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Agenda 3/10/2014
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Slip Quiz 5 AdayPart 1
Gas Law calculations- any questions?
Avogadro’s Principle – notes
Ch 14 Study Guide
Ch 14 Assessment Question: Applying
Scientific Methods The Haber Process
• Homework
• Slip quiz 5 AdayPart 2
5A Day Part 1
1. Which of the following processes involves the
release of energy when a material undergoes
the changes?
A evaporation and melting
B evaporation and freezing
C condensation and freezing
D condensation and evaporation
2. The Combined Gas Law is shown by
A. KE = ½ mv2
B PV = constant
C PV = nRT
D P1V1 = P2V2
T1
T2
5 Aday- Part 1
1. Condensation and freezing involve the release
of energy when a material undergoes the
Gas
changes.
Liquid
energy
release
Solid
2. The Combined Gas Law is shown by …
P1V1 = P2V2
T1
T2
Avogadro’s Principle
Chapter 14
Section 14.2
Starts at Page 430
And Gas Stoichiometry 14.4
(p 440)
Avogadro’s Principle
• Equal volumes of gases at the
same temperature and pressure
contain equal numbers of
particles.
Avogadro’s Principle
• Equal volumes of gases at the
same temperature and pressure
contain equal numbers of
particles.
Ideal Gas Equation PV = nRT
V = nRT
P
Avogadro’s Principle
• Equal volumes of gases at the same temperature and pressure
contain equal numbers of particles.
Ideal Gas Equation PV = nRT
V = nRT
P
If Vgas1 = Vgas2 then ngas1 = ngas2
• Remember one mole of anything
contains 6.02 x 10 23 particles
• Molar volume for a gas is the
volume that one mole occupies at
0.00°C and 1.00 atm pressure.
–These conditions of temperature
and pressure are known as
standard temperature and pressure
(STP)
Avogadro showed experimentally
that:
• One mole of any gas occupies a volume
of 22.4 L at STP
Very useful!!!!
Example: 1 mole oxygen occupies 22.4 L at
STP and contains 6.02 x 1023 molecules of
O2
1 mole oxygen occupies 22.4 L at STP and
contains 6.02 x 1023 molecules of O2
1 mole Carbon dioxide occupies…….L at
STP and
contains………………………molecules CO2
2 moles oxygen - ….. at STP and contains
……………….. molecules O2
3 moles Carbon dioxide - …….L at STP and
contains………………………molecules CO2
1 mole oxygen occupies 22.4 L at STP and
contains 6.02 x 1023 molecules of O2
1 mole Carbon dioxide occupies 22.4 L at
STP and
contains 6.02 x 1023 molecules CO2
2 moles oxygen - 44.8L at STP and
contains 1.20 x 1024molecules O2
3 moles Carbon dioxide - 67.2 L at STP and
contains 1.80 x 1024 molecules CO2
Test type questions:
1) Standard temperature and pressure
(STP) are defined as
Possible answers?
Test type questions:
1) Standard temperature and pressure
(STP) are defined as
Possible answers?
0°C and 1 atm pressure
273 K and 1 atm pressure
0°C and 760 mm Hg pressure
273 K and 760 mm Hg pressure
0°C and 101.325 kPa pressure
Test type questions:
2) Under which of the following sets of
conditions will a 1 mole sample of {any
gas} occupy a volume of 22.4 liters?
Possible answers?
This is the definition of molar volume –
that 1 mole of any gas occupies 22.4 L at
STP – appears on your reference sheet
under Constants
Volume of Ideal Gas at STP 22.4 L mol-1
Test type questions:
2) Under which of the following sets of
conditions will a 1 mole sample of {any
gas} occupy a volume of 22.4 liters?
Possible answers? STP
0°C and 1 atm pressure (typically this)
273 K and 1 atm pressure
0°C and 760 mm Hg pressure
273 K and 760 mm Hg pressure
0°C and 101.325 kPa pressure
Test type questions:
3) Under which of the following sets of
conditions will a 0.5 mole sample of {any
gas} occupy a volume of 11.2 liters?
Possible answers?
Look carefully
If 0.5 mole occupies 11.2 liters at this
T and P
What would 1 mole of this gas occupy?
Test type questions:
3) Under which of the following sets of
conditions will a 0.5 mole sample of {any
gas} occupy a volume of 11.2 liters?
Possible answers?
Look carefully
If 0.5 mole occupies 11.2 liters at this
T and P
What would 1 mole of this gas occupy?
twice as much = 22.4 liters = molar volume
Test type questions:
3) Under which of the following sets of
conditions will a 0.5 mole sample of {any
gas} occupy a volume of 11.2 liters?
Possible answers? Any answer that shows
Standard temperature and pressure (STP)
0°C and 1 atm pressure
Or equivalent in other common units
Test type questions:
4) A sample of carbon dioxide gas occupies
a volume of 20 L at standard temperature
and pressure at (STP). What will be the
volume of a sample of argon gas that has
the same number of moles and pressure
but twice the absolute temperature?
