Reactions that cause change of oxidation numbers are called redox reactions.
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Transcript of Reactions that cause change of oxidation numbers are called redox reactions.
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Reactions that cause change of oxidation numbersare called redox reactions.
Element loses electrons → its oxidation number increases→ element is oxidized → oxidation reaction
Substance that contains the oxidized element is call thereducing agent.
Substance that contains the reduced element is call theoxidizing agent.
Element gains electrons → its oxidation number decreases→ element is reduced → reduction reaction
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1.Write down the oxidation number for each element above its symbol.
2. Which element in what substance is oxidized? Which element in what substance is reduced?
3. What is the reducing agent? What is the oxidizing agent?
2SO2 + O2 → 2SO3
..
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Chapter 5
Gases
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Four variables used to describe a gas
Amount — n — unit: mol
Volume — V — unit: m3, L, mL
Temperature — T — unit: °C, K
Pressure — P — unit:
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1 atm = pressure exerted by the air at 0 °C at sea level
1 atm = 760 mmHg = 760 torr
= 101325 Pa
Pressure = Force / Area
Pa = N / m2
e.g. 125 torr = ? atm = ? Pa
demo
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Four variables used to describe a gas
Amount — n — unit: mol
Volume — V — unit: m3, L, mL
Temperature — T — unit: °C, K
Pressure — P — unit: atm, torr, Pa
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How is the volume of gas defined?
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Volume of a gas equals the volume of the container in which it is held because gas expands to fill the container.
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Cu
Cu
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Orange juice
Orange juice
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Relationships among the four variables
Gas Laws
P, V, T, n: study the relationship betweentwo variables while keep the other twoconstant.
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Boyle’s Law
Relationship between P and V underconstant n and T.
P V T n
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Boyle’s Law
Relationship between P and V underconstant n and T.
PV = k
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(1) (2)
P1V1 = P2V2
Boyle’s Law
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Charles’s Law
Relationship between V and T underconstant n and P.
P V T n
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Plots of V Versus T(°C) for Several Gases
2732− 273 °C
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Plots of V versus T as Before, Except Here the Kelvin Scale is Used for Temperature
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Charles’s Law
Relationship between V and T underconstant n and P.
V = bT
T must be in K
2
2
1
1
T
V
T
Vor
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A sample of gas at 15 °C and 1 atm has a volume of 2.58 L.
What volume will this gas occupy at 38 °C and 1 atm?
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P V T n
Relationship between V and n underconstant P and T.
Avogadro’s Law
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Relationship between V and n underconstant P and T.
Avogadro’s Law
V = an
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V = anV = bTV = k/P
PV = nRT
Ideal Gas Law
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Units in ideal gas law
PV = nRT
P — atm, V — L, n — mol, T — K
Option 1
R = 0.082 atm · L · mol−1 · K−1
P — Pa, V — m3, n — mol, T — KOption 2
R = 8.314 J · mol−1 · K−1
Chem 1211
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Calculate the volume occupied by 0.845 mol
of nitrogen gas at a pressure of 1.37 atm
and a temperature of 315 K.
Example 5.5, page 192
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A sample of diborane gas (B2H6) has a pressure of 345 torr
at a temperature of −15 °C and a volume of 3.48 L. How many
B2H6 molecules are in this sample?
If conditions are changed so that the temperature is 36 °C and
the pressure is 468 torr, what will be the volume?
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What is the volume in liters of exactly 1 mol ideal gas
at exactly 0 °C and 1 atm?
0 °C and 1 atm: standard temperature and pressure (STP)
At STP, the volume of one mole ideal gas is 22.4 L
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For a gas sample, the density is 1.95 g/L at 1.50 atm and
27 °C. What is the molar mass of the gas?
molar mass = dRT
P
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Calculate the density of nitrogen gas at 125 °C and a
pressure of 755 torr.
Example 5.7, page 195
molar mass = dRT
P
d = 0.853 g/L
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What about a gas mixture?
