12.2 DALTON’S LAW OF PARTIAL PRESSURES

CONTEXT

We have looked at how P, T, V are connected We know the ideal gas law: PV=nRT We haven’t really discussed mixtures of

gases If there is a mixture, how much of the total

pressure is due to each gas?

In a mixture of non-reacting gases Each gas particle behaves independently Particles spread themselves out to occupy the space

available

DALTON’S LAW OF PARTIAL PRESSURES

The total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases (the sum of the pressures that each gas would exert if it were alone).

Ptot = P1 + P2 + P3 + …

PARTIAL PRESSURES OF ATMOSPHERIC GASES

Air consists of a mixture of gases:N2 = 78.08% O2 = 20.95%

Ar = 0.93 % CO2 = 0.03%

If Standard atmospheric pressure is 101.325 kPa, determine the partial pressure of each gas.

P N2 = 79.11 kPa P O2 = 21.23 kPa

P Ar = 0.94 kPa P CO2 = 0.03 kPa

PT = PN2 + PO2 + Par + PC02

PT ≅ 101.325 kPa

GAS COLLECTION OVER WATER

Gases can be collected by the downward displacement of water

The gas collected is a mixture of the desired gas and water.

This partial pressure of water vapour will be a component of the total pressure

To find the partial pressure of the desired gas, the vapour pressure of water must be subtracted.

The partial pressure of water vapour depends on the temperature of the water

GAS COLLECTION OVER WATER

Hydrogen gas from the reaction of Zinc with hydrochloric acid was collected by the displacement of water. Atmospheric conditions were 100.5 kPa and 20˚C.

Determine the partial pressure of hydrogen gas.Ptot = Patm = PH2 + PH2O

PH2 = Patm - PH2O

= 100.5 kPa - 2.3 kPa

= 98.2 kPa

EXAMPLE Oxygen gas is collected over water at 297 K and

an atmospheric pressure of 1.0 atm. The volume of gas is 128 mL. Calculate the mass of O2 obtained. Hint: Water vapour at 297 K is 3 kPa.

T = 297 KPT = 1.0 atm = 101.325 kPaPH2O = 3 kPaV = 128 mL = 0.128 Lm = ?

1. PT = PO2 + PH2O

PO2 = 101.325 kPa – 3 kPa PO2 = 98.325 kPa

2. PV =nRT n = PV / RT n = (98.325)(0.128)/(8.314)(297) n = 5.10 x 10-3 mol

3. m = n x M m = (5.10 x 10-3)(32.00) m = 0.16 g

HOMEWORK

P. 460 # 22 -25 Worksheet