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CHM 101 GENERAL CHEMISTRY
FALL QUARTER 2008
Section 2
Lecture Notes – 11/5/2008
(last revised: 11/5/08, 4:00 PM)
5.4
Gas Stoichiometry: Now that we have the ideal gas law in our
toolkit, we are equipped to handle problems in stoichiometry
that involve measurements of the temperatures, pressures,
volumes, and numbers of moles of gaseous reactants and
products.
•
Sample Exercise 5.11 (p. 191): A sample of nitrogen gas (N2)
has a volume of 1.75 L at STP (0 °C & 1 atm). How many moles
of N2 do we have?
o We could use the ideal gas law to solve this problem, and
that would be a perfectly correct approach. But we could
also start from our knowledge that one mole of gas has a
volume of 22.42 L at STP. Here we use the second
approach.
nN 2 =
1.75 L
= 7.81× 10−2 mol
22.42 L
o (If you don’t recall the figure, 22.42 L, you can always use
the ideal gas law directly for a problem like this.)
•
Sample Exercise 5.12 (pp. 191-2): Quicklime (CaO) is
produced by heating calcium carbonate (CaCO3) to drive off
carbon dioxide (CO2). The reaction is written:
CaCO3 (s) ——> CaO (s) + CO2 (g)
Calculate the volume of CO2 at STP produced by the reaction
(decomposition) of 152 g of CaCO3.
o The equation clearly indicates that one mole of CO2 is
generated for each mole of CaCO3 that decomposes. We
can thus calculate the number of moles of CO2 we produce
from the mass and molar mass of CaCO3 that decomposes:
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nCO2 =
mCaCO3
MM CaCO3
=
152 g
= 1.52 mol
100.09 g/mol
o Now we can convert 1.52 moles of CO2 to volume at STP.
This time we’ll use the ideal gas law, even though we could
work from a molar volume of 22.42 L.
V=
•
nCO2 RT
P
=
(1.52 mol)(0.08206 KL⋅⋅atm
mol )(273 K)
= 34.1 L
(1 atm)
Sample Exercise 5.13 (pp. 192-3): A sample of methane gas
(CH4) with a volume of 2.80 L at 25 °C and 1.65 atm is mixed
with a sample of oxygen gas with a volume of 35.0 L at 31 °C
and 1.25 atm. The mixture is ignited to produce carbon dioxide
(CO2) and water (H2O). Calculate the volume of carbon dioxide
(CO2) that forms at a pressure of 2.50 atm and a temperature of
125 °C.
o This is a limiting reactant problem where all the data are
values of P, V, and T, and where none of the conditions
match STP. We must use the ideal gas law all the way. We
start by writing (and balancing) the equation for the
reaction:
CH4 (g) + 2O2 (g) ——> CO2 (g) + 2H2O (g)
o The number of moles of CH4 is:
nCH4 =
PV
(1.65 atm)(2.80 L)
=
= 0.189 mol
RT (0.08206 KL⋅⋅atm
+
)(273
25
K)
mol
o The number of moles of O2 is:
nO2 =
PV
(1.25 atm)(35.0 L)
=
= 1.75 mol
RT (0.08206 KL⋅⋅atm
)(273
31
K)
+
mol
o Since 0.378 moles of O2 will react with the entire 0.189
moles of CH4, we see that CH4 is the limiting reactant, and
there will be 0.189 moles of CO2 produced. We finish by
calculating the volume of CO2 when it is measured at 2.50
atm and 125 °C:
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VCO2
5.5
nRT (0.189 mol)(0.08206 KL⋅⋅atm
mol )(273 + 125 K)
=
=
= 2.47 L
P
(2.50 atm)
Dalton’s Law of Partial Pressures: You will recall that we
”derived” Dalton’s law last time from the postulates of the kinetic
molecular theory (KMT) of gases. We argued that since
individual gas particles do not interact with each other, the
particles of different kinds of gases would behave as if the other
gas were not there. Hence each kind of gas would exert its own
pressure on the walls of the container, and the total pressure on
the container would be the sum of the individual pressures.
Dalton had no knowledge of the KMT. He obtained Dalton’s law
from his careful experimentation on mixtures of gases.
•
The Law: Dalton’s law can be stated: For a mixture of gases in
a container, the total pressure exerted is the sum of the
pressures each gas would exert if it were alone in the container.
