GASES Chemistry – Chapter 14
description
Transcript of GASES Chemistry – Chapter 14
![Page 1: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/1.jpg)
1
GASESChemistry – Chapter 14
![Page 2: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/2.jpg)
2
Importance of Gases• Airbags fill with N2 gas in an
accident. • Gas is generated by the
decomposition of sodium azide, NaN3.
• 2 NaN3 ---> 2 Na + 3 N2
![Page 3: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/3.jpg)
3
THREE STATES OF MATTER
![Page 4: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/4.jpg)
4General Properties of
Gases• There is a lot of “free” space in a gas.
• Gases can be expanded infinitely.
• Gases fill containers uniformly and completely.
• Gases diffuse and mix rapidly.
![Page 5: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/5.jpg)
5
Properties of Gases
Gas properties can be modeled using math. Model depends on—
• V = volume of the gas (L)• T = temperature (K)
–ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions!
• n = amount (moles)• P = pressure
(atmospheres)
![Page 6: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/6.jpg)
6PressurePressure of air is
measured with a BAROMETER (developed by Torricelli in 1643)
Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink)
P of Hg pushing down related to
• Hg density• column height
![Page 7: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/7.jpg)
7
![Page 8: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/8.jpg)
8
Can You Forecast???
The Answer
![Page 9: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/9.jpg)
9PressureColumn height measures Pressure of atmosphere
• 1 standard atmosphere (atm) *
= 760 mm Hg (or torr) *= 29.92 inches Hg *= 101.3 kPa (SI unit is
PASCAL) * HD only= about 34 feet of water!
* Memorize these!
![Page 10: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/10.jpg)
10
Pressure ConversionsA. What is 475 mm Hg expressed in atm?
1 atm 760 mm Hg
B. The pressure of a tire is measured as 3.6 atm. What is this pressure in mm Hg?
760 mm Hg 1 atm = 2736 mm Hg
= 0.625 atm475 mm Hg x
3.6 atm x
![Page 11: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/11.jpg)
11
Pressure ConversionsA. What is 2 atm expressed in torr?
760 torr 1 atm
B. The pressure of a tire is measured as 623 torr. What is this pressure in kPa?
101.3 kPa 760 torr =83.04 kPa
= 1520 torr2 atm x
623 torr x
![Page 12: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/12.jpg)
12
http://wps.prenhall.com/wps/media/objects/3311/3391331/blb1002.html
Vacuum
Pgas= Patm- PHg
![Page 13: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/13.jpg)
13Manometer
![Page 14: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/14.jpg)
14
An open manometer is filled with mercury and connected to a
container of hydrogen gas. The mercury level is 57 mm higher in the arm of the tube connected to the hydrogen. If the atmospheric pressure is 0.985 atm, what is the pressure of the hydrogen gas, in
atmospheres?
•
![Page 15: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/15.jpg)
15
• An open manometer connected to a tank of argon has a mercury level 83 mm higher in the atmospheric arm. If the atmospheric pressure is 76.9 kPa, what is the pressure of the argon in kPa?
![Page 16: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/16.jpg)
16
http://boomeria.org/chemlectures/textass2/firstsemass.html
![Page 17: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/17.jpg)
17
Closed manometer calculations• A closed manometer like the one in Figure
15-3 is filled with mercury and connected to a container of nitrogen. The difference in the height of mercury in the two arms is 690 mm. What is the pressure of the nitrogen in kilopascals?
Ans: 690mm = 90 kPa.
![Page 18: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/18.jpg)
18Boyle’s LawP α 1/V or PV=k (k is a
constant)This means Pressure
and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.
P1V1 = P2 V2
Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.
![Page 19: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/19.jpg)
19Boyle’s Law and Kinetic Molecular
Theory
P proportional to 1/V
![Page 20: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/20.jpg)
20
Boyle’s LawA bicycle pump is a
good example of Boyle’s law.
As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.
![Page 22: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/22.jpg)
22
Charles’s Law
If n and P are constant, then V = bT (b is a constant)
V and T are directly proportional.
V1 V2
=
T1 T2If one temperature goes up,
the volume goes up!
Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.
![Page 23: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/23.jpg)
23
Charles’s original balloon
Modern long-distance balloon
![Page 24: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/24.jpg)
24
Charles’s Law
![Page 26: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/26.jpg)
26
Gay-Lussac’s LawIf n and V are
constant, then P α T
P and T are directly proportional.
P1 P2
=
T1 T2 • If one temperature goes
up, the pressure goes up!
