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Molecular Composition of Gases. Chapter 11. Gay-Lussac’s law of combining volumes of gases. At constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers. Example.

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molecular composition of gases

Molecular Composition of Gases

Chapter 11

Chemistry Chapter 11

gay lussac s law of combining volumes of gases
Gay-Lussac’s law of combining volumes of gases
  • At constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Chemistry Chapter 11

example
Example
  • When 2 L of hydrogen react with 1 L of oxygen 2 L of water vapor are produced.
  • Write the balanced chemical equation:

Chemistry Chapter 11

you try
You try
  • When 1 L of hydrogen gas reacts with 1 L of chlorine gas, 2 L of hydrogen chloride gas are produced.
  • Write the balanced chemical equation:

Chemistry Chapter 11

avogadro s law
Avogadro's Law
  • Equal volumes of gases at the same pressure and temperature contain the same number of molecules
    • Atoms can’t split  diatomic molecules
  • Gas volume is proportional to the number of molecules

Chemistry Chapter 11

molar volume
Molar Volume
  • 1 mole of any gas contains 6.022 x 1023 molecules.
  • According to Avogadro’s law, 1 mole of any gas must have the same volume.
  • Standard molar volume: volume of 1 mole of any gas at STP
    • 22.4 L

Chemistry Chapter 11

example1
Example
  • You are planning an experiment that requires 0.0580 mol of nitrogen monoxide gas. What volume in liters is occupied by this gas at STP?
  • 1.30 L NO

Chemistry Chapter 11

you try1
You try
  • A chemical reaction produces 2.56 L of oxygen gas at STP. How many moles of oxygen are in this sample?
  • 0.114 mol O2

Chemistry Chapter 11

example2
Example
  • Suppose you need 4.22 g of chlorine gas. What volume at STP would you need to use?
  • 1.33 L Cl2

Chemistry Chapter 11

you try2
You try
  • What is the mass of 1.33 x 104 mL of oxygen gas at STP?
  • 19.0 g O2

Chemistry Chapter 11

discuss
Discuss
  • Explain Gay-Lussac’s law of combining volumes
  • State Avogadro’s law and explain its significance.

Chemistry Chapter 11

review
Review
  • Boyles Law:
  • Charles Law:
  • Avogadro’s Law:

Chemistry Chapter 11

slide13
Math
  • A quantity that is proportional to each of several quantities is also proportional to their product. Therefore:

Chemistry Chapter 11

more math
More math
  • Convert a proportionality
  • to an equality by multiplying by a constant

Chemistry Chapter 11

therefore
Therefore
  • We can covert
  • to

Chemistry Chapter 11

more neatly
More neatly

Chemistry Chapter 11

this means
This means….
  • The volume of a gas varies directly with the number of moles and the temperature in Kelvin.
  • The volume varies indirectly with pressure.

Chemistry Chapter 11

what if
What if…
  • n and T are constant?
    • nRT is a constant, k
      • Boyle’s Law
  • n and P are constant?
    • nR/P is a constant, k
      • Charles’s Law

Chemistry Chapter 11

what if1
What if…
  • P and T are constant?
    • RT/P is a constant, k
    • Avogadro’s law

Chemistry Chapter 11

the ideal gas constant
The ideal gas constant
  • R
  • Value depends on units
  • SI units:

Chemistry Chapter 11

other units
Other units

Chemistry Chapter 11

solving ideal gas problems
Solving ideal gas problems
  • Make sure the R you use matches the units you have.
  • Make sure all your units cancel out correctly.

Chemistry Chapter 11

example3
Example
  • A 2.07 L cylinder contains 2.88 mol of helium gas at 22 °C. What is the pressure in atmospheres of the gas in the cylinder?
  • 33.7 atm

Chemistry Chapter 11

you try3
You try
  • A tank of hydrogen gas has a volume of 22.9 L and holds 14.0 mol of the gas at 12 °C. What is the reading on the pressure gauge in atmospheres?
  • 14.3 atm

Chemistry Chapter 11

example4
Example
  • A reaction yields 0.00856 mol of oxygen gas. What volume in mL will the gas occupy if it is collected at 43 °C and 0.926 atm pressure?
  • 240. mL

Chemistry Chapter 11

you try4
You try
  • A researcher collects 9.09 x 10-3 mol of an unknown gas by water displacement at a temperature of 16 °C and 0.873 atm pressure (after the partial pressure of the water vapor has been subtracted). What volume of gas in mL does the researcher have?
  • 247 mL

Chemistry Chapter 11

finding mass
Finding mass
  • Number of moles (n) equals mass (m) divided by molar mass (M).

Chemistry Chapter 11

example5
Example
  • What mass of ethene gas, C2H4, is contained in a 15.0 L tank that has a pressure of 4.40 atm at a temperature of 305 K?
  • 74.0 g

Chemistry Chapter 11

you try5
You try
  • NH3 gas is pumped into the reservoir of a refrigeration unit at a pressure of 4.45 atm. The capacity of the reservoir is 19.4 L. The temperature is 24 °C. What is the mass of the gas in kg?
  • 6.03 x 10-2 kg

Chemistry Chapter 11

example6
Example
  • A chemist determines the mass of a sample of gas to be 3.17 g. Its volume is 942 mL at a temperature of 14 °C and a pressure of 1.09 atm. What is the molar mass of the gas?
  • 72.7 g/mol

Chemistry Chapter 11

density
Density

Chemistry Chapter 11

you try6
You try
  • The density of dry air at sea level (1 atm) is 1.225 g/L at 15 °C. What is the average molar mass of the air?
  • 29.0 g/mol

