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Equilibrium Terms (OCR A Level Chemistry A): Revision Note

Richard Boole

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Mole Fractions & Partial Pressure

Mole fractions and partial pressures are a feature of Kp calculations Put simply, the mole fraction is the fraction of the total number of moles that each chemical in a reaction is responsible for  Partial pressure is the part of the total pressure that each chemical in a reaction is responsible for  The partial pressure of a gas is the pressure it exerts in a mixture of gases if it occupied the container on its own Partial pressure is given the symbol p, so for a gas X, it is written as pX The total pressure is the sum of the partial pressures (this is known as Daltons' Law)

Partial Pressures, downloadable AS & A Level Chemistry revision notes

The relationship between partial pressures in a mixture of gases and the total pressure

  • To find the partial pressure of a gas, you need two pieces of information

    • The total pressure in the container

    • The mole fraction

  • The mathematical relationships are as follows

Partial pressure and mole fraction formulae, downloadable AS & A Level Chemistry revision notes

Partial pressure and mole fraction expressions

Equilibrium Quantities

  • When dealing with equilibrium calculations there are certain calculations that you will be expected to be able to perform

    • Calculating concentrations

      • Concentration (mol dm-3) = fraction numerator number space of space moles space over denominator volume space left parenthesis dm cubed right parenthesis end fraction

    • Calculating equilibrium quantities

    • Calculating mole fractions and partial pressures

Worked Example

Calculating concentrations

Ethanoic acid and ethanol react according to the following equation:

CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)

At equilibrium, 500 cm3 of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.

Calculate the concentration of each chemical at equilibrium.

Answer

  • [CH3COOH] equals space fraction numerator 0.235 over denominator 0.500 end fraction space equals space0.470 mol dm-3 

    • [C2H5OH] equals space fraction numerator 0.035 over denominator 0.500 end fraction space equals0.070 mol dm-3 

    • [CH3COOC2H5] equals space fraction numerator 0.182 over denominator 0.500 end fraction space equals0.364 mol dm-3 

    • [H2O] equals space fraction numerator 0.182 over denominator 0.500 end fraction space equals0.364 mol dm-3 

Calculating equilibrium quantities

  • Some questions give the initial and equilibrium concentrations of the reactants but not the products

  • An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked Example

Ethyl ethanoate is hydrolysed by water:

CH3COOC2H5(I) + H2O(I) ⇌ CH3COOH(I) + C2H5OH(I)

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1dm3. At equilibrium 0.0654 mol of water are present.

Use this data to calculate the equilibrium concentrations of each chemical.

Answer

Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Equilibria Calculating Kc of ethyl ethanoate table

Step 2: Calculate the concentrations of the reactants and products

  • [CH3COOC2H5] equals space fraction numerator 0.0654 over denominator 1.00 end fraction space equals0.0654 mol dm-3 

    • [H2O] equals space fraction numerator 0.0654 over denominator 1.00 end fraction space equals0.0654 mol dm-3 

    • [CH3COOH] equals space fraction numerator 0.0346 over denominator 1.00 end fraction space equals0.0346 mol dm-3 

    • [C2H5OH] equals space fraction numerator 0.0346 over denominator 1.00 end fraction space equals0.0346 mol dm-3 

Calculating mole fractions and partial pressures

  • These are two of the fundamental calculations associated with Kp calculations

Worked Example

Working out mole fractions

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 mol.

N2 (g) + 3H2 (g) ⇌  2NH3 (g)

Calculate the mole fractions of nitrogen, hydrogen and ammonia.

Answer

Write out the equation and record the initial, the change and the equilibrium amounts:

WE Mole Fractions Answer, downloadable AS & A Level Chemistry revision notes

Examiner Tips and Tricks

You can check you have the mole fractions correct by adding them up and making sure they come to 1: 0.195 + 0.585 + 0.220 = 1

Worked Example

The total pressure for the reaction, described above, of nitrogen and hydrogen to form ammonia was 150 kPa.

Calculate the partial pressure of each gas.

Answer

  • Partial pressure of N2 (g) = 0.195 x 150 = 29.25 kPa

    • Partial pressure of H2 (g) = 0.585 x 150 = 87.75 kPa

    • Partial pressure of NH3 (g) = 0.220 x 150 = 33.0 kPa

Examiner Tips and Tricks

  • You can check that your partial pressures are correct by adding them up:

    • 29.25 + 87.75 + 33.0 = 150

  • They should add up to the total pressure - if they don't, then there is at least one calculation wrong somewhere!

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    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.