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E. Chemical Systems and Equilibrium · Chemical Equilibrium

ICE Tables II: Finding Equilibrium Concentrations from K

E2.4 E3.2 A1.12 A1.13About 75 minutes
Learning goals
  • I can use K and initial concentrations to write an equation in x.
  • I can solve perfect-square equilibrium problems by taking the square root.
  • I can solve other equilibrium problems with the quadratic formula and choose the meaningful root.
  • I can check an answer by substituting the equilibrium values back into K.

Words to know

perfect square (in equilibrium problems)
A K expression in which both the numerator and denominator are squares, such as (2x)² ÷ (0.200 − x)², so that taking the square root gives a simple linear equation.
quadratic formula
For ax² + bx + c = 0, x = [−b ± √(b² − 4ac)] ÷ 2a.
extraneous root
A solution of the equation that has no physical meaning, for example one that would make a concentration negative.
percent dissociation
The percentage of the starting amount of a substance that has broken apart (dissociated) at equilibrium.

Predicting instead of measuring

In the last lesson you measured one equilibrium concentration and calculated K. Chemists usually work the other way: K values are published in data tables, and an engineer wants to predict how much product a mixture will give before running the reaction. That means solving for x. Sometimes the algebra is quick (a square root); sometimes you need the quadratic formula you learned in Grade 10 and Grade 11 math.

The method

  1. ICE table in mol/L, with the E row written in terms of x.
  2. Substitute the E row into the K expression to get one equation in x.
  3. Solve:
  • If both sides are perfect squares, take the square root of both sides.
  • Otherwise, rearrange to ax² + bx + c = 0 and use the quadratic formula.
  1. Choose the root that gives no negative concentrations (and no more product than the reactants can make).
  2. Calculate every equilibrium concentration and check by substituting into K.
Write the ICE table in mol/LE row in terms of xSubstitute E row into KIs it a perfect square?yesnoTake the square rootof both sides; solvethe linear equation for xExpand to ax² + bx + c = 0x = [−b ± √(b² − 4ac)] ÷ 2aKeep the root that gives no negativeconcentrations; find every E value; check K
Choosing a solution method. A perfect square saves time; the quadratic formula always works.
Worked example

Example 1: A perfect square

For H₂(g) + I₂(g) ⇌ 2 HI(g), K = 50.0 at a certain temperature. A flask starts with [H₂] = [I₂] = 0.200 mol/L and no HI. Find the equilibrium concentrations.

  1. ICE: I: 0.200, 0.200, 0. C: −x, −x, +2x. E: 0.200 − x, 0.200 − x, 2x.
  2. K = (2x)² ÷ (0.200 − x)² = 50.0. Both sides are squares.
  3. Square root: 2x ÷ (0.200 − x) = √50.0 = 7.071.
  4. 2x = 7.071(0.200 − x) = 1.4142 − 7.071x, so 9.071x = 1.4142 and x = 0.1559.
  5. [HI] = 2x = 0.312 mol/L; [H₂] = [I₂] = 0.200 − 0.1559 = 0.0441 mol/L.
  6. Check: (0.312)² ÷ (0.0441)² = 0.0973 ÷ 0.00194 = 50.1 ✓ (rounding).

[HI] = 0.312 mol/L; [H₂] = [I₂] = 0.0441 mol/L. (The negative square root, −7.071, would give a negative concentration, so it is ignored.)

Worked example

Example 2: The quadratic formula

Same reaction and K = 50.0, but the flask starts with [H₂] = 0.200 mol/L and [I₂] = 0.100 mol/L. Find the equilibrium concentrations.