Chemistry Chapter 11

stoichiometry
Stoichiometry
  • Involves mass relationships between reactants and products in a chemical reaction
  • For gases, the coefficients in the balanced chemical equation show volume ratios as well as mole ratios
    • All volumes must be measured at the same temperature and pressure

Chemistry Chapter 11

volume volume calculations
Volume-Volume calculations
  • From volume of one gas to volume of another gas
  • Use volume ratios just like mole ratios in chapter 9

Chemistry Chapter 11

example7
Example
  • Xenon gas reacts with fluorine gas to produce the compound xenon hexafluoride, XeF6. Write the balanced equation for this reaction.
    • Xe(g) + 3F2(g)  XeF6(g)
  • If a researcher needs 3.14 L of XeF6 for an experiment, what volumes of xenon and fluorine should be reacted?
    • 3.14 L of Xe and 9.42 L of F2

Chemistry Chapter 11

example8
Example
  • Nitric acid can be produced by the reaction of gaseous nitrogen dioxide with water.3NO2(g) + H2O(l)  2HNO3(l) + NO(g)
  • If 708 L of NO2 gas react with water, what volume of NO gas will be produced?
  • 236 L

Chemistry Chapter 11

you try7
You try
  • What volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?
  • 9.10 L

Chemistry Chapter 11

you try8
You try
  • At STP, what volume of oxygen gas is needed to react completely with 2.79 x 10-2 mol of carbon monoxide gas, CO, to form gaseous carbon dioxide?
  • 0.312 L

Chemistry Chapter 11

you try9
You try
  • Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following equation:3F2(g) + 3H2O(l)  6HF(g) + O3(g)
  • What volumes of O3 and HF gas would be produced by the complete reaction of 3.60 x 104 mL of fluorine gas?
  • 1.20 x 104 mL O3 and 7.20 x 104 mL HF

Chemistry Chapter 11

you try10
You try
  • Ammonia is oxidized to make nitrogen monoxide and water4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
  • At STP, what volume of oxygen will be used in a reaction of 125 mol of NH3? What volume of NO will be produced?
  • 3.50 x 103 L O2 and 2.80 x 103 L NO

Chemistry Chapter 11

volume mass and mass volume
Volume-mass and mass-volume
  • Converting from volume to mass or from mass to volume
  • Must convert to moles in the middle
  • Ideal gas law may be useful for finding standard conditions

Chemistry Chapter 11

example9
Example
  • Aluminum granules are a component of some drain cleaners because they react with sodium hydroxide to release both heat and gas bubbles, which help clear the drain clog. The reaction is:2NaOH(aq) + 2Al(s) + 6H2O (l)  2NaAl(OH)4(aq) + 3 H2(g)
  • What mass of aluminum would be needed to produce 4.00 L of hydrogen gas at STP?
  • 3.21 g

Chemistry Chapter 11

example10
Example
  • Air bags in cars are inflated by the sudden decomposition of sodium azide, NaN3 by the following reaction:2NaN3(s)  3N2(g) + 2Na(s)
  • What volume of N2 gas, measured at 1.30 atm and 87 °C, would be produced by the reaction of 70.0 g of NaN3?
  • 36.6 L

Chemistry Chapter 11

you try11
You try
  • What volume of chlorine gas at 38°C and 1.63 atm is needed to react completely with 10.4 g of sodium to form NaCl?
  • 3.54 L Cl2

Chemistry Chapter 11

example11
Example
  • A sample of ethanol burns in O2 to form CO2 and H2O according to the following reaction.C2H5OH + 3O2 2CO2 + 3H2O
  • If the combustion uses 55.8 mL of oxygen measured at 2.26 atm and 40.°C, what volume of CO2 is produced when measured at STP?
  • 73.3 mL CO2

Chemistry Chapter 11

you try12
You try
  • Dinitrogen pentoxide decomposes into nitrogen dioxide and oxygen. If 5.00 L of N2O5 reacts at STP, what volume of NO2 is produced when measured at 64.5 °C and 1.76 atm?
  • 7.02 L NO2

Chemistry Chapter 11

review1
Review
  • Diffusion: the gradual mixing of gases due to their random motion
  • Effusion: gases in a container randomly pass through a tiny opening in the container

Chemistry Chapter 11

rate of effusion
Rate of effusion
  • Depends on relative velocities of gas molecules.
  • Velocity varies inversely with mass
    • Lighter particles move faster

Chemistry Chapter 11

kinetic energy
Kinetic energy
  • Depends only on temperature
  • Equals
  • For two gases, A and B, at the same temperature
  • Each M stands for molar mass

Chemistry Chapter 11

algebra time
Algebra time

Chemistry Chapter 11

rate of effusion1
Rate of effusion
  • Depends on relative velocities of gas molecules.

Chemistry Chapter 11

graham s law of effusion
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.

Chemistry Chapter 11

graham s law
Graham’s law
  • Graham experimented with densities of gases, not molar masses.
  • Density and molar mass are directly proportional
  • So we can replace molar mass with density in the equation

Chemistry Chapter 11

use of graham s law
Use of Graham’s law
  • Finding the molar mass
    • Compare rates of effusion of a gas with known molar mass and a gas with unknown molar mass
    • Use Graham’s law equation to solve for the unknown M
  • Used to separate isotopes of uranium

Chemistry Chapter 11

example12
Example
  • Compare the rates of effusion of hydrogen and helium at the same temperature and pressure.
  • Hydrogen diffuses about 1.41 times faster

Chemistry Chapter 11

example13
Example
  • Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element at the same conditions. Estimate the other element’s molar mass and determine its probable identity.
  • 81 g/mol, krypton

Chemistry Chapter 11

you try13
You try
  • Estimate the molar mass of a gas that effuses at 1.6 times the effusion rate of carbon dioxide.
  • 17 g/mol

Chemistry Chapter 11