Post on 27-Dec-2015
transcript
Chapter 11
Molecular Composition of Gases
Gay Lussac’s Law of Combining Volumes
N2
1 volume
+
+
3 H2
3 volumes
→
→
2 NH3
2 volumes
2
2
H
N
V 3=
V 13
2
NH
H
V 2=
V 33
2
NH
N
V 2=
V 1
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
Avogadro’s Law
Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
1 volume 1 volume 2 volumes
1 molecule 1 molecule 2 molecules
1 mol 1 mol 2 mol
hydrogen + chlorine → hydrogen chloride
Each molecule of hydrogen and each molecule of chlorine contains 2 atoms.
H2 + Cl2 → 2 HCl
Mole-Mass-Volume Mole-Mass-Volume RelationshipsRelationships
• 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.
Mole-Mass-Volume Relationships
Find the mass in grams of 2.80 L of carbon dioxide?
• (2.8 L CO2)(1 mol CO2/22.4 L CO2)(44 g CO2 /1 mol CO2 ) = 5.5 g CO2
Density of Gases
md =
vliters
grams
Density of Gases
md =
vdepends
on T and P
64.07 gd =
mol
1 mol22.4 L
g = 2.86
L
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
V PnTnRT
V = P
PV = nRT
Ideal Gas Equation
V PnTnRT
V = P
PV = nRT
atmospheres
V PnTnRT
V = P
PV = nRT
liters
V PnTnRT
V = P
PV = nRT
moles
V PnTnRT
V = P
PV = nRT
Kelvin
V PnTnRT
V = P
PV = nRT
Ideal Gas Constant
L-atm0.0821
mol-K
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
Step 2. Write and solve the ideal gas equation for the unknown.
Step 3. Substitute the given data into the equation and calculate.
A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is .987 atm. What is the balloon’s volume?
nRTV =
PPV = nRT
(0.987 atm)
(5.00 mol)V =
(0.0821 L×atm/mol×K)
(298 K)
= 124 L
Determination of Molecular Weights Using the Ideal Gas Equation
gmolar mass =
mol
gRTM =
PV
gmol =
molar mass
M = molar massg
n = mol = M
PV = nRTg
PV = RTM
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.
gRTM =
PV
V = 250 mL = 0.250 L g = 0.020 g
T = 305 K P = 0.045 atm
(0.020 g)M =
(0.082 L × atm/mol × K)
(305 K)
(0.250 L)
(0.045 atm)g
= 44 mol
Determination of Density Using the Ideal Gas Equation
• Density = mass/volume
gRTM =
PV
D = MP/ RT
The density of a gas was measured at 1.50 atm and 27 ºC and found to be 1.95 g/L.
Calculate the molar mass of the gas.
• (1.95g/L)(0.08206 L atm/K mol)(300K)
1.50 atm
M = dRT P
= 32.0 g/mol
• All calculations are done at STP.
• Gases are assumed to behave as ideal gases.
• A gas not at STP is converted to STP.
Gas Stoichiometry
deals with the quantitative relationships among reactants and products in a chemical reaction.
Gas Stoichiometry
Primary conversions involved in stoichiometry.
• 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?
2
3
3 mol O2 mol KClO
• Step 3. Calculate the moles of O2, using the mole-ratio method.
3(0.500 mol KClO )
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
• Step 4. Convert moles of O2 to liters of O2
2 = 0.750 mol O
2(0.750 mol O )22.4 L1 mol
2= 16.8 L O
2 KClO3 2KCl + 3 O2
The problem can also be solved in one continuous calculation.
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
2 KClO3 2KCl + 3 O2
3(0.500 mol KClO ) 2
3
3 mol O2 mol KClO
22.4 L1 mol
2= 16.8 L O
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?
50.0 g Al1 mol Al
26.98 g Al
23 mol H2 mol Al
2 = 2.78 mol H
• Convert oC to K: 30.oC + 273 = 303 K
2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g)
Step 2 Calculate liters of 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?
What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
PV = nRT
nRTV =
P
• Solve the ideal gas equation for V
(0.921 atm)
2(2.78 mol H )V =
(0.0821 L-atm)
(303 K)
(mol-K) 2 = 75.1 L H
H2(g) + Cl2(g) 2HCl(g)
1 mol H2 1 mol Cl2 2 mol HCl
22.4 LSTP
22.4 LSTP
2 x 22.4 LSTP
1 volume 1 volume 2 volumes
For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.
What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed?
N2(g) + 3H2(g) 2NH3(g)
2600. ml H 2
2
1 vol N3 vol H
2= 200. mL N
2600. ml H 3
2
2 vol NH3 vol H
3= 400. mL NH
Mass- volume problem
Graham’s Law of Effusion
• Effusion – the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.
• Rate (like diffusion) depends on the relative velocities of gas molecules.
• Lighter molecules move faster than heavier molecules at the same temperature.
Graham’s Law of Effusion
• The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.
• Rate of effusion of A = MB
Rate of effusion of B MA
Calculate the ratio of the effusion rates of H2 & UF6, a gas used in the enrichment process
to produce fuel for nuclear reactors.•Rate of effusion of H2 Rate of effusion of UF
•Square root of molar mass of UF6
Square root of molar mass of H2
Square root of 352.02/2.016 =
13.213.2