Joseph Louis Gay-Lussac (1778-1850)
![Page 27: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/27.jpg)
27Gas Pressure, Temperature, and Kinetic Molecular
Theory
P proportional to T
![Page 28: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/28.jpg)
28Gay Lussac’s LawThe pressure and temperature of a gas aredirectly related, provided that the volume
remains constant.
Temperature MUST be in KELVINS!
€
P1T1=P2T2
![Page 29: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/29.jpg)
29A Graph of Gay-Lussac’s Law
![Page 30: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/30.jpg)
30
Combined Gas Law• The good news is that you don’t
have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V1 P2 V2
= T1 T2
No, it’s not related to R2D2
![Page 31: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/31.jpg)
31
Combined Gas LawIf you should only need one of the other gas
laws, you can cover up the item that is constant and you will get that gas law!
= P1 V1
T1
P2 V2
T2
Boyle’s LawCharles’ LawGay-Lussac’s
Law
![Page 32: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/32.jpg)
32
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data TableP1 = 0.800 atm V1 = 180 mL T1 = 302 KP2 = 3.20 atm V2= 90 mL T2 = ??
![Page 33: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/33.jpg)
33
CalculationP1 = 0.800 atm V1 = 180 mL T1 = 302 KP2 = 3.20 atm V2= 90 mL T2 = ??
P1 V1 P2 V2 = P1 V1 T2 = P2 V2 T1 T1 T2
T2 = P2 V2 T1 P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
T2 = 604 K - 273 = 331 °C
= 604 K
![Page 34: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/34.jpg)
34
Learning Check
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?
![Page 35: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/35.jpg)
35
Solution
T1 = 308 K T2 = ?
V1 = 675 mL V2 = 0.315 L = 315 mL
P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg
T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL
= 178 K - 273 = - 95°C
![Page 36: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/36.jpg)
36
One More Practice Problem
A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
![Page 37: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/37.jpg)
37
SolutionComplete the following setup:Initial conditions Final conditionsV1 = 785 mL V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
Since P is constant, P cancels out of the equation. V1 V2 V1 T2 = V1T2 = T1V2 = V2 T1 T2 T1
= 728 mLCheck your answer: If temperature decreases,
V should decrease.
![Page 38: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/38.jpg)
38And now, we pause for this commercial message from STP
OK, so it’s really not THIS kind of STP…
STP in chemistry stands for Standard Temperature and
Pressure
Standard Pressure = 1 atm (or an equivalent)Standard
Temperature = 0 deg C (273 K)
STP allows us to compare amounts of
gases between different pressures and temperatures
![Page 39: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/39.jpg)
39
Try This OneA sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
P1 = 1.0 atm V1 = 15 L T1 = 273 K P2 = 2.0 atm V2 = ?? T2 = 248 K
V2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K
![Page 40: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/40.jpg)
40Avogadro’s Hypothesis
Equal volumes of gases at the same T and P have the same number of molecules.
V = n (RT/P) = knV and n are directly related.
twice as many molecules
![Page 41: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/41.jpg)
41Avogadro’s Hypothesis and Kinetic Molecular
Theory
P proportional to n
The gases in this experiment are all measured at the same T and V.
![Page 42: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/42.jpg)
42
• At STP one mole of gas fills 22.4 Liters
• KNOW THIS
![Page 43: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/43.jpg)
43IDEAL GAS LAW
Brings together gas properties.
Can be derived from experiment and theory.
BE SURE YOU KNOW THIS EQUATION!
P V = n R T
![Page 44: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/44.jpg)
44
Using PV = nRTP = PressureV = VolumeT = TemperatureN = number of moles
R is a constant, called the Ideal Gas ConstantInstead of learning a different value for R for all the
possible unit combinations, we can just memorize one value and convert the units to match R.
R = 0.0821
L • atm
Mol • K
![Page 45: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/45.jpg)
45
Using PV = nRTHow much N2 is required to fill a small room
with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?
Solution1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg)
= 0.98 atmAnd we always know R, 0.0821 L atm / mol K
![Page 46: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/46.jpg)
46
Using PV = nRTHow much N2 is req’d to fill a small room with a volume of 960
cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
Solution2. Now plug in those values and solve for
the unknown.PV = nRT
n = (0.98 atm)(2.7 x 10 4 L)(0.0821 L • atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
RT RT
![Page 47: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/47.jpg)
47
Learning Check
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?
![Page 48: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/48.jpg)
48
SolutionSet up data for 3 of the 4 gas
variables Adjust to match the units of R
V = 20.0 L
T = 23°C + 273 = 296 K
n = 2.86 mol
P = ?
