AP Chemistry
AP Chemistry 4: States of Matter Name __________________________
A. Gas State (10.2 to 10.9)
1. tend to be small, non-polar molecules
2. form homogenous mixtures
3. distributed throughout the entire container
(molecules typically occupy 0.1 % of the volume)
4. kinetic theory for gases (ideal gas)
a. molecules in continuous, chaotic motion, which is proportional to temperature
1. kinetic energy, Kmole = 1/2(MM)v2 = 3/2RT
a. R = 8.31 J/mol•K
b. MM (molar mass) in kg
c. T in kelvin (TK = ToC + 273)
2. root-mean-square speed, v = (3RT/MM)½
3. Graham’s law
a. rate r of effusion (leaking) or diffusion (spreading out) is proportional to speed
b. rA/rB = (MMB/MMA)½
c. time is inversely proportional to rate ∴ TB/TA = rA/rB
b. molecular volume is insignificant compared to container volume (approximation—see real gas)
c. collisions produce pressure w/o loss of total kinetic energy
d. Bonding between molecules is insignificant (approximation—see real gas)
5. gas laws
a. Ideal gas Law equation: PV = nRT
1. molecules generate pressure via collisions
a. pressure = force/area
b. 1 atm = 101 kPa = 760 mm Hg (torr)
c. measuring tools
1. barometer: atmospheric pressure
2. manometer: enclosed gas pressure
Pgas = Patm ± h (in mm Hg)
2. gas pressure is affected by:
a. n: each molecule exerts pressure ∴ more molecules exert more pressure: P ∝ n
b. T: hotter molecules move faster and collide with greater force ∴ generate more pressure: P ∝ T
c. V: molecules spread out which reduces collision per surface area ∴ generate less pressure: P ∝ 1/V
3. ideal gas law constant R (V in L, T in K)
a. 8.31 J/mol•K (P in kPa)
b. 0.0821 atm•L/mol•K (P in atm)
4. molar volume at STP = 22.4 L/mol
(standard T = 0oC, standard P = 1 atm)
5. derived equations
a. P1V1/T1 = P2V2/T2
b. MM = mRT/PV = dRT/P
b. Dalton’s law (P ∝ n)
1. Ptot = PA + PB
2. PA = XAPtot , where XA = molA/(molA + molB)
6. real gases
a. Van der Waals:(Preal + n2a/V2)(Vreal – nb) = nRT
b. "a" corrects for molecular bonding
1. "low" temperature (close to boiling point) molecules clump and collide less often, which generates less pressure ∴ Preal < Pideal
2. a is proportional to molecular polarity
c. "b" corrects for molecular volume
1. high pressure is generated by crowded molecules where the volume of empty space (Videal) is significantly less than 100 % of the total volume (Vreal)∴ Vreal > Videal
2. b is proportional to molar mass
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B. Phase Change (11.1 to 11.2, 11.4 to 11.6)
1. molecular-level comparison of gas, liquid, and solid states of a substance
[pic]
|Characteristic |Gas |Liquid |Solid |
|Energy |Highest |Middle |Lowest |
|Disorder |Greatest |Middle |Least |
|Occupied space |Whole |Bottom |Own |
|Compressibility |Yes |No |No |
|Flow rate |Fast |Middle |No |
|Diffusion |Rapid |Slow |No |
2. cohesive forces (van der Waals forces)
a. attraction between molecules
(covalent bonds hold atoms together in molecule)
b. dipole-dipole forces
1. polar molecules
2. δ+ of one molecule attracts δ− of a neighbor
3. strength ∝ to polarity, if all else is even
c. London dispersion forces
1. attraction between nuclei of one molecule's atoms for the electrons in a neighboring molecule causes temporary polarization throughout the liquid or solid (polarizability)
2. generalization
a. operates between all molecules (stronger than dipole-dipole for large molecules, i.e. large nonpolar > small polar)
b. only force for nonpolar (strength ∝ to mass: Xe > He, I2 > F2, C3H8 > CH4)
d. hydrogen bonding
1. super strong dipole-dipole force (stronger than dispersion forces)
2. H bonded to N, O or F
a. H ≈ +1 charge and N, O or F ≈ –1 charge because of extreme electronegative difference and small radius
b. bonding is ionic like (E ∝ Q1Q2/d)
3. explains unusual properties of water
a. each water molecule bonds to 4, which makes a 3-d structure with open cavities
b. high melting and boiling temperatures
c. low vapor pressure (low volatility)
3. cooling profile for water from 110oC to -10oC
| |A | | | | | |
|100oC | | | | | | |
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|0oC | | | |D E| | |
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| | | | | | |F |
| Heat Removed → (J) |
