Dalton’s Law of Partial Pressures and Gas Law Stoichiometry Honors Chemistry

Dalton’s Law of Partial Pressures The pressure of each gas in a mixture is called the partial pressure of that gas. Dalton’s Law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each of the component gases. Ptotal = P1 + P2 + P3 + . . .

Ptotal = P1 + P2 + P3 + . . . Ptotal = 2.4 + 6.0 = 8.4 atm PH2 = 2.4 atm PHe = 6.0 atm Ptotal = 2.4 + 6.0 = 8.4 atm

If the first three containers are all put into the fourth, we can find the pressure in that container 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 equal to the mole fraction (X) of each gas times the total pressure Moles gas x Ptotal = Pgas Total Moles Mole fraction is like a percent

Sample Problem #1 A mixture of He, Ne and Ar gases have a total pressure of 790 mmHg. If there is 15% Ar, 60% He and 25% Ne, what is the partial pressure of 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 #2 The partial pressure of CO2 in a mixture of gases is 0.8 atm. If the total pressure is 1.05 atm, what is the mole fraction of CO2 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 Displacement A gas collected by water displacement is not pure, but always mixed with water vapor. Some molecules at the surface of water always evaporate and exert a 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 #3 50mls of CH4 was collected over water at 25oC and 748mmHg. What is the volume of the gas at STP? (The vapor pressure of water is 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 Stoichiometry What volume of oxygen will form at 25oC and 1.3atm when 50 grams of KClO3 decomposes according to the following equation? 2KClO3  2KCl + 3O2 First you find out how many moles of oxygen gas will form using stoichiometry = 0.61 moles O2 50 g KClO3 1 mole KClO3 3 moles O2 122.55 g KClO3 2 moles KClO3 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