Mathematically, the law can be expressed:
Ptotal = P1 + P2 + P3 +…
Here P1 stands for the partial pressure of gas 1, P2 stands for the
partial pressure of gas 2 and so on, and Ptotal is the total
pressure of the mixture. We can calculate these partial pressures
from the ideal gas law:
P1 =
n1 RT
V
,
P2 =
n2 RT
V
,
P3 =
n3 RT
V
,
…
Then the total pressure of the mixture is:
Ptotal = P1 + P2 + P3 +… =
n1 RT n2 RT n3 RT
+
+
+…
V
V
V
Ptotal = (n1 + n2 + n3 +…)
Ptotal = ntotal
RT
V
RT
V
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Here, ntotal is the sum of the numbers of moles of the individual
gases. This result indicates that the total pressure of a mixture
of gases depends on the total numbers of moles of gas particles
but not on the identities of the gases. This is illustrated in Figure
5.12:
•
Sample Exercise 5.15 (p. 195): A 5.0 L scuba tank is charged
with 46 L of helium gas (He) and 12 L of oxygen gas (O2), both
measured at 25 °C and 1.0 atm. Calculate the partial pressure of
each gas and the total pressure in the tank at 25 °C.
First we calculate the number of moles of each gas:
nHe =
nO2 =
PV
(1.0 atm)(46 L)
=
= 1.9 mol
L atm
)(298
K)
RT (0.08206 mol
K
PV
(1.0 atm)(12 L)
=
= 0.49 mol
L atm
RT (0.08206 mol
K )(298 K)
Next, we calculate the partial pressures of each gas in the 5.0 L
tank (at 25 °C):
L atm
nHe RT (1.9 mol)(0.08206 mol
K )(298 K)
PHe =
=
= 9.3 atm
V
(5.0 L)
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PO2 =
nO2 RT
V
L atm
(0.49 mol)(0.08206 mol
K )(298 K)
=
= 2.4 atm
(5.0 L)
We finish by adding the two partial pressures to obtain the total
pressure:
Ptotal = PHe + PO2 = 9.3 atm + 2.4 atm = 11.7 atm
Notice that we could have avoided a bunch of arithmetic. Since
the numbers of moles of the two gases remain constant, and the
temperature remains at 25 °C, we could have applied Boyle’s
law:
P1V 1 = P 0V 0
P 0V 0
P =
V1
1
The superscript, 0, refers to the starting conditions, the
superscript, 1, refers to the conditions in the scuba tank. Here
are the direct calculations of the partial pressures:
1
He
P
P 0VHe 0 (1.0 atm)(46 L)
=
=
= 9.3 atm
V1
(5.0 L)
PO2 =
1
•
P 0VO2 0
V1
=
(1.0 atm)(12 L)
= 2.4 atm
(5.0 L)
Collecting a Gas Over Water: This is an important procedure
you will probably apply on numerous occasions in chemistry lab.
If you run a reaction that generates a gas that you wish to
collect, you can pipe it through a collection tube and into a
collection vessel, an inverted container initially filled with water,
as illustrated in Figure 5.13, where the mixture of potassium
chlorate (KClO3) and manganese (IV) oxide (MnO2) in the test
tube will generate oxygen gas (O2) when heated:
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Before the reaction starts to run, the collection vessel remains
full of water, held in place by the pressure of the atmosphere
bearing down on the surface of the water in the large pan.
When heat is applied to the test tube, the KClO3 decomposes to
produce O2, which flows through the glass pipe and into the
collection vessel, where it displaces liquid water and creates a
vapor space.
Eventually, when all the KClO3 has decomposed and no more O2
is being produced, the volume of the vapor space in the
collection vessel can be measured. Before this measurement can
be used in a stoichiometric calculation, one major correction
needs to be applied. One must compensate for the water vapor
present in the vapor space and in equilibrium with the liquid
water in the system.
Where does the water vapor come from? It comes from
evaporation of liquid water in the system. This water will also
condense out of the vapor space at a rate proportional to the
partial pressure of water vapor in the vapor space. When the
rate of condensation matches the rate of evaporation, an
equilibrium is established, and the vapor pressure of water in the
vapor space remains at a steady value, called the equilibrium
vapor pressure. Extensive tables of equilibrium vapor pressures
versus temperature have been compiled, and one can look up
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the value that needs to be applied in order to correct the
measured volume for the volume of water vapor it contains.
