350 likes | 649 Views
11-1 Volume-Mass Relationships of Gases Avogadro’s Law, STP, molar volume 11-2 The Ideal Gas Law PV=nRT (R = 0.0821 (L∙atm)/(mol∙K)), finding molar mass and density 11-3 Stoichiometry of Gases 11-4 Effusion and Diffusion Graham’s law (√M B )/(√M A ) (assuming constant T and P).
E N D
11-1 Volume-Mass Relationships of GasesAvogadro’s Law, STP, molar volume 11-2 The Ideal Gas LawPV=nRT (R = 0.0821 (L∙atm)/(mol∙K)), finding molar mass and density 11-3 Stoichiometry of Gases 11-4 Effusion and DiffusionGraham’s law (√MB)/(√MA) (assuming constant T and P) Chapter 11 Molecular Composition of Gases
Avogadro’s Law (principle) Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
hydrogen + chlorine → hydrogen chloride 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol Each molecule of hydrogen and each molecule of chlorine contains 2 atoms. H2 + Cl2→ 2 HCl
Mole-Mass-Volume Relationships • Volume of one mole of any gas at STP = 22.4 L. • 22.4 L at STP is known as the molar volume of any gas.
Find the mass in grams of 2.8 L of carbon dioxide at STP? • 2.8 L CO2 x1 mol CO2 x 44 g CO2 22.4 L CO2 1 mol CO2 • = 5.5 g CO2
Density of Gases grams liters
Density of Gases depends on T and P
The molar mass of SO2 is 64.07 g/mol. Determine the density of SO2 at STP. 1 mole of any gas occupies 22.4 L at STP
nT V a P Ideal Gas Equation
atmospheres nT V a P
liters nT V a P
moles nT V a P
Kelvin nT V a P
nT Ideal Gas Constant V a P
A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 0.987 atm. What is the balloon’s volume? Step 1. Organize the given information. Convert temperature to kelvins. K = oC + 273 K = 25oC + 273 = 298K
A balloon filled with 5.00 moles of helium gas is at a temperature of 25.0oC. The atmospheric pressure is .987 atm. What is the balloon’s volume? Step 2. Write and solve the ideal gas equation for the unknown. Step 3. Substitute the given data into the equation and calculate.
Determination of Molecular Weights Using the Ideal Gas Equation
Calculate the molar mass of an unknown gas, if 0.020 g occupies 250. mL at a temperature of 305 K and a pressure of 0.045 atm. V = 250. mL = 0.250 L g = 0.020 g P = 0.045 atm T = 305 K
Determination of Density Using the Ideal Gas Equation • Density = mass/volume D = MP/ RT
The density of a gas was found to be 1.95 g/L at 1.50 atm and 27.0 ºC. Calculate the molar mass of the gas. • (1.95g/L)(0.0821 L atm/K mol)(300K) 1.50 atm M = dRT P = 32.0 g/mol
Gas Stoichiometry: deals with the quantitative relationships among reactants and products in a chemical reaction. Gases are assumed to behave as ideal gases
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? • Step 1 Write the balanced equation 2 KClO3 2 KCl + 3 O2 • Step 2 The starting amount is 0.500 mol KClO3. The conversion is moles KClO3 moles O2 liters O2
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? • Step 3. Calculate the moles of O2, using the mole-ratio method. 2 KClO3 2KCl + 3 O2 • Step 4. Convert moles of O2 to liters of O2
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? The problem can also be solved in one continuous calculation. 2 KClO3 2KCl + 3 O2
What volume of hydrogen, collected at 30.oC and 0.921 atm, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g) Step 1 Calculate moles of H2. grams Al moles Al moles H2
What volume of hydrogen, collected at 30.oC and 0.921 atm, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g) Step 2 Calculate liters of H2. • Convert oC to K: 30.oC + 273 = 303 K
What volume of hydrogen, collected at 30.0oC and .921 atm, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? • Solve the ideal gas equation for V PV = nRT
Graham’s Law • Diffusion • Spreading of gas molecules throughout a container until evenly distributed. • Effusion • Passing of gas molecules through a tiny opening in a container.
Graham’s Law • Rate of diffusion/effusion • Rate depends on speed & pressure of molecules. • At the same temp (ave KE) heavier molec.move more slowly. • Larger m smaller v KE = ½mv2
Graham’s Law • Rate of diffusion of a gas is inversely related to the square root of its molar mass (same temp & pressure.). Unit: amount/time • The equation shows the ratio of Gas A’s speed(v A) to B’s (v B).
Graham’s Law • Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Krdiffuses 1.381 times faster than Br2.
Graham’s Law • A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as “Gas A”.
Graham’s Law • An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign.
PRACTICE! • Work the following problems in your book. Check your work using the answers provided in the margin. • p. 324 • SAMPLE PROBLEM 10-6 • PRACTICE 1 & 2 • p. 355 • SAMPLE PROBLEM 11-10 • PRACTICE 1, 2, & 3