Unit 7 · Topic 7.7 Beta

Calculating Equilibrium Concentrations

To find equilibrium concentrations from K, compare Q with K to get the direction, build an ICE table with changes of ±x times each coefficient, substitute the Equilibrium row into K and solve for x.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

A factory making ammonia needs to know how much product it will get from a tank of nitrogen and hydrogen before it builds the reactor. Nobody can wait for the gases to settle and measure. Instead, engineers take the equilibrium constant, the starting amounts and a three-row table, and calculate exactly what the mixture will hold at equilibrium.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In an ICE table for A ⇌ 2 B, if A changes by −x, what is the change in B?

  1. +2x
  2. +x
  3. −2x
  4. +x/2
Show the answer

Changes follow the coefficients: 1 A used makes 2 B.

  • Correct: +2x:
  • +x:
  • −2x:
  • +x/2:

2. A mixture has Q > K. Which way does the net reaction go?

  1. In reverse, toward reactants
  2. Forward, toward products
  3. Neither; it is at equilibrium
  4. It depends on the temperature only
Show the answer

Q > K means too much product, so the reaction runs in reverse until Q = K.

  • Correct: In reverse, toward reactants:
  • Forward, toward products:
  • Neither; it is at equilibrium:
  • It depends on the temperature only:

3. For a reaction with K = 1 × 10−8, starting from reactants, how much reactant is left at equilibrium?

  1. Almost all of it
  2. About half
  3. Almost none
  4. Exactly none
Show the answer

A tiny K means a reactant-favored equilibrium: very little reacts.

  • Correct: Almost all of it:
  • About half:
  • Almost none:
  • Exactly none:

Part 4 · See it

See it first

A flow chart. Step 1, compare Q with K to find the direction. Step 2, write an ICE table with each change equal to plus or minus x times the coefficient. Step 3, substitute the equilibrium row into the K expression. Then choose a way to solve: if both sides are perfect squares, take the square root; if K is small compared with the starting concentration, drop x from the initial minus x terms; otherwise use the quadratic formula. Step 4, check: x should be under 5% of the starting concentration if the approximation was used, and the answers should give Q = K.
Direction, ICE table, substitute, then solve by square root, small-x approximation or the quadratic formula, and check. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Q is compared with K for the starting mixtureyou know which side is used up and which forms
  2. The ICE table writes each change as ±x times a coefficientevery equilibrium concentration is written with one unknown
  3. The Equilibrium row is substituted into the K expressionyou get one equation in x
  4. K is small compared with the starting concentrationx can be dropped from (initial − x), and a 5% check confirms it
  5. x is solved and checkedevery equilibrium concentration follows from Initial + Change

Part 6 · Key ideas

Key ideas

  • Always find the direction first by comparing Q with K; it sets the signs in the Change row.
  • Changes are ±x × coefficient. Equilibrium = Initial + Change. Use molarities.
  • Perfect square: take the square root of both sides. Small K: drop x only from (initial − x), never x standing alone.
  • Check the small-x approximation: the change must be under about 5% of the starting value; otherwise solve exactly.

Part 7 · Misconception

A common mistake

The wrong idea: If K is small, x is small, so every x in the equation can be set to zero.

What actually happens: Only x that is added to or subtracted from a larger number can be dropped: 0.50 − 2x ≈ 0.50. The x terms standing alone on top of the expression stay, or there is nothing left to solve. Then check the 5% rule.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

An ICE table for ammonia

A sealed 1.00 L vessel starts with 0.500 M N2 and 0.800 M H2 at a high temperature: N2(g) + 3 H2(g) ⇌ 2 NH3(g). At equilibrium, [NH3] = 0.150 M. A student starts an ICE table.

Partly completed ICE table (concentrations in M)
[N2][H2][NH3]
Initial0.5000.8000
Change−x?+2x
Equilibrium0.500 − x?0.150

1. What belongs in the Change row for H2?

  1. −3x
  2. −x
  3. +3x
  4. −x/3
Show the answer

The Change row is x times each coefficient, negative for species used up. H₂ has coefficient 3: −3x.

  • Correct: −3x: Right: three H₂ react for every N₂, so H₂ falls three times as fast.
  • −x: This ignores the coefficient 3 on H₂.
  • +3x: H₂ is a reactant in a net forward reaction, so it is used up: the sign is negative.
  • −x/3: The coefficient multiplies x; it does not divide it.

2. Calculate [H2] at equilibrium.

Type a number and its unit.

Show the answer

2x = 0.150 M, so x = 0.0750 M. [H₂] = 0.800 − 3(0.0750) = 0.800 − 0.225 = 0.575 M.

