Revision NotesExam QuestionsPast Papers
International A LevelChemistryEdexcelRevision Notes4: Rates, Equilibria & Further Organic ChemistryChemical EquilibriaEquilibrium Constant, Kp

Equilibrium Constant, Kp (Edexcel International A Level Chemistry): Revision Note

Stewart Hird

Last updated

Kp Expressions - Deduction

We have seen previously that equilibrium reactions can be quantified by reference to an equilibrium expression and equilibrium constant The equilibrium expression links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account So, for a given reaction:

aA + bB ⇌ cC + dD

Kc is defined as follows:

Equilibria Equilibrium Expression, downloadable AS & A Level Chemistry revision notes

Equilibrium expression linking the equilibrium concentration of reactants and products at equilibrium

  • Solids are ignored in equilibrium expressions

  • The Kc of a reaction is constant and only changes if the temperature of the reaction changes

Homogeneous reactions

  • In the generic example above, if all the substances are gases, we can show the equation with that state symbol

aA (g) + bB (g) ⇌ cC (g) + dD (g)

  • We can write a different equilibrium expression in terms of the partial pressure of the gases

  • This equilibrium constant is called Kp and is defined as follows

Kp Expressions, downloadable AS & A Level Chemistry revision notes

Equilibrium expression linking the partial pressures of reactants and products at equilibrium

Heterogeneous reactions

  • For heterogenous reactions, solids and liquids are ignored in Kp equilibrium expressions

  • The Kp of a reaction is constant and only changes if the temperature of the reaction changes

Worked Example

Write a Kp expression for the following equilibria and deduce the units of Kp :

  1. N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

  2. CaCO3 (s)  ⇌ CaO (s) + CO2 (g)

Answer 1

Worked Example Answer 1, downloadable AS & A Level Chemistry revision notes

Answer 2

Kp = pCO2(g)

Kp Expressions - Calculations

Calculations involving Kp

  • In the equilibrium expression, the p represents the partial pressure of the reactants and products in Pa

  • The units of Kp therefore depend on the form of the equilibrium expression

Worked Example

Calculating Kp of a gaseous reaction

Sulfur dioxide and oxygen react according to the following equation:

2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

The equilibrium partial pressures at constant temperature 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

   begin mathsize 14px style K subscript p space equals space fraction numerator p squared space SO subscript 3 over denominator p squared space SO subscript 2 space cross times p straight O subscript 2 end fraction end style

  • Step 2: Substitute the equilibrium concentrations into the expression

   begin mathsize 14px style K subscript p space equals space fraction numerator left parenthesis 8.0 space cross times 10 to the power of 6 right parenthesis squared over denominator left parenthesis 1.0 space cross times 10 to the power of 6 right parenthesis squared space cross times left parenthesis 7.0 space cross times 10 to the power of 6 right parenthesis end fraction end style

   Kp = 9.1 x 10-6

  • Step 3: Deduce the correct units of Kp

   begin mathsize 14px style K subscript italic p space equals space fraction numerator P italic a squared over denominator P italic a squared space cross times P italic a end fraction end style

The units of Kp are Pa-1

Therefore, Kp = 9.1 x 10-6 Pa-1

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

  • The number of moles of each gas should be used to first calculate the mole fractions

  • The mole fractions are then used to calculate the partial pressures

  • The values of the partial pressures are then substituted in the equilibrium expression

Worked Example

Calculating Kp of hydrogen iodide equilibrium reaction

The equilibrium between hydrogen, iodine and hydrogen iodide, at 600 K, is as follows:

H2 (g) + I2 (g) ⇌ 2HI (g)

At equilibrium, the number of moles present 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

   begin mathsize 14px style straight H subscript 2 equals 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 equals 0.0810 straight I subscript 2 space equals space fraction numerator 2.91 space cross times 10 to the power of negative 3 end exponent over denominator 2.112 space cross times 10 to the power of negative 2 end exponent end fraction equals space 0.1378 HI space equals space fraction numerator 1.65 space cross times 10 to the power of negative 2 end exponent over denominator 2.112 space cross times 10 to the power of negative 2 end exponent end fraction equals space 0.7813 end style

  • 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

   begin mathsize 14px style K subscript p space equals fraction numerator p squared HI over denominator p straight H subscript 2 space cross times p straight I subscript 2 end fraction end style

  • Step 5: Substitute the values into the equilibrium expression

   begin mathsize 14px style K subscript p space equals space fraction numerator 78.13 squared over denominator 8.10 space cross times 13.78 end fraction end style

   Kp = 54.7

  • Step 6: Deduce the correct units for Kp

   begin mathsize 14px style K subscript p space equals space fraction numerator P a squared over denominator P a space cross times P a end fraction end style

   All units cancel out

   Therefore, Kp = 54.7

  • Other questions related to equilibrium expressions may involve calculating quantities present at equilibrium given appropriate data

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Equilibrium Constant, KcNext:Equilibrium Constant & Conditions
    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.