a. slope C-D < slope E-F ∴ more heat is removed when one mole H2O(l) is cooled 1oC compared to one mole H2O(s)
b. length B-C > length D-E ∴ more heat is removed when one mole H2O(g) → H2O(l) compared to one mole H2O(l) → H2O(s)
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c. calculations
|Step |Process |Formula |Constants |
|B-C |Condensation |Q = nΔHvap |ΔHvap = 40.7 kJ/mol |
| |(100oC) | | |
|C-D |Cooling liquid |Q = nClΔT |Cl = 75.3 J/mol•K |
|D-E |Freezing |Q = nΔHfus |ΔHfus = 6.01 kJ/mol |
| |(0oC) | | |
|E → |Cooling solid |Q = nCsΔT |Cs = 37.8 J/mol•K |
4. phase diagram
|Press| |
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| |Temperature |
a. point A: triple point (three phases at equilibrium)
1. below triple point: sublimation
2. above triple point: melting and vaporization
b. line A-B: equilibrium vapor-pressure curve for liquid (normal boiling point occurs at 1 atm pressure)
c. B: critical point, where there is no distinction between liquid and vapor (no liquid-vapor surface)
d. line A-C: equilibrium vapor-pressure curve for solid
e. line A-D: melting point of solid at various pressures
(normal freezing point occurs at 1 atm)
1. positive slope when solid is the densest phase (melting point increases with pressure)
2. negative slope when liquid is the densest phase (melting point decrease with pressure)
5. vapor
a. some surface molecules in the condensed phase have enough kinetic energy (speed) to escape surface (evaporate) below boiling point
b. as temperature increases more molecules have sufficient kinetic energy ∴ more vapor molecules
[pic]
c. cooling process (hottest evaporate first, leaving cooler molecules behind)
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d. equilibrium between liquid and vapor
1. evaporation rate = condensation rate in a closed container
2. concentration of vapor measured as Pvap
3. independent of container size until no liquid
4. Pvap increases at higher temperature because
a. more molecules are in vapor phase
b. vapor molecules exert greater pressure
e. boiling occurs when Pvap = Patm ∴ boiling point decreases with elevation (lower air pressure, Patm)
f. high Pvap indicates volatility—tendency to evaporate
C. Crystalline Solids (11.8)
1. ions, atoms or molecules fit into a regular geometric pattern (crystal lattice)
2. minimum energy state—maximum bond energy
3. intermolecular forces (attraction between δ+ and δ-) or bonds (covalent, ionic or metallic) hold particles together
4. 4 types of solids
a. metallic—metals only
1. attraction between cations and delocalized valence electrons (electron sea model)
[pic]
2. melting point: variable (∝ bond strength)
3. conductivity: free electrons ∴ yes
4. malleable: non-directional bond ∴ yes
5. water solubility: no molecular interactions ∴ no
6. examples: Cu, Ag, etc.
b. covalent network—nonmetals w/o H or halogen
1. atoms covalently bond throughout w/o size limit (different than large molecule)
2. melting point: strong bonds ∴ high
3. conductivity: no free electrons ∴ no
4. malleability: bond highly directional ∴ brittle
5. water solubility: no molecular interactions ∴ no
6. three examples
a. diamond and graphite—C
1. allotropes (2 forms in the same state)
2. diamond: covalently bonded 3-d structure—good abrasive
3. graphite: covalent bonded planar sheets linked by dispersion forces
a. separate easily—good lubricant
b. electron flow—good conductivity
b. quartz—SiO2
1. 3-d structure similar to diamond
2. softens when heated until liquid
3. fast cooling = non-crystalline glass
c. molecular—nonmetals often with H and/or halogen
1. attraction between +δ of one molecule with
–δ of another
2. melting point: weak bonds ∴ low
3. conductivity: no free electrons ∴ no
4. malleability: non-directional ∴ yes
(H-bonding in H2O(s) is somewhat directional)
5. water solubility ("like dissolves like") ∴ yes/no
6. examples: H2O, C6H12O6, etc.
d. Ionic—metal plus nonmetals
1. attraction between cations and anions
2. melting point: strong bonds ∴ high
3. conductivity: no free electrons ∴ no
(fused or dissolved state is conducting)
4. malleability: bond highly directional ∴ brittle
5. water solubility: ion-dipole interaction ∴ yes
6. examples: NaCl, CaCO3, etc.
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Gas State
1. What features of the kinetic theory of gases
a. describe all gas molecules?
|Molecules are in continuous, chaotic motion. |
|Collisions produce pressure w/o loss of total energy. |
b. describe ideal gas molecules only?
|Molecules have zero volume. |
|Molecules don't interact with each other. |
2. Consider one mole of Ne gas at 274 K. Determine
a. the total kinetic energy.
|Kmole = 3/2RT |
|Kmole = 3/2(8.31 J/mol•K)274 K = 3420 J |
b. the average speed.
|u = (3RT/MM)½ |
|u = [(3)(8.31)(274)/(20.2/1000)]½ ∴ u = 582 m/s |
3. Consider the drawing below.
[pic]
a. A and B are He and O2, at 25oC, which is which? Explain
A is O2 and B is He. O2 is heavier and would have slower speeds compared to He. u = (3RT/MM)½
b. A and B are at 100 K and 200K, which is which? Explain
B is the higher temperature. Hotter gas has greater speed than cooler gas. u = (3RT/MM)½
4. a. What alkane effuses at 1/5 the rate of He?
rA/rB = (MMB/MMA)½
5/1 = (MMB/4)½ → MMB = 100 g mol-1 ∴ (C7H16)
b. How many times faster does C2H2 diffuse compared to the alkane?
TB/TA = (MMB/MMA)½
TB/TA = (100/26)½ = 2 times faster
5. Determine the pressure of 1.22 atm in the following units.
|mm Hg |kPa |torr |
|1.22 atm x 760 = 927 mm|1.22 atm x 101 = 123 |1.22 atm x 760 = 927 |
|Hg |kPa |torr |
6. If the atmospheric pressure is 749 mm Hg, what is the pressure of the enclosed gas in each case below?
[pic]
| 749 mm Hg |749 mm Hg |103 mm Hg |
|–520 mm Hg |+67 mm Hg | |
|229 mm Hg |816 mm Hg | |
7. A gas is confined inside a container with a movable piston held down by a fixed pressure.
[pic]
a. What affect would doubling the number of gas molecules at the same temperature have on the system? Explain.
|The piston would move upward to double the volume because the greater |
|number of molecules exerts twice the pressure. |
b. What affect would doubling the Kelvin temperature of the gas have on the system? Explain.
|The piston would move upward to double the volume because the faster |
|moving molecules exert twice the pressure. |
c. What affect would doubling pressure by the piston at the same temperature have on the system? Explain.
|The piston would move downward to halve the volume so that molecules are|
|more crowded in order to produce enough collisions to match the |
|pressure. |
8. Complete the following table for an ideal gas:
|P |V |n |T |
|2.00 atm |1.00 L |1.500 mol |16.2 K |
|30.3 kPa |1.250 L |0.0152 mol |27oC |
|650 torr |11.2 L |0.333 mol |350 K |
|10.4 atm |585 mL |0.250 mol |295 K |
9. Oxygen gas in a 10.0-L container has a pressure of 94.6 kPa and temperature of 25oC.
a. How many moles of oxygen gas are in the container?
|PV = nRT |
|(94.6)(10.0) = n(8.31)(25 + 273) ∴ n = 0.382 mol |
b. How many grams of oxygen gas are in the container?
|0.382 mol O2 x 32.0 g/1 mol = 12.2 g |
10. A sample of gas occupies 350 mL at 15oC and 750 torr. What temperature will the gas have at the same pressure if its volume increases to 450 mL?
|P1V1/T1 = P2V2/T2 |
|(350)/(15 + 273) = (450)/T2 ∴ T2 = 370 K (97oC) |
11. Determine the molar mass of an unknown gas given the data.
|Mass |Volume |Temperature |Pressure |
|4.93 g |1.00-L |400. K |1.05 atm |
MM = mRT/PV
MM = (4.93)(0.0821)(400)/(1.05)(1.00) = 154 g/mol
12. Calculate the density of ammonia, NH3, at STP.
|MM = dRT/P |
|(17.0) = d(0.0821)(0 + 273)/(1) ∴ d = 0.758 g/L |
13. Consider the samples of gases.
[pic]
I II III
The samples are at the same temperature. Rank them with respect to the following (1 is highest).
| |I |II |III |
|Total Pressure |1 |3 |1 |
|Partial Pressure of He |1 |2 |3 |
|Density |2 |3 |1 |
|Average Kinetic energy per molecule |1 |1 |1 |
|Total Kinetic energy |1 |3 |1 |
14. Each bulb contains a gas at the pressure and volume shown and temperature of 25oC. Determine
[pic]
a. the number of moles of each gas.
|N2 |n = PV/RT = (265/760)(1.0)/(0.0821)(298) |
| |n = 0.014 mol |
|Ne |n = PV/RT = (800/760)(1.0)/(0.0821)(298) |
| |n = 0.043 mol |
|H2 |n = PV/RT = (532/760)(0.5)/(0.0821)(298) |
| |n = 0.014 mol |
b. the total pressure after all stopcocks are opened.
|P = nRT/V = (0.071)(0.0821)(298)/(2.5) |
|P = 0.695 atm (528 torr) |
c. the partial pressure of each gas.
|N2 |PN2 = XN2Ptot = (0.014/0.071)(528 torr) |
| |PN2 = 104 torr |
|Ne |PNe = XNePtot = (0.043/0.071)(528 torr) |
| |PNe = 320 torr |
|H2 |PH2 = XH2Ptot = (0.014/0.071)(528 torr) |
| |PH2 = 104 torr |
15. 2.00 L of Hydrogen gas is collected over water at 30.0oC. The total pressure is 740 torr (PH2O = 32 torr).
a. What is the partial pressure of the hydrogen gas?
|Ptot = PH2 + PH2O |
|740 = PH2 + 32 ∴ PH2 = 708 torr |
b. How many moles of hydrogen gas are collected?
|PV = nRT |
|(708/760)(2.00) = n(0.0821)(30 + 273) ∴ n = 0.0747 mol |
16. A 20-L flask holds 0.20 mol O2 and 0.40 mol NO2 at 27oC.
a. What is the pressure of the mixture in kPa?
|PV = nRT |
|(P)(20) = (0.60)(8.31)(27 + 273) ∴ P = 75 kPa |
b. What is the partial pressure of oxygen in kPa?
|PA = XAPtot |
|PO2 = (0.020/0.060)(75 kPa) = 25 kPa |
17. Which gas, SO2 or CO2, should be least ideal at STP? Explain
|SO2, it is both heavier and more polar than CO2. |
18. a. Why do gases under high pressure deviate from ideal behavior?
|high pressure is generated by crowded molecules where the volume of |
|empty space (Videal) is significantly less than 100 % of the total |
|volume (Vreal) ∴ Vreal > Videal |
b. Why do gases at temperatures near their boiling point deviate from ideal behavior?
|close to boiling point, molecules clump and collide less often, which |
|generates less pressure ∴ Preal < Pideal |
Molar Mass of a Gas Lab
19. Measure the mass and volume of butane released from a lighter, use the data to determine the molar mass and compare the experimental value with known molar mass.
Mass a butane lighter (m1). Fill a 50 mL graduated cylinder with water and place it upside down in a trough filled with water. Release butane into the graduated cylinder until the water level in the cylinder equals the water level in the trough. Record the volume (V). Dry the butane lighter thoroughly and then mass it (m2). Record the room temperature (T), room pressure (Plab) and vapor pressure (PH2O).
a. Record the collected data.
|m1 |m2 |V |T |Plab |PH2O |
| | | | | | |
b. Complete the following calculations to determine the molar mass and percent difference from known value.
|Pgas |T |V |m |
|(atm) |(K) |(L) |(g) |
| | | | |
|Experiment MM |Known MM |% Δ |
|(g/mol) |(g/mol) | |
| | | |
c. How would the following affect the molar mass value?
(1) The butane lighter was not thoroughly dried.
|m2 used in the calculation would be greater than it should be. This |
|would make the calculated mass of gas too smaller ∴ the molar mass would|
|be less. |
(2) The vapor pressure of water was not subtracted from the room pressure.
|P used in the calculation would be greater than it should be. This |
|would make the molar mass too small since P is in the formula's |
|denominator. |
(3) The water level in the inverted graduated cylinder was higher than the level in the trough.
|P used in the calculation would be smaller than it should be. This |
|would make the molar mass too large since P is in the formula's |
|denominator. |
Phase Change
20. Which letter illustrates the types of molecular forces?
[pic]
|Dipole-Dipole |d |Dispersion |b |H-bond |a |
21. For each pair, highlight the molecule with the higher boiling point and then justify your choice.
|Pair |Justification |
|H2O & H2S |H-bonding > dipole-dipole forces |
|Ne & Kr |Greater atomic mass |
|Cl2 & SO2 |More polar |
22. Explain the boiling points for the two isomers.
[pic]
|Propanol molecules H-bond, which is a stronger attraction then ethyl |
|methyl ether's dipole forces. |
23. In addition to dispersion, what type of force would you expect between the following molecules?
|H2 |H2S |CHF3 |NH3 |
|none |dipole |dipole |H-bond |
24. Consider the heating profile for water in your notes.
a. What can you conclude about the value of Cl compared to Cs based on the slope of line C-D compared to E-F?
|The steeper slope for ice, E-F, means Cs < Cl. |
b. What can you conclude about ΔHfus compared to ΔHvap based on the length of line B-C compared to D-E?
|The longer boiling line, B-C, means that ΔHvap > ΔHfus. |
25. Calculate the amount of heat needed to warm 2 mole of H2O between the given temperatures or phase change.
|-10oC → 0oC |Q = nCsΔT = (2 mol)(37.8 J/mol•K)(10 K) |
| |Q = 756 J (0.756 kJ) |
|melt |Q = nΔHfus |
| |Q = (2 mol)(6.01 kJ/mol) = 12.02 kJ |
|0oC → 100oC |Q = nClΔT = (2 mol)(75.3 J/mol•K)(100 K) |
| |Q = 15,060 J (15.1 kJ) |
|boil |Q = nΔHvap |
| |Q = (2 mol)(40.7 kJ/mol) = 81.4 kJ |
|Total |Q = 0.756 + 12.02 + 15.1 + 81.4 |
| |Q = 109.3 kJ |
26. Consider the phase diagram in your notes.
a. How does melting point change when pressure increases?
|The melting point increases in pressure increases. |
b. How would the diagram differ for water?
|The solid/liquid line would have a negative slope. |
27. Answer the questions based on the phase diagram.
[pic]
|Normal melting point |270 K |
|Normal boiling point |370 K |
|Most dense phase |Solid |
|Phase at 150 K, 0.2 atm |Gas |
|Phase at 100 K, 0.8 atm |Solid |
|Phase at 300 K, 1.0 atm |Liquid |
|Phase change: 50 K → 150 K at 0.2 atm |Sublimation |
|Phase change: 1.0 atm → 0.2 atm at 300 K |Vaporization |
|Phase change: 200 K → 300 K at 1.0 atm |Melting |
28. Answer the questions based on the vapor pressure curve.
[pic]
|Vapor pressure at 30oC |400 mm Hg |
|Temperature where pressure = 300 mm Hg |27oC |
|Normal boiling point |48oC |
29. Explain how pure water can boil at room temperature when placed in an evacuated bell jar.
|Water boils when vapor pressure equals atmospheric pressure. As air is |
|pumped out of the bell jar, its pressure decreases until it reaches the |
|vapor pressure. |
30. Explain why baking takes longer at high elevations.
|Water boils at a lower temperature at high altitudes, which means that |
|water based foods cook slower because they cook at the boiling point. |
31. 0.010 moles of water is added to a 5.0-L container filled with dry air at 20oC (vapor pressure = 20 torr). The container is then sealed and equilibrium is established.
a. How many moles of water evaporate?
|PV = nRT |
|(20/760)(5.00) = n(0.0821)(20 + 273) ∴ n = 0.0055 mol |
b. What percentage of the water evaporates?
|0.0054/0.010 x 100 = 54 % |
32. Explain why water droplets form on a cold water bottle.
|Water vapor in the air in contact with the cold surface of the bottle |
|condenses because the low temperature vapor pressure is lower than the |
|pressure exerted by the water vapor in the room. |
Crystalline Solids
33. Use the electron-sea model of metals to explain
a. malleability.
|Cations are not in a fixed position. Thus, metal will bend when stress |
|is placed on it. |
b. conductivity.
|Valence electrons are free to move throughout the solid. |
34. What are allotropes?
|Allotropes are different structural forms of the same substance in the |
|same physical state. |
35. In what way is SiO2 like diamond; unlike diamond?
|SiO2 is like diamond because it has covalent bonding throughout its 3-d |
|structure. |
|SiO2 is unlike diamond because it does not melt at a constant |
|temperature. |
36. What two factors affect ionic bond strength (lattice energy)?
|Ionic charge and distance between ions. |
37. Complete the chart for each type of solid.
| |Metallic |Covalent |Molecular |Ionic |
| | |Network | | |
|Structural Unit |ion |atom |molecule |ion |
|Bond name |metallic |covalent |molecular |ionic |
|Bond strength |variable |strong |weak |strong |
|Melting point |variable |high |low |high |
|Solubility |low |low |variable |high |
|Conductivity |high |low |low |low |
|Malleability |high |low |variable |low |
|Example |copper |diamond |water |salt |
38. Explain the following observations. You must discuss both of the substances in your explanation.
a. SO2 melts at 201 K and SiO2 melts at 1,883 K.
|SO2 forms London dispersion and dipole forces between distinct molecules|
|whereas, SiO2, a covalent network solid, forms covalent bonds |
|throughout. The much stronger covalent bonds, which are broken during |
|melting of SiO2, require much more energy (higher temperature) to break.|
b. Cl2 boils at 238 K and HCl boils at 188 K.
|Liquid Cl2 is held together by London dispersion forces, which although |
|weak increase in strength as the number of electrons increases. Liquid |
|HCl is held together by dipole forces in addition to London dispersion |
|forces, but the addition of dipole forces between HCl molecules must not|
|make up for the fewer electrons around the HCl molecule compared to Cl2.|
c. KCl melts at is 776oC and NaCl melts at 801oC.
|The stronger ionic bond in NaCl is due to the smaller Na+ ion compared |
|to K+, which allows the Cl- ion to get closer and strengthens the |
|attraction between ions, making the NaCl bond stronger and the melting |
|point higher than KCl. |
d. Si melts at 1,410oC and Cl2 melts at -101oC.
|Si is a covalent network solid with strong covalent bonds between atoms.|
|Cl2 has discrete molecules with weak London dispersion forces between |
|molecules. Therefore, melting Si requires a higher temperature than |
|Cl2. |
Practice Quiz
Multiple Choice (no calculator)
Briefly explain why the answer is correct in the space provided.
|1 |2 |3 |4 |5 |
|Boiling Point, oC |+19 |-85 |-67 |-35 |
1. The relatively high boiling point of HF can be correctly explained by which of the following?
(A) HF gas is more ideal.
(B) HF molecules have a smaller dipole moment.
(C) HF is much less soluble in water.
(D) HF molecules tend to form hydrogen bonds.
|HF forms H-bonds (F = small, high electronegativity), which are stronger|
|than dipole forces ∴ high BP. |
2. The increasing boiling points for HCl, HBr and HI can be best explained because of the increase in
(A) dispersion force (B) dipole moment
(C) valence electrons (D) hydrogen bonding
|Bonding is similar, except I has more electrons = greater dispersion |
|forces ∴ higher boiling temp. |
3. A sample of an ideal gas is cooled from 50oC to 25oC in a sealed container of constant volume. Which of the following values for the gas will decrease?
I. The average kinetic energy of the molecules
II. The average distance between the molecules
III. The average speed of the molecules
(A) I only (B) II only (C) III only (D) I and III
|Kinetic energy (3/2RT) & speed (3RT/MM)½ decrease with temperature, but |
|spacing is unchanged. |
Questions 4-7 refer to the phase diagram of a pure substance.
[pic]
4. Which phase is most dense?
(A) solid (B) liquid
(C) gas (D) can't determine
|The solid/liquid line has a positive slope. Increasing pressure favors |
|densest phase ∴ solid is densest. |
5. Which occurs when the temperature increases from 0°C to 40°C at a constant pressure of 0.5 atm?
(A) Sublimation (B) Condensation
(C) Freezing (D) Fusion
|0.5 atm is below the triple point, so the solid sublimates rather than |
|melts. |
6. Which occurs when the pressure increases from 0.5 to 1.5 atm at a constant temperature of 60°C?
(A) Sublimation (B) Condensation
(C) Freezing (D) Fusion
|As the pressure increases at 60oC, the substance crosses the |
|vapor/liquid line ∴ condensation. |
7. The normal boiling point of the substance is closest to
(A) 20oC (B) 40oC (C) 70oC (D) 100oC
|Normal boiling occurs at the intersection of the vapor/ liquid line and |
|1 atm pressure ∴ 70oC. |
Questions 8-9The graph shows the temperature of a pure solid substance as it is heated at a constant rate to a gas.
[pic]
8. Pure liquid exists at time
(A) t1 (B) t2 (C) t3 (D) t4
|Pure state exists during the temperature increase phases. The first |
|rise is solid, second rise is liquid. |
9. Which of the following best describes what happens to the substance between t4 and t5?
(A) The molecules are leaving the liquid phase.
(B) The solid and liquid phases coexist in equilibrium.
(C) The vapor pressure of the substance is decreasing.
(D) The average intermolecular distance is decreasing.
|Phase change occurs along the plateaus. The first plateau is solid to |
|liquid and second is liquid to gas. |
10. Which actions would be likely to change the boiling point of a sample of a pure liquid in an open container?
(A) Placing it in a smaller container
(B) Increasing moles of the liquid in the container
(C) Moving the container to a higher altitude
(D) Increase the setting on the hot plate
|Boiling occurs when vapor pressure = air pressure. Higher altitude = |
|lower air pressure ∴ lowers BP. |
11. Gas in a closed rigid container is heated until its absolute temperature is doubled, which is also doubled?
(A) The density of the gas
(B) The pressure of the gas
(C) The average speed of the gas molecules
(D) The number of molecules per liter
|Rigid container = constant V ∴ P ∝ T (PV = nRT). Density, |
|molecules/liter are unchanged. Speed x √2. |
12. A 2.00-L of gas at 27oC is heated until its volume is 5.00 L. If the pressure is constant, the final temperature is
(A) 68oC (B) 120oC (C) 477oC (D) 677oC
|P1V1/T1 = P2V2/T2 |
|(2.00 L)/(300 K) = (5.00 L)/T2 ∴ T2 = 750 – 273 = 477oC |
13. Under the same conditions, which of the following gases effuse at approximately half the rate of NH3?
(A) O2 (B) He (C) CO2 (D) Cl2
|rateNH3/rateA = (MMA/17)½ |
|22 = MMA/17 ∴ MMA = 4(17) = 68 MMCl2 = 71 |
14. What is the partial pressure (in atm) of N2 in a gaseous mixture, which contains 7.0 moles N2, 2.5 moles O2, and 0.50 mole He at a total pressure of 0.90 atm.
(A) 0.13 (B) 0.27 (C) 0.63 (D) 0.90
|PN2 = XN2Ptot = 7.0/(7.0 + 2.5 + 0.50)(0.90 atm) |
|PN2 = (0.70)(0.90 atm) = 0.63 atm |
Questions 15-16 refer to the following gases at 0°C and 1 atm.
(A) Ne (B) Xe (C) O2 (D) CO
15. Has an average atomic or molecular speed closest to that of N2 molecules at 0°C and 1 atm
|Speed is related to MM (u = (3RT/MM)½) ∴ MM similar to N2 will have the |
|same speed. MMN2 = 28 = MMCO |
16. Has the greatest density
|at STP density is proportional to MM (MM = dRT/P). The molecule/atom |
|with the greatest MM is Xe. |
17. A 2-L container will hold about 4 g of which of the following gases at 0oC and 1 atm?
(A) SO2 (B) N2 (C) CO2 (D) C4H8
|At STP one mole of gas = 22.4 L. |
|MM/4 g = 22.4 L/2 L MM = 44.8 g/mol ∴ CO2 (MM = 44) |
18. Which is the same for the structural isomers C2H5OH and CH3OCH3? (Assume ideal behavior.)
(A) Gaseous densities at STP
(B) Vapor pressures at the same temperature
(C) Boiling points
(D) Melting points
|C2H5OH (H-bond) is more polar than CH3OCH3 (dipole) ∴ (B), (C), (D) |
|would be different. density = STP. |
19. As the temperature is raised from 20oC to 40oC, the average kinetic energy of Ne atoms changes by a factor of
(A) ½ (B) (313/293)½
(C) 313/293 (D) 2
|k = 3/2RT, 40oC + 273 = 313 K, 20oC + 273 = 293 K |
|∴ E313/E293 = T313/T293 = 313/293 |
20. The system shown above is at equilibrium at 28°C. At this temperature, the vapor pressure of water is 28 mm Hg.
[pic]
The partial pressure (in mm Hg) of O2(g) in the system is
(A) 28 (B) 56 (C) 133 (D) 161
|PO2 + 28 mm Hg = 161 mm Hg ∴ PO2 = 133 mm Hg |
21. The partial pressure of toluene is 22 mm Hg and that of benzene is 75 mm Hg in a mixture of these two gases. What is the mole fraction of benzene in the gas mixture?
