Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

and Gas Law Stoichiometryand Gas Law Stoichiometry

Honors ChemistryHonors Chemistry

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

The pressure of each gas in a mixture is The pressure of each gas in a mixture is called the called the partial pressurepartial pressure of that gas. of that gas.

Dalton’s Law statesDalton’s Law states that the total pressure that the total pressure of a mixture of gases is equal to the sum of of a mixture of gases is equal to the sum of the partial pressures of each of the the partial pressures of each of the component gases.component gases.

PPtotaltotal = P = P11 + P + P22 + P + P33 + . . . + . . .

PPtotaltotal = P = P11 + P + P22 + P + P33 + . . . + . . .

PH2 = 2.4 atm PHe = 6.0 atmPtotal = 2.4 + 6.0 =

8.4 atm

If the first three containers are all put into the If the first three containers are all put into the fourth, we can find the pressure in that container fourth, we can find the pressure in that container by adding up the pressure in the first 3:by adding up the pressure in the first 3:

2 atm + 1 atm + 3 atm = 6 atm

1 2 3 4

The partial pressure of each gas is The partial pressure of each gas is equal to the mole fraction (X) of each equal to the mole fraction (X) of each gas times the total pressuregas times the total pressure

Moles gas x PMoles gas x Ptotaltotal = P = Pgasgas

Total MolesTotal Moles

Mole fraction is like a percent

Sample Problem #1Sample Problem #1A mixture of He, Ne and Ar gases have a total A mixture of He, Ne and Ar gases have a total pressure of 790 mmHg. If there is 15% Ar, 60% pressure of 790 mmHg. If there is 15% Ar, 60% He and 25% Ne, what is the partial pressure of He and 25% Ne, what is the partial pressure of each gas?each gas?

Ptotal = PHe + PNe + PAr = 790 mmHg

15% Ar = (0.15)(790 ) = 118.5 mmHg +

60% He = (0.60)(790) = 474 mmHg +

25% Ne = (0.25)(790) = 197.5 mmHg

P total = 790 mmHg

Sample Problem #2Sample Problem #2

The partial pressure of COThe partial pressure of CO22 in a mixture of gases in a mixture of gases is 0.8 atm. If the total pressure is 1.05 atm, what is 0.8 atm. If the total pressure is 1.05 atm, what is the mole fraction of COis the mole fraction of CO22 in the mixture? in the mixture?

PCO2 = 0.8atm Ptotal = 1.05atm

(Mole Fraction )Ptotal = PCO2

XCO2 (1.05atm) = 0.8 atm

XCO2 = 0.8 atm = 0.76 1.05 atm

Gas Collected by Water Gas Collected by Water DisplacementDisplacement

AA gas collected by water displacement is not pure, but gas collected by water displacement is not pure, but always mixed with water vapor. Some molecules at the always mixed with water vapor. Some molecules at the surface of water always evaporate and exert a pressure surface of water always evaporate and exert a pressure called a vapor pressure.called a vapor pressure.

•Therefore, you must subtract the water –vapor pressure from the total pressure to get the pressure just from the gas.

Sample Problem #3Sample Problem #3

50mls of CH50mls of CH44 was collected over water at 25 was collected over water at 25ooC C

and 748mmHg. What is the volume of the gas and 748mmHg. What is the volume of the gas at STP? (The vapor pressure of water is at STP? (The vapor pressure of water is 23.8mmHg)23.8mmHg)

Ptotal = PCH4 + PH2O

748mmHg = PCH4 + 23.8mmHg

PCH4 = 724.2 mmHg (dry gas)

P1V1 = P2V2 (50mls)(724.2mmHg) = (760mmHg)V2

T1 T2 298K 273K

Solving for V2 = 43.6 mL

Gas Law StoichiometryGas Law StoichiometryWhat volume of oxygen will form at 25What volume of oxygen will form at 25ooC and 1.3atm C and 1.3atm

when 50 grams of KClOwhen 50 grams of KClO33 decomposes according to decomposes according to

the following equation?the following equation?

2KClO2KClO33 2KCl + 3O 2KCl + 3O22

First you find out how many moles of oxygen gas will form using stoichiometry

=

Then you use the Ideal gas law to get the volume. You cannot use the 22.4 L per 1 mole relationship because you are not at STP!!!!

(1.3atm)V = (0.61 moles) (0.0821 L.atm/K.mole) (298K)

V = 11.5 Liters

50 g KClO3 1 mole KClO3

122.55 g KClO3 2 moles KClO3

3 moles O20.61 moles O2