Topic 13 · page 7 of 12

Calculating Kp

For Kp, extend the table by two rows. Total the moles of gas, turn each amount into a mole fraction, multiply by the total pressure, then substitute the partial pressures.

15 min read13.3

Read first: Partial pressures and Kp, Kc from equilibrium amounts

The same table, two more rows 13.3

The method is the table from page 5 with a different ending. The amounts at equilibrium are the same. Instead of dividing by the volume, you turn them into partial pressures.

MethodKp from amounts of gas

Write the balanced equation and the expression for $K_\text{p}$ .
Make the table: start, change, equilibrium, in moles.
Add up the equilibrium moles of gas to get the total.
Mole fraction of each gas = its moles ÷ the total.
Partial pressure = mole fraction × total pressure.
Check that the mole fractions add up to 1 and the partial pressures to the total pressure.
Substitute, keeping the unrounded partial pressures in your calculator. Give the units from page 4.

Rounding the partial pressures early can change the third significant figure of $K_\text{p}$ , so the working below shows rounded values but the answers use the unrounded ones.

Worked exampleDissociation of N₂O₄

1.00 mol of N₂O₄ is heated and 20.0% dissociates. The total pressure at equilibrium is 2.00 atm. Calculate $K_\text{p}$ for $\ce{N2O4(g) <=> 2NO2(g)}$ .

	N₂O₄	2NO₂	Total
Start / mol	1.00	0
Change / mol	−0.20	+0.40
Equilibrium / mol	0.80	0.40	1.20
Mole fraction	0.80 ÷ 1.20 = 0.667	0.40 ÷ 1.20 = 0.333	1
Partial pressure / atm	0.667 × 2.00 = 1.33	0.333 × 2.00 = 0.667	2.00

20.0% of 1.00 mol = 0.20 mol of N₂O₄ dissociates.
The ratio is 1 : 2, so 0.20 mol of N₂O₄ makes 0.40 mol of NO₂.
$K_\text{p} = \dfrac{p_{\ce{NO2}}^{\,2}}{p_{\ce{N2O4}}} = \dfrac{0.667^2}{1.33}$ = 0.333 atm

Worked exampleDissociation of PCl₅

1.00 mol of PCl₅ is 40.0% dissociated at a total pressure of 1.50 atm. Calculate $K_\text{p}$ for $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ .

PCl₅ dissociated = 0.400 mol. At equilibrium: PCl₅ = 0.600 mol, PCl₃ = 0.400 mol, Cl₂ = 0.400 mol. Total = 1.400 mol.
$p_{\ce{PCl5}}$ = (0.600 ÷ 1.400) × 1.50 = 0.643 atm
$p_{\ce{PCl3}}$ = $p_{\ce{Cl2}}$ = (0.400 ÷ 1.400) × 1.50 = 0.429 atm
Check: 0.643 + 0.429 + 0.429 = 1.50 atm.
$K_\text{p} = \dfrac{p_{\ce{PCl3}}\;p_{\ce{Cl2}}}{p_{\ce{PCl5}}} = \dfrac{0.429 \times 0.429}{0.643}$ = 0.286 atm

When the ratio is not 1 : 1 13.3

The change row follows the coefficients, as on page 5. The units of $K_\text{p}$ then follow the count of species.

Worked exampleAmmonia at high pressure

1.00 mol of N₂ and 3.00 mol of H₂ react at a total pressure of 100 atm. At equilibrium there is 0.600 mol of NH₃ (illustrative). Calculate $K_\text{p}$ for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ .

NH₃ formed = 0.600 mol. N₂ used = 0.300 mol. H₂ used = 0.900 mol.
Equilibrium: N₂ = 0.700 mol, H₂ = 2.10 mol, NH₃ = 0.600 mol. Total = 3.40 mol.
$p_{\ce{N2}}$ = (0.700 ÷ 3.40) × 100 = 20.6 atm. $p_{\ce{H2}}$ = (2.10 ÷ 3.40) × 100 = 61.8 atm. $p_{\ce{NH3}}$ = (0.600 ÷ 3.40) × 100 = 17.6 atm.
$K_\text{p} = \dfrac{p_{\ce{NH3}}^{\,2}}{p_{\ce{N2}}\;p_{\ce{H2}}^{\,3}} = \dfrac{17.6^2}{20.6 \times 61.8^3}$ = 6.42 × 10⁻⁵ atm⁻²

Heterogeneous systems 13.3

When a solid is part of the equilibrium, leave it out of the expression and out of the total moles of gas.

