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STATES OF MATTER - properties of gases and liquids (fluids) and solids

22. Dalton's Law of Partial Pressures and calculations and use in solving gaseous equilibrium expression problems

Helpful for UK advanced level chemistry students aged ~16-18, IB courses and US grades 11-12 K12 honors.

22. Dalton's law of partial pressures and calculations

• Dalton's Law of partial pressures states that at constant temperature the total pressure exerted by a mixture of gases in a definite volume is equal to the sum of the individual pressures which each gas would exert if it alone occupied the same total volume.

• For a mixture of gases 1, 2, 3 ...

• Ptot = p1 + p2 + p3 etc.

• where p1, p2 etc. represent the partial pressures.

• The partial pressure ratio is the same as the % by volume ratio and the same as the mole ratio of gases in the mixture.

• This means for a component gas z, the partial pressure is expressed as:

• pz = Ptot x %z/100

• for % of volume of gases or in terms of moles of gases:

• pz = Ptot x mol z/total mol

• pz= Ptot x mol fraction z

• Examples of partial pressure calculations

• Ex. Q4d.1

• In the manufacture of ammonia a mixture of nitrogen : hydrogen in a 1 : 3 ratio is passed over an iron/iron oxide catalyst at high temperature and high pressure.

• N2(g) + 3H2(g) 2NH3(g)

• What are the partial pressures of nitrogen and hydrogen if the total pressure of the gases is 200 atm prior to reaction? So Ptot = 200 atm.

• The 1 : 3, N2 :H2 ratio means that nitrogen forms 1/4 of the mixture, therefore

• pN21/4 x 200 = 50 atm and

• pH2 = ptot – pN2 = 150 atm (or from 3/4 x 200)

• Ex. Q4d.2

• Methanol can be synthesised by combining carbon monoxide and hydrogen in a 1 : 2 ratio.

• CO(g) + 2H2(g) CH3OH(g)

• In an experimental reactor experiment, 300oC at a total pressure of 400kPa, the final equilibrium gaseous mixture contained 10% carbon monoxide.

• (a) Calculate the % of hydrogen gas and % methanol vapour in the final mixture.

• Whatever hydrogen is left, its % must be double that of carbon monoxide since they were both mixed and react in a 1 : 2 ratio, so there will 20% hydrogen left in the equilibrium mixture.

• Therefore there will be 100 – 10 – 20 = 70% methanol vapour in the final mixture.

• (b) Calculate the partial pressures of the three gases in the mixture.

• pCO = 0.1 x 400 = 40 kPa

• pH2 = 0.2 x 400 = 80 kPa

• pCH3OH = 0.7 x 400 = 280 kPa

• (useful to check 40 + 80 + 280 = 400)

• (c) From the partial pressure data in (b) calculate the value of the equilibrium constant, Kp, under these reaction conditions (use Pa pressure units).

•  Kp = pCH3OH ––––––––––– pCO pH22
•  Kp = 280 000 –––––––––––––––––   = 1.09 x 10–9 Pa–2 40 000 x 80 0002
• Note that although the equilibrium constant seems small for the 70% methanol, its to do with the relatively large numbers on the bottom line and a power of 2 as well.

Learning objectives for Dalton's Law of partial pressures

Be able to quote and use in calculations Dalton's Law of partial pressures.

Know how to calculate partial pressures from the % of gases in a mixture.

Know how to calculate partial pressures from the complete ratio of gases in a mixture.

Know how to calculate partial pressures from the molar composition of gases in a mixture.

Know how to calculate partial pressures and use them to solve gaseous equilibrium problems,

either given the equilibrium constant Kp and known partial pressures apart from one unknown,

or given all the partial pressures to calculate the equilibrium constant Kp.

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