Download Chemistry Chapter 11

Chemistry Chapter 11
Molecular Composition of Gases
Volume and mass
• Gay-Lussac examined gas volume in reactions
• Noted: 2 L H2 and 1 L O2 can form 2 L water
vapor
• 2:1:2 volume relationship of H:O: water
• Simple definite proportions hold true for other
gases in reactions
• this lead to ….
Gay-Lussac’s Law of
Combining Gas Volumes
• At constant temperature and pressure the
volumes of gaseous reactants and products
can be expressed as ratios of small whole
numbers
Avogadro (again)
• Combining volumes seemed to challenge the
indivisibility of the atom…
• Avogadro posited that some molecules might
contain more than one atom (ex. O2 explains
the 2:1:2 H:O:water ratio)
• Avogadro's Law: Equal volumes of gases at the
same temperature and pressure contain equal
numbers of molecules
Implications of Avogadro
• At the same temperature and pressure the
volume of a gas varies directly with the number
of molecules
• Avogadro believed that some elements must exist
in diatomic form (H2, O2, N2)
Avogadro continued
• H, O and water illustrate this well
• 2 volumes H2 + 1 volume O2 = 2 volumes H2O
Leads to the balanced equation…
2H2 + O2  2H2O
• Which confirms the diatomic molecule hunch
nicely!
Avogadro in algebra
• Gas volume is directly proportional to the
amount of gas (number of particles) at a given
temperature and pressure give us:
V = kn
• Where
– V is volume
– n is the amount of gas (in moles)
– k is a constant
Molar Volumes
• One mole of gas at STP will occupy 22.4 L
• 1 mole/ 22.4 L of ________gas can be used as
a conversion factor to find number of
particles, mass, or volume of a gas at STP
Practice!
• Problems 1-3 page 337
11-2
Ideal Gas Law
• A mathematical relationship among pressure,
volume, temperature and number of moles
• To derive (see p. 341)
• Ideal gas law:
• V= nRT/P OR PV = nRT
• V is volume, P is pressure, T is temperature, n
is number of moles, and R is a constant
The Ideal Gas Constant
• R= (1 atm) (22.4 L)/(1 mol) (273.15 K)
Or
• R= 0.08205784 L x atm / mol x K
• (round to .0821)
• USE ONLY when units are appropriate!
• For any other units see chart on p. 342
• Practice Problems! p. 345
Finding Molar Mass or Density
• Use V= nRT/P but remember that n (number of
moles) is equal to mass (m) /molar mass (M)
Substituting gives:
PV= mRT/M
or
M= mRT/PV
• Density is just mass (m) per unit volume (V)
Substituting gives:
M= DRT/P or D= MP/RT
Practice problems! 1-4 page 346
11-3
Stoichiometry of gases
• Volume ratios of gases in reactions can be
used exactly as mole ratios are in standard
mass-mole, mass-mass, mole mole etc.
problems
• Practice! 1-2 pg. 348
More Stoich!
• When given a volume for a reactant and a
mass for a reactant…
• Go through moles
• Need conditions (temp, pressure, etc) for each
gas
• Ideal gas law works well for this
• Practice problems! #1-2, p.349 # 1-2 p.350
Effusion and Diffusion
• Rates of either can be calculated!
• Remember! KE= ½ m v2
• And… for any two gases (A & B, lets say) at the same
temp KEA = KEB so…
½ MA vA2 = ½ MBvB2
• Where M is molar mass and v is molecular velocity
• You can multiply by 2 to clean up and get
MA vA2 = MBvB2
Effusion and diffusion (con’t)
•
•
•
•
Recall: MA vA2 = MB vB2
Rearrange: vA2 / vB2 = MB / MA
Take square roots: vA / vB = √MB / √MA
Because rate of effusion is directly
proportional to molecular velocity we can say
that:
rate of effusionA / rate of effusionB = √MB / √MA
Grahams Law of effusion
• Rates of effusion of gases at the same
temperature and pressure are inversely
proportional to the square roots of their molar
masses
• So? Density varies directly with molar mass
so…
rate of effusionA / rate of effusionB = √DB / √DA
(Where D is density)
Grahams…
• Can be used to find density or molar mass of
gases effusing.
• Practice problems! #1-3 p.355
That’s all folks!