Using
PV = nRT
For the problem we have same number
of moles of CO2 and Argon = n, P
doesn’t change
PV = nRT or P = T
nR
V
If T is doubled V must double to keep
the ratio the same
OR
Combined gas law – is for a fixed amount of
gas – a fixed number of moles of gas
P1V1 = P2V2 = constant
T1
T2
P.V = (Constant).T
OR
4) A sample of carbon dioxide gas occupies
a volume of 20 L at standard temperature
and pressure at (STP). What will be the
volume of a sample of argon gas that has
the same number of moles and pressure
but twice the absolute temperature?
Use Avogadro’s principle – If equal volumes
of gas contain equal numbers of particles
under same conditions of T and P, then
Equal numbers of moles of gas will occupy
the same volume at the same T and P.
Test type questions:
4) A sample of carbon dioxide gas occupies
a volume of 20 L at standard temperature
and pressure at (STP). What will be the
volume of a sample of argon gas that has
the same number of moles and pressure
but twice the absolute temperature?
CO2 V = 20 L at STP (0°C and 1 atm)
Ar (same number of moles as the CO2) so
would also be same volume at STP
V1 = 20 L at STP
Test type questions:
4) …What will be the volume of a sample of
argon gas that has the same number of
moles and pressure but twice the absolute
temperature?
Ar
V1 = 20 L at STP
P1= 1atm
T1= 273K
V2 =
P2 = 1 atm
T2 = 2 x 273K = 546 K
Test type questions:
4) cont.
Ar
V1 = 20 L at STP
V2 =
P1= 1atm
P2 = 1 atm
T1= 273K
T2 = 2 x 273K = 546 K
1atm. 20L = 1atm.V2
273 K
546 K
V2 = 1atm. 20L. 546 K
1atm. 273 K
= 40L
Gas Stoichiometry
CH4(g) + 2 O2(g)
1 mole
 CO2(g) + 2 H2O (g)
2 moles
1 volume 2 volumes
1 mole
2 moles
1volume 2 volumes
CH4(g) + 2 O2(g)
 CO2(g) + 2 H2O (g)
1 volume 2 volumes
1volume 2 volumes
Ex. Qu.: What volume of oxygen gas is needed for the complete combustion of
4.00 L of methane gas? (Assume constant temperature and pressure.)
Look at relevant ratios in equation
1 : 2 or 1 volume of CH4 will need 2 volumes O2
1L
:
2L
so
4L
:
2L x 4 = 8L
Answer: 4 L of methane will require 8.00 L of
oxygen for complete combustion.
Ex. Qu. 2: What volume of oxygen gas is
needed for the complete combustion of
4.00 L of ethane gas? (Assume constant
temperature and pressure.)
2 C2H6(g) + 7O2(g)
 4 CO2(g) + 6 H2O (g)
2 volumes :7 volumes
Or 1 vol : 7/2 volumes
1L : 7/2 L
4L : 7/2 x 4
Ex. Qu. 2: What volume of oxygen gas is
needed for the complete combustion of
4.00 L of ethane gas? (Assume constant
temperature and pressure.)
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O (g)
2 volumes : 7 volumes
Or 1 vol : 7/2 volumes
1.00L : 7/2 L
4.00L : 7/2 x 4 L
Answer: 4.00 L of ethane will require 14.00 L of
oxygen for complete combustion.
Ex. Qu. 3: What volume of oxygen gas at
STP is needed for the complete reaction of
107.92g of aluminum in the reaction below:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
4 moles: 3 moles
Look up atomic mass of aluminum
26.98 amu – remember 1 mole is atomic mass in g
107.92 g x 1 mol = 4 mol
26.98g
Which is what we need according to the equation
(yeah!)
Ex. Qu. 3: What volume of oxygen gas at
STP is needed for the complete reaction of
107.92g of aluminum in the reaction below:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
4 moles: 3 moles
107.92g Al needs 3 moles oxgyen to react
completely
At STP Molar volme = 22.4 L/mol
3 mol x 22.4 L = 67.2 L
1mol
Ex. Qu. 3: What volume of oxygen gas at
STP is needed for the complete reaction of
107.92g of aluminum in the reaction below:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
4 moles: 3 moles
Answer: 67.2 L of oxygen at STP will be needed
for complete reaction with 107.92 g of aluminum.
Ch 14 Assessment Sheet
Applying Scientific Methods
1. Write a balanced chemical equation for
the chemical reaction that occurs in the
Haber process. Show the state of each
reactant and product.