One also needs to measure the pressure in the vapor space, in
order to use the volume for any gas law calculations. Notice,
however, that the apparatus in effect is a manometer, and if one
adjusts the position of the collection vessel so that the water
level inside it matches the water level in the large pan, then the
pressure in the vapor space is the same as the measured
barometric pressure.
Another factor needs to be considered. some of the gas in the
vapor space of the collection vessel is actually air that was
present in the test tube and the collection pipe before the start
of the reaction. However, this volume of air is equal to the
volume of oxygen from the reaction that now remains behind in
the test tube and collection pipe. Thus there is no need to
correct for it.
•
Sample Exercise 5.18 (pp. 198-9): Now let us apply the
foregoing discussion to a stoichiometric problem.
A sample of potassium chlorate (KClO3) is heated in an
apparatus like Figure 5.13. It decomposes according to the
following reaction:
2KClO3 (s) ——> 2KCl (s) + 3O2 (g)
(The MnO2 shown in the diagram acts as a catalyst, but does not
take part in the net reaction.) The oxygen was displaced over
water at 22 °C at a total pressure of 754 torr. The volume of gas
collected was 0.650 L, and the vapor pressure of water at 22 °C
is known to be 21 torr. Calculate the partial pressure of O2 in the
collected gas, and calculate the mass of KClO3 that was
decomposed.
o First, we calculate the partial pressure of O2, using Dalton’s
law:
Ptotal = PO2 + PH2O
PO2 = Ptotal − PH2O = 754 torr − 21 torr = 733 torr
PO2 =
733 torr
= 0.964 atm
760 torr/atm
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o Next, we find the number of moles of O2 by using the ideal
gas law:
nO2 =
PO2V
RT
=
(0.964 atm)(0.650 L)
= 2.59 × 10−2 mol
L⋅atm
(0.08206 K⋅mol )(273 + 22 K)
o Next, we consider the stoichiometry of the reaction to
calculate the number of moles of KClO3:
1
1
nKClO3 = nO2
2
3
nKClO3 =
2
2
nO2 = × 2.59 ×10−2 mol = 1.73 ×10−2 mol
3
3
o Finally, we convert the number of moles of KClO3 to mass:
mKClO3 = nKClO3 × MM KClO3 = (1.73 ×10−2 mol)(122.6 g/mol) = 2.12 g
•
The Mole Fraction: We define the mole fraction as the ratio of
the number of moles of a given component in a mixture to the
total number of moles of all components in the mixture. We use
the Greek letter, chi (χ), as a symbol for mole fraction. Thus, the
mole fraction of component 1 in a mixture is:
χ1 =
n1
n1
=
ntotal n1 + n2 + n3 +…
Mole fractions can be defined for all components of any
homogeneous mixture, whether the state of the mixture is a
solid, liquid or gas. In the special case of a mixture of ideal
gases (where Dalton’s law is true) we can write:
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χ1 =
n1
P1 (V / RT )
=
n1 + n2 + n3 +… P1 (V / RT ) + P2 (V / RT ) + P3 (V / RT ) +…
χ1 =
P1 (V / RT )
P1
=
( P1 + P2 + P3 +…)(V / RT ) ( P1 + P2 + P3 +…)
χ1 =
•
P1
Ptotal
Sample Exercise 5.16 (p. 197): The partial pressure of oxygen
(O2) is found to be 156 torr in an atmosphere with a total
pressure of 743 torr. What is the mole fraction of the oxygen in
this atmosphere?
o We can use the equation:
χO =
2
PO2
Ptotal
o All we need to do is plug in our data:
χO =
2
PO2
Ptotal
=
(156 torr)
= 0.210
(743 torr)
o Notice that the mole fraction does not have any units; it is
dimensionless. Also notice that we can use any convenient
pressure unit, so long as the partial and the total pressure
are expressed in the same unit of pressure.
•
Sample Exercise 5.17 (p. 198): The mole fraction of nitrogen
(N2) in air is 0.7808. What is the partial pressure of in air with
an atmospheric pressure of 760. torr?
o We can use the equation:
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χN =
2
PN 2
Ptotal
o Only this time we need to rearrange it to:
PN 2 = χ N 2 × Ptotal
o Now we can plug in our data:
PN 2 = χ N 2 × Ptotal = 0.7808 × 760. torr = 593. torr
5.10 Chemistry in the Atmosphere
•
Composition of the Atmosphere: The composition of the
atmosphere (actually dry, clean air near sea level) is shown in
Table 5.4 from the text.