Worked exampleCarbon and carbon dioxide

Excess carbon is heated with CO₂. At equilibrium the gas is 80.0% CO by volume, at a total pressure of 2.00 atm. Calculate $K_\text{p}$ for $\ce{C(s) + CO2(g) <=> 2CO(g)}$ .

Percentage by volume is percentage by moles, so $x_{\ce{CO}}$ = 0.800 and $x_{\ce{CO2}}$ = 0.200.
$p_{\ce{CO}}$ = 0.800 × 2.00 = 1.60 atm. $p_{\ce{CO2}}$ = 0.200 × 2.00 = 0.400 atm.
$K_\text{p} = \dfrac{p_{\ce{CO}}^{\,2}}{p_{\ce{CO2}}} = \dfrac{1.60^2}{0.400}$ = 6.40 atm. Carbon is a solid, so it is not in the expression.

Check yourself

At equilibrium a mixture of 1.20 mol SO₂, 0.60 mol O₂ and 8.20 mol SO₃ is at a total pressure of 2.0 atm. Calculate $K_\text{p}$ for $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ , with units.
0.50 mol of H₂ and 0.50 mol of I₂ react in a sealed flask at a total pressure of 3.0 atm. At equilibrium there is 0.78 mol of HI. Calculate $K_\text{p}$ for $\ce{H2(g) + I2(g) <=> 2HI(g)}$ .
1.00 mol of N₂O₄ is 30.0% dissociated at a total pressure of 1.00 atm. Calculate $K_\text{p}$ for $\ce{N2O4(g) <=> 2NO2(g)}$ .
1.00 mol of N₂ and 3.00 mol of H₂ react at a total pressure of 50.0 atm. At equilibrium there is 0.800 mol of NH₃ (illustrative). Calculate $K_\text{p}$ for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , with units.
A student finds the mole fractions in an equilibrium mixture to be 0.50, 0.40 and 0.30. What is wrong?
Over excess carbon, the equilibrium gas for $\ce{C(s) + CO2(g) <=> 2CO(g)}$ has $p_{\ce{CO2}}$ = 0.40 atm and $p_{\ce{CO}}$ = 1.60 atm. Calculate $K_\text{p}$ and give its units.

Answers cover these until you have tried the questions

Total = 10.0 mol. $p_{\ce{SO2}}$ = 0.24 atm, $p_{\ce{O2}}$ = 0.12 atm, $p_{\ce{SO3}}$ = 1.64 atm. $K_\text{p} = \dfrac{1.64^2}{0.24^2 \times 0.12} = \dfrac{2.690}{6.912 \times 10^{-3}}$ = 389 atm⁻¹
H₂ used = I₂ used = 0.39 mol, so at equilibrium H₂ = I₂ = 0.11 mol. Total = 0.11 + 0.11 + 0.78 = 1.00 mol. $p_{\ce{H2}}$ = $p_{\ce{I2}}$ = 0.11 × 3.0 = 0.33 atm, $p_{\ce{HI}}$ = 0.78 × 3.0 = 2.34 atm. $K_\text{p}$ = 2.34² ÷ (0.33 × 0.33) = 5.476 ÷ 0.1089 = 50, no units. The pressure cancels, so it equals $K_\text{c}$ here.
Dissociated = 0.300 mol. Equilibrium: N₂O₄ = 0.700 mol, NO₂ = 0.600 mol, total = 1.300 mol. $p_{\ce{N2O4}}$ = (0.700 ÷ 1.300) × 1.00 = 0.538 atm. $p_{\ce{NO2}}$ = (0.600 ÷ 1.300) × 1.00 = 0.462 atm. $K_\text{p}$ = 0.462² ÷ 0.538 = 0.396 atm
N₂ used = 0.400 mol, H₂ used = 1.20 mol. Equilibrium: N₂ = 0.600 mol, H₂ = 1.80 mol, NH₃ = 0.800 mol. Total = 3.20 mol. $p_{\ce{N2}}$ = 9.375 atm, $p_{\ce{H2}}$ = 28.125 atm, $p_{\ce{NH3}}$ = 12.5 atm. $K_\text{p}$ = 12.5² ÷ (9.375 × 28.125³) = 156.25 ÷ 2.086 × 10⁵ = 7.49 × 10⁻⁴ atm⁻²
The mole fractions of all the gases add up to 1. These add up to 0.50 + 0.40 + 0.30 = 1.20, so at least one is wrong, probably from a wrong total number of moles.
$K_\text{p} = \dfrac{p_{\ce{CO}}^{\,2}}{p_{\ce{CO2}}} = \dfrac{1.60^2}{0.40}$ = 6.4 atm. The top has a power of 2 and the bottom a power of 1, so the units are atm. Carbon is a solid and is left out.