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Dalton’s Law of partial pressure
The total pressure of a mixture of gases equals the sum of
the pressures that each gas would exert if it were present alone.
pressure that each gas would exert if it were present alone
= partial pressure of that gas
Pt = P1 + P2 + P3 + · · ·
= ntRT/V
Each component and the whole gas mixture obey ideal gas law.
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6.00 g O2 and 9.00 g CH4 are placed in a 15.0 L vessel
at 0 °C. What is the partial pressure of each gas and what
is the total pressure?
PO2 = 0.280 atm
PCH4 = 0.837 atm
Ptotal = 1.117 atm
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How to describe the composition of a mixture?
mass of a component1) mass percent = x 100%
mass of the whole sample
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321
1
t
11 nnn
n
n
nχ
moles of a component2) mole fraction
total moles of the mixture
1
t
P
P
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6.00 g O2 and 9.00 g CH4 are placed in a 15.0 L vessel
at 0 °C. What is the partial pressure of each gas and what
is the total pressure?
What is the mole fraction of O2 in the mixture?
PO2 = 0.280 atm
PCH4 = 0.837 atm
Ptotal = 1.117 atm
1 11
t t
P
P
n
n
A similar example on page 200. practice.
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Air: 78 % N2, 21 % O2, 1 % other gases.
What is the partial pressure of N2 in torr?
1 11
t t
P
P
n
n
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The air on Pandora consists of N2, O2, CO2,Xe, CH4, and H2S. The mole fractions of CO2
and H2S are 18 % and 1.0 %, respectively. If the partial pressure of CO2 is 164 torr, what ispartial pressure of H2S in Pa?
1 11
t t
P
P
n
n
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The partial pressure of CH4(g) is 0.175 atm and that of O2(g) is 0.250 atm in a mixture of the two gases.
a)What is the mole fraction of each gas in the mixture?
b)If the mixture occupies a volume of 10.5 L at 65 °C, calculatethe total number of moles of gas in the mixture.
c) Calculate the mass of each gas in the mixture.
1 11
t t
P
P
n
n
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How is ideal gas law relatedto chemical reactions?
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PV = nRT
Link between a gas and chemical reaction
aA + bB cC + dD
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Methanol (CH3OH) can be synthesized by the following reaction:
CO(g) + 2H2(g) CH3OH(g)
What volume (in liters) of hydrogen gas, at a temperature of
355 K and a pressure of 738 torr, is required to synthesize
35.7 g of methanol?
Example 5.12, page 203
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The metabolic oxidation of glucose, C6H12O6, in our bodies
produces CO2, which is expelled from our lungs as a gas:
C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)
Calculate the volume of CO2 produced at body temperature
(37 °C) and 0.970 atm when 24.5 g of glucose is consumed
in this reaction.
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FOR PRACTICE 5.12, page 204
In the following reaction, 4.58 L of O2 was formed at P = 745
torr and T = 308 K. How many grams of Ag2O must have
decomposed?
2Ag2O(s) 4Ag(s) + O2(g)
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FOR PRACTICE 5.13, page 205
How many liters of oxygen (at STP) are required to form 10.5 g
of H2O?
2H2(g) + O2(g) 2H2O(g)
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What is the theory behindideal gas law?
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Kinetic Molecular Theory
The size of gas molecules is assumed to be zero
Gas molecules do not attract or repel each other
Gas molecules constantly move, collisions with the wallof a container cause pressure
Average kinetic energy of gas molecules is directly proportional to absolute temperature
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Kinetic Molecular Theory
The size of gas molecules is assumed to be zero
Gas molecules do not attract or repel each other
Gas molecules constantly move, collisions with the wallof a container cause pressure
Average kinetic energy of gas molecules is directly proportional to absolute temperature
Remember these four assumptions
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Problems for Chapter 5 Work Exercises: 11, 15, 18, 20, 23, 29a-b, 35, 41, 43, 49, 65, 67, 69, 75, 77, & 79.