Nitrogen is the most abundant component, followed by oxygen.
But a gas does not need to be very abundant to be important.
Consider carbon dioxide. Its mole fraction seems negligible, but
it is an absolutely necessary reactant for photosynthesis. On the
other hand, its concentration in the atmosphere has roughly
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doubled since the beginning of the Industrial Revolution, and
most scientists consider this increase to be a major cause of
global warming.
•
Photochemical Smog: Clean air doesn’t necessarily stay that
way. Our industrial civilization dumps significant quantities of
reactive chemicals into our atmosphere. For example, internal
combustion engines convert fuel into mechanical energy, but the
combustion process also generates oxides of nitrogen and leaves
unburned residues of fuel, both of which end up released into the
atmosphere where they undergo a series of chemical
transformations to produce what is collectively known as smog.
See Figure 5.31.
•
Electricity from Coal: The most abundant fossil fuel we have in
the USA is coal, and much of our electric power is generated in
coal-burning power plants. The primary component of coal is
carbon, but coal always contains various amounts of
contaminants. (I have heard it said that every element in the
periodic table can be found in coal. I have also heard that a
typical coal-fired power plant releases more radioactivity than a
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nuclear generating plant of the same power generating
capacity.)
One coal contaminant of special concern is sulfur. The sulfur in
coal burns to form sulfur dioxide:
S (s, in coal) + O2 (g) ——> SO2 (g)
And the sulfur dioxide is released to the atmosphere along with
carbon dioxide.
•
Acid Rain: We have just seen that internal combustion engines
tend to produce oxides of nitrogen (sometimes abbreviated NOx)
and that sulfur dioxide is a by-product of power generation from
coal. But where does acid rain come from? And how do NOx and
SO2 fit in?
In fact, rainwater is naturally acidic, but its acidity is very small.
The source is atmospheric CO2. When raindrops form inside
clouds, CO2 from the air tends to dissolve in them and ionize:
H2O (l) + CO2 (g) ——> H+ (aq) + HCO3— (aq)
But, the acidity of acid rain is very small, since carbonic acid
(aqueous carbon dioxide) is a very weak acid.
If, however, raindrops form in air that has been contaminated
with oxides of sulfur and nitrogen, the acidity of the rain is much
higher. Nitrogen dioxide (a principal component of NOx) reacts in
water to form nitrous and nitric acids:
2NO2 (g) + H2O (l) ——> HNO2 (aq) + HNO3 (aq)
While nitrous acid is weak, albeit much stronger than carbonic
acid, nitric acid is one the strong mineral acids that ionizes
completely in aqueous solution.
Sulfur dioxide will dissolve in water to produce sulfurous acid, a
weak acid, but like nitrous acid, much stronger than carbonic
acid:
SO2 (g) + H2O (l) ——> H2SO3 (aq)
However, airborne sulfur dioxide can and will react with
atmospheric oxygen to produce sulfur trioxide:
2SO2 (g) + O2 (g) ——> 2SO3 (g)
And when sulfur trioxide reacts with water, it generates sulfuric
acid, another strong, mineral acid that totally ionizes in water:
SO3 (g) + H2O (l) ——> H2SO4 (aq) ——> H+ (aq) + HSO4— (aq)
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•
Some Consequences of Acid Rain: When acid rain falls to
earth, it can cause considerable damage.
o It can acidify lakes to the point where some forms of
aquatic life can no longer survive.
o It can neutralize the lime content of soil so as to interfere
with plant growth.
o An especially visible consequence of acid rain is the
destruction of outdoor marble statues and the exteriors of
limestone buildings. The chemistry is simple: marble and
limestone are forms of calcium carbonate, which can react
with sulfuric acid to form calcium sulfate:
CaCO3 (s) + H2SO4 (aq) ——> Ca2+ (aq) + SO42— (aq) + H2O (l) + CO2
(g)
Calcium sulfate is a more-or-less water soluble salt that
can be washed away by rainwater. If it is protected from
excessive amounts of water, it will crystallize as gypsum:
Ca2+ (aq) + SO42— (aq) + 2H2O (l) ——> CaSO4 · 2 H2O (s)
The two waters in the formula for gypsum are called
waters of hydration. Gypsum is much softer than marble or
limestone, and it easily binds soot and dust. The
consequence is that limestone and marble become eroded
and discolored when attacked by acid rain.
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