  • Answer: 0.575 M

Experimental setup

When x is not small

PCl5 decomposes on heating: PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.040 at a certain temperature. A sealed flask starts with 0.200 M PCl5 and nothing else. A student writes x2 / (0.200 − x) = 0.040, assumes x is small compared with 0.200, and gets x = 0.089 M.

3. Is the student’s small-x approximation valid?

  1. No: 0.089 M is about 45% of 0.200 M, far more than 5%.
  2. Yes: x comes out less than 0.200 M.
  3. Yes: K = 0.040 is less than 1.
  4. No: the approximation is not used when the products start at zero.
Show the answer

Check the approximation: x / [initial] × 100% = 0.089 / 0.200 × 100% ≈ 45%. Over about 5%, x cannot be dropped and the equation must be solved exactly.

  • Correct: No: 0.089 M is about 45% of 0.200 M, far more than 5%.: Right: the approximation is only safe when x is a few percent of the starting concentration or less.
  • Yes: x comes out less than 0.200 M.: Being smaller than 0.200 is not enough; x must be negligible next to it (under about 5%).
  • Yes: K = 0.040 is less than 1.: K < 1 alone does not guarantee x is small. Here K is only moderately small compared with the concentration.
  • No: the approximation is not used when the products start at zero.: Starting with no products is the usual case for the approximation. It fails here because K is not small enough.

4. Solve x2 / (0.200 − x) = 0.040 exactly (with the quadratic formula) and give [Cl2] at equilibrium.

Type a number and its unit.

Show the answer

Rearrange: x² = 0.040(0.200 − x), so x² + 0.040x − 0.0080 = 0. x = [−0.040 + √((0.040)² + 4(0.0080))] / 2 = (−0.040 + √0.0336) / 2 = 0.0717 M ≈ 0.072 M.

  • Answer: 0.072 M

5. For H2(g) + I2(g) ⇌ 2 HI(g), Kc = 50.0 at 450 °C. A flask starts with 0.100 M H2 and 0.100 M I2. Calculate [HI] at equilibrium.

Type a number and its unit.

Show the answer

(2x)² / (0.100 − x)² = 50.0. Both sides are perfect squares: 2x / (0.100 − x) = √50.0 = 7.071. So 2x = 0.7071 − 7.071x, 9.071x = 0.7071, x = 0.0780 M. [HI] = 2x = 0.156 M.

  • Answer: 0.156 M

6. For 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g), Kc = 1.6 × 10−5. A flask starts with 0.50 M NOCl. Using the small-x approximation, calculate [Cl2] at equilibrium.

Type a number and its unit.

Show the answer

(2x)²(x) / (0.50)² = 1.6 × 10⁻⁵ → 4x³ = 4.0 × 10⁻⁶ → x³ = 1.0 × 10⁻⁶ → x = 0.010 M. Check: 2x = 0.020 M is 4% of 0.50 M, under 5%. [Cl₂] = 0.010 M.

  • Answer: 0.010 M

7. For CO(g) + H2O(g) ⇌ CO2(g) + H2(g), Kc = 1.56. A flask starts with 0.0500 M CO, 0.0500 M H2O, 0.200 M CO2 and 0.200 M H2. How should the Change row be written?

  1. +x for CO, H2O; −x for CO2, H2
  2. −x for CO, H2O; +x for CO2, H2
  3. −x for each of the four species
  4. 0 for each species, since the four start at nonzero values
Show the answer

Compare Q with K before writing the Change row. Q > K means net reverse: the reactant side gains +x and the product side loses −x.

  • Correct: +x for CO, H2O; −x for CO2, H2: Right: Q = (0.200)² / (0.0500)² = 16.0 > K, so the net reaction goes in reverse: reactants form, products are used.
  • −x for CO, H2O; +x for CO2, H2: This assumes a forward reaction. Check Q first: Q = 16.0 is greater than K, so the reaction runs in reverse.
  • −x for each of the four species: Every species cannot decrease; atoms must go somewhere. One side forms while the other is used.
  • 0 for each species, since the four start at nonzero values: Having all species present does not mean equilibrium. Q = 16.0 ≠ K = 1.56.

Part 9 · Summary

Summary

To find equilibrium concentrations from K, compare Q with K to get the direction, build an ICE table with changes of ±x times each coefficient, substitute the Equilibrium row into K and solve for x. A perfect-square expression is solved by a square root. When K is small compared with the starting concentration, x can be dropped from (initial − x) but never set to zero elsewhere, and the result must pass the 5% check; otherwise use the quadratic formula.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections