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Kp Calculations (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Partial Pressure

  • For reactions involving mixtures of gases, the equilibrium constant Kp is used

    • This is because it is easier to measure the pressure than the concentration for gases

  • The partial pressure of a gas is the pressure that the gas would have if it was in the container all by itself

  • The total pressure is the sum of all the partial pressures:

Ptotal = PA + PB + PC + .......

  • Where:

    • Ptotal = total pressure

    • PA, PB, PC = partial pressures

How partial pressures contribute to total pressure

Equilibria Partial Pressures
Partial pressures can be added together to calculate the total pressure

Mole fraction

  • The mole fraction of a gas is the ratio of moles of a particular gas to the total number of moles of gas present

bold Mole bold space bold fraction bold equals fraction numerator bold number bold space bold of bold space bold moles bold space bold of bold space bold a bold space bold particular bold space bold gas over denominator bold total bold space bold number bold space bold of bold space bold moles bold space bold of bold space bold all bold space bold the bold space bold gases bold space bold in bold space bold the bold space bold mixture end fraction

  • This equation can be used to calculate the partial pressure of each gas:

bold Partial bold space bold pressure bold equals bold mole bold space bold fraction bold cross times bold total bold space bold pressure bold space

Examiner Tips and Tricks

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

Worked Example

A sample of 0.25 mole of nitrogen and 0.75 mole of hydrogen were 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

Kp Calculations

  • Kp calculations are a step-by-step process

  • Some questions may give the equilibrium partial pressure of all the gases

  • This means that the only remaining steps are:

    • Write the Kp expression

    • Calculate the value of Kp

Worked Example

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

2SO2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 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 Kp for this reaction.

Answer:

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

    • Kp = fraction numerator p squared space SO subscript 3 over denominator p squared space SO subscript 2 cross times p space straight O subscript 2 end fraction

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

    • Kp = fraction numerator open parentheses 8.0 cross times 10 to the power of 6 close parentheses squared over denominator open parentheses 1.0 cross times 10 to the power of 6 close parentheses squared cross times open parentheses 7.0 cross times 10 to the power of 6 close parentheses end fraction

    • Kp = 9.1 x 10–6

  • Step 3: Deduce the correct units of Kp:

    • Kp = fraction numerator Pa squared over denominator Pa squared cross times Pa end fraction

    • So, the units of Kp are Pa-1

    • Therefore, Kp = 9.1 x 10-6 Pa-1

  • Some questions give the number of moles of gases present and the total pressure

  • So, the steps to complete these problems are:

    • Calculate the mole fractions

    • Calculate the partial pressures

    • Write the Kp expression

    • Calculate the value of Kp

Worked Example

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

H2 (g) + I2 (g) rightwards harpoon over leftwards harpoon 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 Kp 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 = fraction numerator 1.71 cross times 10 to the power of negative 3 end exponent over denominator 2.112 cross times 10 to the power of negative 2 end exponent end fraction = 0.0810

    • I2 = fraction numerator 2.91 cross times 10 to the power of negative 3 end exponent over denominator 2.112 cross times 10 to the power of negative 2 end exponent end fraction = 0.1378

    • HI = begin mathsize 14px style fraction numerator 1.65 cross times 10 to the power of negative 2 end exponent over denominator 2.112 cross times 10 to the power of negative 2 end exponent end fraction end style = 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:

    • Kp = fraction numerator p squared space HI over denominator p space straight H subscript 2 cross times p space straight I subscript 2 end fraction

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

    • Kp = begin mathsize 14px style fraction numerator 78.13 squared over denominator 8.10 cross times 13.78 end fraction end style

    • Kp = 54.7

  • Step 6: Deduce the correct units for Kp:

    • Kp = fraction numerator Pa squared over denominator Pa cross times Pa end fraction

    • All units cancel out

    • Therefore, Kp = 54.7

  • A full Kp question will give:

    • Initial number of moles of the gases present

    • The equilibrium number of moles of at least one of the gases present

    • The total pressure

  • So, the steps to complete these problems are:

    • Determine the number of equilibrium moles of all gases

    • Calculate the mole fractions

    • Calculate the partial pressures

    • Write the Kp expression

    • Calculate the value of Kp

Worked Example

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 Kp.

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 Kp expression is:

    • Kp = fraction numerator p squared space HBr space stretchy left parenthesis straight g stretchy right parenthesis over denominator p space straight H subscript 2 stretchy left parenthesis straight g stretchy right parenthesis cross times p space Br subscript 2 stretchy left parenthesis straight g stretchy right parenthesis end fraction

  • The value of Kp is:

    • Kp = fraction numerator 42 squared over denominator 49 cross times 49 end fraction = 0.735

  • There are no units for this Kp value

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

Examiner Tips and Tricks

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

You also need to be able to work backwards to deduce partial pressures and equilibrium moles from given information.

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    Author
    Reviewer
    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.