Partial Pressure (AQA A Level Chemistry): Revision Note

You can check that your mole fractions and partial pressures are correct:

• The sum of the mole fractions should add up to 1

• The sum of the partial pressures should add up to the total pressure

A sample of 0.25 mole of nitrogen and 0.75 mole of hydrogen was reacted together to form ammonia. The equilibrium amount of nitrogen was 0.16 mole at a pressure of 75 kPa.

N₂ (g) + 3H₂ (g) ⇌  2NH₃ (g)

1. Calculate the mole fractions of nitrogen, hydrogen, and ammonia.

2. Calculate the partial pressures of nitrogen, hydrogen, and ammonia.

Answer:

1. The mole fractions of nitrogen, hydrogen, and ammonia:

   • Complete an ICE table:

     	N2 (g)	3H2 (g)	2NH3 (g)
     Initial	0.25 mol	0.75 mol	0.00 mol
     Change	0.25 - 0.16 = -0.09	-3 x 0.09 = -0.27	+2 x 0.09 = +0.18
     Equilibrium	0.16 mol	0.75 - 0.27 = 0.48 mol	0.18 mol

   • Calculate the total number of moles:

     • 0.16 + 0.48 + 0.18 = 0.82

   • Calculate the mole fractions:

     • N₂ (g) = 0.16 / 0.82 = 0.195

     • H₂ (g) = 0.48 / 0.82 = 0.585

     • NH₃ (g) = 0.18 / 0.82 = 0.220

2. The partial pressures of nitrogen, hydrogen, and ammonia:

   • Partial pressure = mole fraction x total pressure

     • N₂ (g) = 0.195 x 75 = 14.625 kPa

     • H₂ (g) = 0.585 x 75 = 43.875 kPa

     • NH₃ (g) = 0.220 x 75 = 16.5 kPa

Check

Mole fractions = 0.195 + 0.585 + 0.220 = 1

Partial pressures = 14.625 + 43.875 + 16.5 =75

The equilibrium between sulfur dioxide, oxygen, and sulfur trioxide is as follows:

2SO₂ (g) + O₂ (g) $\rightleftharpoons$ 2SO₃ (g)

At constant temperature, the equilibrium partial pressures are:

• SO₂ = 1.0 x 10⁶ Pa

• O₂ = 7.0 x 10⁶ Pa

• SO₃ = 8.0 x 10⁶ Pa

Calculate the value of K_p for this reaction.

Answer:

• Step 1: Write the equilibrium constant for the reaction in terms of partial pressures:

  • K_p = $\frac{p^2{\mathrm{SO}}_3}{p^2{\mathrm{SO}}_2\times p{\mathrm{O}}_2}$

• Step 2: Substitute the equilibrium concentrations into the expression:

  • K_p = $\frac{{\left(8.0\times {10}^6\right)}^2}{{\left(1.0\times {10}^6\right)}^2\times \left(7.0\times {10}^6\right)}$

  • K_p = 9.1 x 10^–6

• Step 3: Deduce the correct units of K_p:

  • K_p = $\frac{{\mathrm{Pa}}^2}{{\mathrm{Pa}}^2\times \mathrm{Pa}}$

  • So, the units of K_p are Pa^-1

  • Therefore, K_p = 9.1 x 10^-6 Pa^-1

The equilibrium between hydrogen, iodine and hydrogen iodide is as follows:

H₂ (g) + I₂ (g) $\rightleftharpoons$ 2HI (g)

At constant temperature, the equilibrium moles are:

• H₂ = 1.71 x 10^–3

• I₂ = 2.91 x 10^–3

• HI = 1.65 x 10^–2 

The total pressure is 100 kPa.

Calculate the value of K_p for this reaction.

Answer:

• Step 1: Calculate the total number of moles:

  • Total number of moles = 1.71 x 10^-3 + 2.91 x 10^-3 + 1.65 x 10^-2

  • Total number of moles = 2.112 x 10^-2

• Step 2: Calculate the mole fraction of each gas:

  • H₂ = $\frac{1.71\times {10}^{-3}}{2.112\times {10}^{-2}}$ = 0.0810

  • I₂ = $\frac{2.91\times {10}^{-3}}{2.112\times {10}^{-2}}$ = 0.1378

  • HI = $\frac{1.65\times {10}^{-2}}{2.112\times {10}^{-2}}$ = 0.7813

• Step 3: Calculate the partial pressure of each gas:

  • H₂ = 0.0810 x 100 = 8.10 kPa

  • I₂ = 0.1378 x 100 = 13.78 kPa

  • HI = 0.7813 x 100 = 78.13 kPa

• Step 4: Write the equilibrium constant in terms of partial pressure:

  • K_p = $\frac{p^2\mathrm{HI}}{p{\mathrm{H}}_2\times p{\mathrm{I}}_2}$

• Step 5: Substitute the values into the equilibrium expression:

  • K_p = $\frac{78.{13}^2}{8.10\times 13.78}$

  • K_p = 54.7

• Step 6: Deduce the correct units for K_p:

  • K_p = $\frac{{\mathrm{Pa}}^2}{\mathrm{Pa}\times \mathrm{Pa}}$

  • All units cancel out

  • Therefore, K_p = 54.7

Hydrogen and bromine were mixed in a flask in a 1 : 1 ratio and allowed to reach equilibrium at 450 K. When equilibrium had been achieved, the total pressure in the flask was 140 kPa, and the mole fraction of bromine was 0.35.

The equation for the reaction is:

H₂ (g) + Br₂ (g)  ⇌ 2HBr (g)

Determine the partial pressures for each gas at equilibrium and the value of K_p.

Answer:

• H₂ and Br₂ are in a 1 : 1 ratio

  • So, the mole fraction of Br₂ = the mole fraction of H₂

  • Mole fraction of hydrogen = 0.35

• The mole fraction of hydrogen bromide is the remainder of the mole fraction

  • This is because the total mole fraction adds up to 1

  • So, the mole fraction of HBr = 1 - the mole fraction of H₂ - the mole fraction of Br₂

  • Mole fraction of hydrogen bromide = 1 - 0.35 - 0.35 = 0.3

• The partial pressures are calculated by mole fraction x total pressure:

  • Partial pressure H₂ = 0.35 x 140 = 49 kPa

  • Partial pressure Br₂ = 0.35 x 140 = 49 kPa

  • Partial pressure HBr = 0.30 x 140 = 42 kPa

• The K_p expression is:

  • K_p = $\frac{p^2\mathrm{HBr}\left(\mathrm{g}\right)}{p{\mathrm{H}}_2\left(\mathrm{g}\right)\times p{\mathrm{Br}}_2\left(\mathrm{g}\right)}$

• The value of K_p is:

  • K_p = $\frac{{42}^2}{49\times 49}$ = 0.735

• There are no units for this K_p value

  • The kPa² term on top is cancelled by the kPa x kPa on the bottom

You need to be able to start from initial moles to calculate equilibrium moles and continue to calculate K_p values.

AQA sometimes gives you the value of Kp and asks you to work backwards to find the total pressure or an equilibrium amount. These rearrangement questions have a high failure rate — practise algebraic rearrangement of the K_p expression, not just forward substitution.

Never use square brackets in a K_p expression — [ ] means concentration. Use p or pp to denote partial pressure, e.g., p(SO3) or ppSO3.

When substituting into the Kp expression, remember to raise each partial pressure to the power of its stoichiometric coefficient. Forgetting to square p(H₂) in a reaction with 2H₂ is a common way to lose marks.

Don't put mole fractions directly into the K_p expression — you must first multiply each mole fraction by the total pressure to get the partial pressure.

Always deduce the units of K_p by cancelling the pressure units in the expression. If you end up with kPa^-1 or Pa^-2, write the full unit including the negative power — don't just write kPa.