets you replace:

When is the small-K approximation valid?

x² ÷ 0.20 = 5.0 × 10⁻⁷. Using the approximation, x is:

For a calculation, x = 0.0080 and the initial concentration is 0.10 mol/L. You should:

Why is the approximation so useful for 2 NOCl ⇌ 2 NO + Cl₂?

The 5 % check should be made against:

For parents and teachers

Big idea (E2.4): when K is tiny compared with the starting concentrations, (c − x) ≈ c, which turns hard equations into simple roots. Always check: the change must be under 5 % of the starting concentration, or else solve exactly.

How to help: after each problem, ask "Did you check the 5 % rule?" The spreadsheet activity shows clearly when the shortcut is safe.

Common difficulties:

Dropping x everywhere, including the product terms (2x)².
Skipping the check, or checking against K instead of the initial concentration.
Forgetting that coefficients matter: in 0.500 − 2x, the change is 2x.

Looking ahead: in Acid-Base and Solubility Equilibria, nearly every weak-acid and weak-base calculation uses this approximation.