PV = nRT
Rearrange ideal gas law for unknown P
P = nRT V
Substitute values of n, R, T and V and solve for P
![Page 49: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/49.jpg)
49P = (2.86 mol)(0.0821 L atm/mol K)(296 K)
(20.0 L)
= 3.48 atm
3.48 atm 760 mm Hg 1 atm
= 2.64 x 103 mm Hg
![Page 50: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/50.jpg)
50
Learning Check
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
![Page 51: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/51.jpg)
51
Solution
Solve ideal gas equation for n (moles)n = PV
RT
= (0.967 atm) (5.0 L) (0.0821 L atm/mol K)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
1 mol O2
![Page 52: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/52.jpg)
52Deviations from Ideal Gas Law
• Real molecules have volume.
The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.
• There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.– Otherwise a gas could not
condense to become a liquid.
![Page 53: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/53.jpg)
53
Gases in the AirThe % of gases in air Partial pressure (STP)
78.08% N2 593.4 mm Hg
20.95% O2 159.2 mm Hg
0.94% Ar 7.1 mm Hg
0.03% CO2 0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg 2 2 2
Total Pressure 760 mm Hg
![Page 54: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/54.jpg)
54Dalton’s Law of Partial Pressures
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...Therefore,
Ptotal = PH2O + PO2 = 0.48 atm
Dalton’s Law: total P is sum of PARTIAL pressures.
2 H2O2 (l) ---> 2 H2O (g) + O2 (g) 0.32 atm 0.16 atm
![Page 55: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/55.jpg)
55
Dalton’s Law
John Dalton1766-1844
![Page 56: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/56.jpg)
56Health NoteWhen a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.
![Page 57: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/57.jpg)
57Collecting a gas “over water”
• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
![Page 58: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/58.jpg)
58Table of Vapor Pressures for Water
![Page 59: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/59.jpg)
59
Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the gas?
768 torr – 17.5 torr = 750.5 torr
![Page 60: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/60.jpg)
60
GAS DIFFUSION AND EFFUSION
• diffusion is the gradual mixing of molecules of different gases.
• effusion is the movement of molecules through a small hole into an empty container.
HONORS only
![Page 61: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/61.jpg)
61
Diffusion
![Page 62: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/62.jpg)
62GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules.
Thomas Graham, 1805-1869. Professor in Glasgow and London.
Rate of effusion is inversely proportional to its molar mass.
M of AM of B
Rate for BRate for A
HONORS only
![Page 63: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/63.jpg)
63GAS DIFFUSION AND
EFFUSIONMolecules effuse thru holes in a
rubber balloon, for example, at a rate (= moles/time) that is
• proportional to T• inversely proportional to M.Therefore, He effuses more rapidly
than O2 at same T.
He
HONORS only
![Page 64: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/64.jpg)
64
Gas Diffusionrelation of mass to rate of
diffusion• HCl and NH3 diffuse
from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
HONORS only
![Page 65: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/65.jpg)
65
![Page 66: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/66.jpg)
66
GAS DENSITY
High density
Low density
22.4 L of ANY gas AT STP = 1 mole
![Page 67: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/67.jpg)
67Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the volume of O2 at STP?
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
![Page 68: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/68.jpg)
68Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L.
What is the volume of O2 at STP?
Solution1.1 g H2O2 1 mol H2O2 1 mol O2 22.4 L O2
34 g H2O2 2 mol H2O2 1 mol O2
= 0.36 L O2 at STP
![Page 69: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/69.jpg)
69
What if it’s NOT at STP?
• 1. Do the problem like it was at STP. (V1)
• 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)
![Page 70: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/70.jpg)
70
Try this one!How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
![Page 71: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/71.jpg)
71How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
Find mole of O2
28.0 g NH3 x 1 mol NH3 x 5 mol O2 x 22.4 L O2
17.0 g NH3 4 mol NH3 1 mol O2
At STP, V = 46.1 L V1, P1, and T1
P1 V1 P2 V2
= P1 V1 T2 = P2 V2 T1
T1 T2
V2 = P1 V1 T2 (1 atm) (46.1 L) (297K) P2 T1 (0.950 atm) (273 K)
= 52.8 L
= 46.1 LAT STP
![Page 72: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/72.jpg)
72
Remembering the Ideal gas law and the following relationship you can answer most problems.
P1V1 = P2V2
n1T1 n2T2
![Page 73: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/73.jpg)
73
• If the number of moles and temperature are constant, the statement just becomes Boyle’s Law.
• If the number of moles and the pressure are constant, then we have Charles’ Law.
• If the pressure and temperature are constant, then we have Avogadro’s Law
![Page 74: GASES Chemistry – Chapter 14](https://reader033.fdocuments.net/reader033/viewer/2022061604/568164d3550346895dd704d9/html5/thumbnails/74.jpg)
74
•Now,
•PRACTICE