3 H2(g) + N2(g) 2 NH3(g)
3x2 = 6 H
2N
2N
2x3=6 H (balanced)
Ch 14 Assessment Sheet
Applying Scientific Methods
2. Heated nitrogen and hydrogen gases are
reduced in volume in the compressor. What
effect do these changes in temperature and
volume have on the pressure of the gas.
As the molecules are heated they will move faster (higher
average kinetic energy) and as they are compressed into a
smaller volume they are closer together and will collide with
each other and the walls of the container more often.
These 2 effects will both serve to increase the pressure of
the gas.
Ch 14 Assessment Sheet
Applying Scientific Methods
3. The return pump moves any unreacted hydrogen and nitrogen gases
from the cooler back into the compressor. What effect will removing
gases from the cooler have on the pressure of the gases in the cooler?
Explain.
The pressure of the gases in the cooler will decrease as
the unreacted gases are removed since there will be fewer
gas molecules left in the cooler and pressure is directly
proportional to the number of gas particles (ideal gas
equation PV=nRT, where P is pressure, V is volume, n is
number of moles of gas, R is the gas constant, and T the
temperature in Kelvins).
Ch 14 Assessment Sheet
Applying Scientific Methods
4. Ammonia is a real gas. What will happen to it if the pressure
continues to be increased and the temperature continues to be
decreased.
At high pressures and low temperatures real gases deviate more and
more from ideal behavior as the intermolecular forces start to influence
the behavior of the real gas under these conditions. Ammonia has
hydrogen atoms bonded to a small and highly electronegative atom
(nitrogen) which also has a lone pair of electrons available. This means
that one ammonia molecule can form a hydrogen bond with another
molecule of ammonia. The intermolecular attractions between ammonia
molecules will result in it liquefying at high pressures and low
temperatures.
Ch 14 Assessment Sheet
Applying Scientific Methods
5. What volume of ammonia gas will be produced from 2400 L of
hydrogen gas at the same temperature and pressure?
From the balanced equation we know
3 volumes of H2 produce 2 volumes of NH3
1 vol.
: 2/3 vol.
2400 L
: 2/3 (2400 l)
= 1 600 L NH3(g) produced from 2 400
L of H2(g)
Ch 14 Assessment Sheet
Applying Scientific Methods
6. The Haber process occurs at a pressure of approximately 800.0 atm
and a temperature of 400.0°C. Assume a gas sample occupies 25.0 L
at these conditions. What volume will the sample occupy at STP?
P1 = 800.00 atm
P2 = 1atm
V1= 25.0L
V2= ?
T1= 400.00 + 273
T2= 273K
= 673K
Using combined gas law P1V1 = P2V2
T1
T2
Solving for V2 V2 = T2P1V1 = 800atm 25L 273K
T1P2 673K 1atm
V2 = 8 110 L (3 sig fig)
Ch 14 Assessment Sheet
Applying Scientific Methods
7. If 126 L of ammonia is produced at 800.0 atm and a temperature of
400.0°C in the Haber process, what mass of hydrogen was used in
the reaction.
Thinking ahead to needing to convert volume to mass, we want to know
the volume of gas at STP so we can compare it to the molar volume
(22.4 l/mol).
So first let’s find the volume that 126L of ammonia would occupy at
STP
P1 = 800 atm
P2 = 1atm
V1= 126 L
V2= ?
T1= 400.00 + 273 = 673K
T2= 273K
Use combined gas law P1V1 = P2V2
T1
T2
Ch 14 Assessment Sheet
Applying Scientific Methods
7. If 126 L of ammonia is produced at 800.0 atm and a temperature of
400.0°C in the Haber process, what mass of hydrogen was used in the
reaction.
Solving for V2 V2 = T2P1V1 = 800 atm 126 L 273K
T1P2 673K 1atm
V2 = 40 900L (3 sig fig)
40 900L x 1mol = 1 830 mol
22.4L
126 L of ammonia 800 atm and 400°C is 1 830 mol at STP
From equation 2 mol NH3 : 3 mol H2
Ch 14 Assessment Sheet
Applying Scientific Methods
7. If 126 L of ammonia is produced at 800.0 atm and a temperature of
400.0°C in the Haber process, what mass of hydrogen was used in the
reaction.
126 L of ammonia 800 atm and 400°C is 1 830 mol at STP
From equation 2 mol NH3 : 3 mol H2
1 mol: 3/2 mol
Now convert mol of H2 to mass using atomic mass of
H = 1amu
1830 mol: 3/2 mol x 1830 x 2g
1mol
= 5490g
The answer will depend on how many steps you combine before
rounding your answer to 3 sig figs, and whether you use 1amu or 1.01
for atomic mass of H.
5 Adays Part 2
1. The temperature at which all molecular
motion stops is
A 273 K
B 273°C
C 0K
D freezing pt.
And this temperature is called
___________. (2 words)
2. 1 mole of any gas occupies…at STP.
A. 6.03 x 1023L
B 22.4 L
C PV = nRT
C 1L