Topic 13 · page 5 of 12

Kc from equilibrium amounts

To calculate Kc, build a table from the starting amounts to the equilibrium amounts, convert to concentrations by dividing by the volume, then substitute. Use equilibrium values only.

15 min read13.3

Read first: Writing Kc, Units of K

The table method 13.3

An expression is useful once you can put numbers in it. The numbers must be equilibrium concentrations. Experiments usually give you the amounts you started with, and one measurement at equilibrium. The balanced equation links the two.

MethodKc from amounts of substance

Write the balanced equation and the expression for $K_\text{c}$ .
Make a table with three rows: start, change, equilibrium, in moles.
Fill in what you know. Use the mole ratio of the equation to fill the change row, with a minus sign for reactants used up and a plus sign for products made.
Work out the equilibrium amounts: start + change.
Divide each by the volume in dm³ to get concentrations.
Substitute into the expression. Give the units from page 4.

Worked examplePhosphorus pentachloride

0.200 mol of PCl₅ is heated in a sealed 2.00 dm³ flask. At equilibrium 0.050 mol of PCl₅ remains. Calculate $K_\text{c}$ for $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ .

	PCl₅	PCl₃	Cl₂
Start / mol	0.200	0	0
Change / mol	−0.150	+0.150	+0.150
Equilibrium / mol	0.050	0.150	0.150
Concentration / mol dm⁻³	0.025	0.075	0.075

PCl₅ used up = 0.200 − 0.050 = 0.150 mol. The ratio is 1 : 1 : 1, so 0.150 mol each of PCl₃ and Cl₂ form.
Concentrations: 0.050 ÷ 2.00 = 0.025, and 0.150 ÷ 2.00 = 0.075 mol dm⁻³.
$K_\text{c} = \dfrac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]} = \dfrac{0.075 \times 0.075}{0.025}$ = 0.225 mol dm⁻³

The volume did not cancel here, because there are two concentrations on the top and one on the bottom. Leaving out the division by 2.00 would give 0.45, which is wrong.

When the volume cancels 13.3

If the powers on the top and bottom add up to the same number, every term has the same factor of 1/V, and it cancels. Moles can go straight into the expression. This happens when $K_\text{c}$ has no units.

Worked exampleHydrogen iodide

0.50 mol of H₂ and 0.50 mol of I₂ are sealed in a flask and heated. At equilibrium there is 0.78 mol of HI. Calculate $K_\text{c}$ for $\ce{H2(g) + I2(g) <=> 2HI(g)}$ .

HI formed = 0.78 mol. The ratio H₂ : HI is 1 : 2, so H₂ used = 0.78 ÷ 2 = 0.39 mol. The same amount of I₂ is used.
At equilibrium, H₂ = I₂ = 0.50 − 0.39 = 0.11 mol.
$K_\text{c} = \dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} = \dfrac{0.78^2}{0.11 \times 0.11}$ = 50, no units.

Starting from concentrations 13.3

Sometimes the starting information is a concentration, and the table is in mol dm⁻³ from the start. There is no division by volume.

Worked exampleDinitrogen tetroxide

The initial concentration of N₂O₄ is 0.0400 mol dm⁻³, with no NO₂. At equilibrium [NO₂] = 0.0120 mol dm⁻³. Calculate $K_\text{c}$ for $\ce{N2O4(g) <=> 2NO2(g)}$ .

NO₂ formed = 0.0120 mol dm⁻³. The ratio N₂O₄ : NO₂ is 1 : 2, so N₂O₄ used = 0.0060 mol dm⁻³.
[N₂O₄] at equilibrium = 0.0400 − 0.0060 = 0.0340 mol dm⁻³.
$K_\text{c} = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \dfrac{0.0120^2}{0.0340}$ = 4.24 × 10⁻³ mol dm⁻³

When the ratio is not 1 : 1 13.3

Fill in the change row from the coefficients. Every entry in that row is the same multiple of its coefficient.

Worked exampleAmmonia

1.00 mol of N₂ and 3.00 mol of H₂ are sealed in a 2.00 dm³ vessel. At equilibrium there is 0.300 mol of NH₃. Calculate $K_\text{c}$ for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ .

	N₂	3H₂	2NH₃
Start / mol	1.00	3.00	0
Change / mol	−0.150	−0.450	+0.300
Equilibrium / mol	0.850	2.55	0.300
Concentration / mol dm⁻³	0.425	1.275	0.150

NH₃ formed = 0.300 mol. Using the ratio 1 : 3 : 2, N₂ used = 0.150 mol and H₂ used = 0.450 mol.
Concentrations: 0.850 ÷ 2.00 = 0.425; 2.55 ÷ 2.00 = 1.275; 0.300 ÷ 2.00 = 0.150 mol dm⁻³.
$K_\text{c} = \dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} = \dfrac{0.150^2}{0.425 \times 1.275^3}$ = 0.0255 mol⁻² dm⁶

Check yourself

0.800 mol of PCl₅ is sealed in a 4.00 dm³ vessel. At equilibrium 0.200 mol of PCl₅ remains. Calculate $K_\text{c}$ for $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ , with units.
2.00 mol of SO₂ and 1.00 mol of O₂ are sealed in a 5.00 dm³ vessel. At equilibrium there is 1.60 mol of SO₃. Calculate $K_\text{c}$ for $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ , with units.
1.00 mol of H₂ and 1.00 mol of I₂ react to equilibrium, and 1.56 mol of HI is present. Calculate $K_\text{c}$ for $\ce{H2(g) + I2(g) <=> 2HI(g)}$ , and say why the volume of the flask is not needed.
The initial concentration of N₂O₄ is 0.0500 mol dm⁻³. At equilibrium [NO₂] = 0.0160 mol dm⁻³. Calculate $K_\text{c}$ for $\ce{N2O4(g) <=> 2NO2(g)}$ , with units.
0.500 mol of N₂ and 1.50 mol of H₂ are sealed in a 1.00 dm³ vessel. At equilibrium there is 0.100 mol of NH₃. Calculate $K_\text{c}$ for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , with units.
Explain why the volume cancels in question 3 but not in question 1.

Answers cover these until you have tried the questions

PCl₅ used = 0.800 − 0.200 = 0.600 mol, so PCl₃ = Cl₂ = 0.600 mol. Concentrations: PCl₅ 0.200 ÷ 4.00 = 0.0500; PCl₃ and Cl₂ 0.600 ÷ 4.00 = 0.150 mol dm⁻³. $K_\text{c}$ = 0.150² ÷ 0.0500 = 0.450 mol dm⁻³
SO₃ formed = 1.60 mol, so SO₂ used = 1.60 mol and O₂ used = 0.80 mol. Equilibrium: SO₂ = 0.40 mol, O₂ = 0.20 mol. Concentrations: SO₂ 0.0800, O₂ 0.0400, SO₃ 0.320 mol dm⁻³. $K_\text{c}$ = 0.320² ÷ (0.0800² × 0.0400) = 0.1024 ÷ 2.56 × 10⁻⁴ = 400 mol⁻¹ dm³
HI formed = 1.56 mol, so H₂ used = I₂ used = 0.78 mol. Equilibrium H₂ = I₂ = 1.00 − 0.78 = 0.22 mol. $K_\text{c}$ = 1.56² ÷ (0.22 × 0.22) = 2.4336 ÷ 0.0484 = 50 (50.3), no units. There are two terms on the top and two on the bottom, so every factor of 1/V cancels.
N₂O₄ used = 0.0160 ÷ 2 = 0.0080 mol dm⁻³. [N₂O₄] = 0.0500 − 0.0080 = 0.0420 mol dm⁻³. $K_\text{c}$ = 0.0160² ÷ 0.0420 = 6.10 × 10⁻³ mol dm⁻³
NH₃ formed = 0.100 mol, so N₂ used = 0.050 mol and H₂ used = 0.150 mol. Equilibrium: N₂ = 0.450 mol, H₂ = 1.35 mol, NH₃ = 0.100 mol. The volume is 1.00 dm³, so these are the concentrations in mol dm⁻³. $K_\text{c}$ = 0.100² ÷ (0.450 × 1.35³) = 0.0100 ÷ 1.107 = 9.03 × 10⁻³ mol⁻² dm⁶
In question 3 the expression has two terms on the top and two on the bottom, so the power of the volume is the same on both and cancels (no units). In question 1 there are two on the top and one on the bottom, so one factor of 1/V is left over. The concentrations must be used, and $K_\text{c}$ has units.