  1. E row: 0.200 − x, 0.100 − x, 2x. K = (2x)² ÷ [(0.200 − x)(0.100 − x)] = 50.0. Not a perfect square.
  2. 4x² = 50.0(0.0200 − 0.300x + x²) = 1.00 − 15.0x + 50.0x².
  3. Rearrange: 46.0x² − 15.0x + 1.00 = 0. So a = 46.0, b = −15.0, c = 1.00.
  4. x = [15.0 ± √(225 − 184)] ÷ 92.0 = (15.0 ± 6.403) ÷ 92.0, so x = 0.2326 or x = 0.09344.
  5. x = 0.2326 would make [I₂] = 0.100 − 0.2326 < 0: reject it. Keep x = 0.09344.
  6. [HI] = 0.187 mol/L; [H₂] = 0.200 − 0.09344 = 0.107 mol/L; [I₂] = 0.100 − 0.09344 = 0.0066 mol/L.

[HI] = 0.187 mol/L, [H₂] = 0.107 mol/L, [I₂] = 0.0066 mol/L. Almost all the I₂ (the limiting reactant) reacted, because K is large. Notice that [I₂] is a small difference of two numbers, so keep all the digits of x until the end.

00.020.040.060.080.10050100150x (mol/L)value of [HI]² ÷ ([H₂][I₂])K = 50.0x = 0.0934at x = 0.100all the I₂ is gone
Why one root is rejected. As x grows, the ratio [HI]²/([H₂][I₂]) climbs and crosses K = 50.0 at x = 0.0934. At x = 0.100 all the I₂ is used up, so x can never reach the other root, 0.2326.
Worked example

Example 3: A small K

For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), K = 0.0150 at a certain temperature. 0.100 mol of PCl₅ is placed in a 1.00 L flask. Find the equilibrium concentrations and the percent dissociation of PCl₅.

  1. E row: 0.100 − x, x, x. K = x² ÷ (0.100 − x) = 0.0150.
  2. x² = 0.00150 − 0.0150x, so x² + 0.0150x − 0.00150 = 0.
  3. x = [−0.0150 ± √(0.000225 + 0.00600)] ÷ 2 = (−0.0150 ± 0.07890) ÷ 2. Positive root: x = 0.03195.
  4. [PCl₃] = [Cl₂] = 0.0320 mol/L; [PCl₅] = 0.100 − 0.03195 = 0.0681 mol/L.
  5. Check: (0.03195)² ÷ 0.06805 = 0.0150 ✓.
  6. Percent dissociation = 0.03195 ÷ 0.100 × 100 % = 32.0 %.

[PCl₅] = 0.0681 mol/L; [PCl₃] = [Cl₂] = 0.0320 mol/L; 32.0 % dissociated.

Calculator tips

  • Write a, b and c with their signs before using the formula.
  • Compute b² − 4ac first and store it.
  • If both roots are positive, test each one in the E row; only one keeps every concentration positive.
  • Keep at least 4 significant digits in x, especially when an E value is a difference like 0.100 − 0.0934.
Just for fun

A chemist was asked which of her two roots she liked better. She said, "The positive one. The other one is just too negative."

Hands-on activity

Graph your way to equilibrium

You need: a free graphing tool (such as Desmos on a phone or computer) or graph paper and a calculator.

  1. For Example 2, graph y = (2x)² ÷ ((0.2 − x)(0.1 − x)) for x from 0 to 0.1, and the horizontal line y = 50.
  2. Find where they cross. Compare with the quadratic-formula answer, x = 0.0934.
  3. Change the line to y = 5 and y = 500 (pretend K were smaller or larger). Where do they cross now? What happens to [I₂] at equilibrium as K grows?
  4. Without a graphing tool: make a table of the ratio for x = 0.02, 0.04, 0.06, 0.08, 0.09, 0.095 and find between which two values it passes 50.

What to look for: the curve crosses y = 50 near x = 0.093. For y = 5 the crossing is near x = 0.070 and for y = 500 it is near x = 0.099: as K grows, x gets closer to 0.100 and [I₂] gets closer to zero, so the reaction gets closer to completion. The table method gives ratios of about 0.1, 0.7, 2.6, 10.7, 29.5 and 68.8, so the ratio passes 50 between x = 0.09 and x = 0.095.

Your turn

1

For H₂ + I₂ ⇌ 2 HI, K = 64.0. A flask starts with [H₂] = [I₂] = 0.100 mol/L. Find the equilibrium concentrations.

2

For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), K = 4.00. Start: [CO] = [H₂O] = 0.200 mol/L. Find all equilibrium concentrations.

3

For 2 HI(g) ⇌ H₂(g) + I₂(g), K = 0.0200. A flask starts with [HI] = 0.500 mol/L. Find the equilibrium concentrations.

4

For N₂O₄(g) ⇌ 2 NO₂(g), K = 4.6 × 10⁻³ at 25 °C. Start: [N₂O₄] = 0.100 mol/L. Find [NO₂] at equilibrium.

5

Set up (do not solve) the equation in x for A(g) + 2 B(g) ⇌ C(g), K = 10.0, starting with [A] = 0.50 mol/L and [B] = 1.00 mol/L.

6

In Example 2, why can't x be bigger than 0.100?

More practice: warm-up

1

Solve x² = 0.0400 for the root that makes sense as a concentration.

2

Is x² ÷ (0.50 − x)² = 9.0 a perfect square?

3

Solve x ÷ (0.50 − x) = 3.0.

4

For 2x² − 3x + 1 = 0, what are a, b and c?

5

How do you check an equilibrium answer?

More practice: on your own

1

For H₂ + I₂ ⇌ 2 HI, K = 50.0. Start: [H₂] = 0.300 mol/L, [I₂] = 0.150 mol/L. Find [HI] at equilibrium.

2

For A(g) ⇌ 2 B(g), K = 0.50. Start: [A] = 1.00 mol/L. Find [A] and [B] at equilibrium.

3

For CO + H₂O ⇌ CO₂ + H₂, K = 4.00. Start: [CO] = 0.100 mol/L and [H₂O] = 0.300 mol/L. Find [CO₂] at equilibrium.

4

For the same reaction (K = 4.00), a flask starts with only [CO₂] = [H₂] = 0.100 mol/L. Find [CO] at equilibrium.

5

For PCl₅ ⇌ PCl₃ + Cl₂ with K = 0.0150, find the percent dissociation if the flask starts with 0.0500 mol/L PCl₅.

Challenge

1

Compare Example 3 (0.100 mol/L PCl₅, 32.0 % dissociated) with the last question (0.0500 mol/L, 41.8 %). Explain why a more dilute sample dissociates by a larger percentage.

2

For N₂O₄ ⇌ 2 NO₂, K = 4.6 × 10⁻³ at 25 °C. Find the percent dissociation of N₂O₄ starting at 0.0500 mol/L.

3

For H₂ + I₂ ⇌ 2 HI, K = 50.0. A flask starts with [H₂] = [I₂] = [HI] = 0.100 mol/L. Find the equilibrium concentrations.

Check your understanding

Quiz

1.

For H₂ + I₂ ⇌ 2 HI starting with equal [H₂] and [I₂] and no HI, the K equation is best solved by:

2.

Solving 46x² − 15x + 1 = 0 gives x = 0.233 or x = 0.0934. If [I₂]ᵢ = 0.100 mol/L, which x is correct?

3.

For A ⇌ B with K = 3.0, starting with [A] = 0.40 mol/L and no B, [B] at equilibrium is:

4.

x² ÷ (0.20 − x) = 0.050. Rearranged, this is:

5.

After solving an equilibrium problem, the best check is to:

6.

If K is very large and one reactant is limiting, its equilibrium concentration will be:

For parents and teachers

Big idea (E2.4): with K and the starting concentrations, an ICE table gives an equation in x. Perfect squares are solved with a square root; other equations with the quadratic formula. Reject any root that makes a concentration negative, and check by substituting back into K.

How to help: this lesson is mostly algebra. If your teen is stuck, have them write a, b and c with signs before using the formula, and test each root in the E row. A graphing app (see the activity) gives a visual check.

Common difficulties:

  • Sign errors when expanding (0.200 − x)(0.100 − x).
  • Keeping the root that makes a concentration negative.
  • Rounding x too early, which wrecks small values like [I₂] = 0.0066.
  • Forgetting that a cubic (like Your turn question 5) is beyond hand methods; the next lesson shows an approximation that helps when K is small.