(A) 0.23 (B) 0.29 (C) 0.50 (D) 0.77
|Pbenzene = XbenzenePtot |
|Xbenzene = 75 mm Hg/(75 mm Hg + 22 mm Hg) ≈ .75 |
22. In which of the processes are covalent bonds broken?
(A) I2(s) → I2(g) (B) CO2(s) → CO2(g)
(C) NaCl(s) → NaCl(l) (D) C(diamond) → C(g)
|(D), covalent network = covalent bonds throughout. (A), (B) = |
|molecular, (C) = ionic. |
23. Of the following compounds, which is the most ionic?
(A) SiCl4 (B) BrCl (C) PCl3 (D) CaCl2
|Ionic compounds are cation + anion. Metals form cations, not nonmetals.|
|Ca is the only metal ∴ CaCl2. |
24. Which of the following oxides is a gas at 25°C and 1 atm?
(A) Rb2O (B) N2O (C) Na2O2 (D) SiO2
|Ionic (A), (C) and covalent network (D) are solids, Molecular (B) can be|
|gaseous. |
25. Which of the following has the highest melting point?
(A) S8 (B) I2 (C) SiO2 (D) SO2
|Covalent network (C) has high melting points, but molecular (A), (B), |
|(D) tend to have low melting points. |
Free Response (calculator)
1. Explain why a helium filled balloon expands in each case.
a. Additional helium is added.
|Additional atoms strike the walls more frequently, increasing the |
|internal pressure, which causes the balloon to expand until pressure is |
|equalized. |
b. The temperature is increased.
|Hotter atoms move faster and strike the walls with more force and more |
|often. This increases the internal pressure, which causes the balloon |
|to expand until pressure is equalized. |
2. 5.00 g of dry ice (solid CO2) is placed in a flexible balloon and allowed to warm to room temperature (22oC).
a. How many moles of CO2 are in the balloon?
|5.00 g CO2 x 1 mol/44.0 g CO2 = 0.114 mol CO2 |
b. What is the volume of CO2 at 1.04 atm pressure?
|PV = nRT |
|(1.04)V = (0.114)(0.0821)(22.0 + 273) ∴ V = 2.65 L |
3. The vapor pressure of solid iodine, I2, at 30oC is 0.47 torr.
a. How many moles of iodine will sublime into an evacuated 1.0-L flask?
|PV = nRT |
|(0.47/760)(1.0) = n(0.0821)(30 + 273) ∴ n = 2.5 x 10-5 mol |
b. If 2.0 x 10-5 mol of I2 is used, what is the final pressure?
|PV = nRT |
|P(1.0) = (2.0 x 10-5)(0.0821)(30 + 273) = 5.0 x 10-4 atm |
c. If 3.0 x 10-5 mol of I2 is used, what will the final pressure be? Explain your answer.
|0.47 torr (maximum vapor pressure at 30oC). |
4. a. 3.327 g of an unknown gas occupies 1.00-L at 25oC and 103 kPa. What is the molar mass of the gas?
|MM = mRT/PV |
|MM = (3.327)(8.31)(298)/(103)(1.00) = 80.0 g/mol |
b. What is the density of this gas at STP (standard temperature—0oC, and pressure—1 atm)?
|MM = dRT/P |
|80.0 g/mol = d(0.0821)(273 K)/1 atm ∴ d = 3.56 g/L |
c. Which noble gas would have twice the effusion rate?
|rA/rB = (MMB/MMA)½ |
|2/1 = (80/MMA)½ → MMA = 20 g/mol ∴ Ne |
5. N2 with a volume of 200. mL, pressure of 99.7 kPa, and temperature of 27.0oC is mixed with O2 and transferred to a 750.-mL container at 27.0oC. The total pressure of the mixture is found to be 90.4 kPa, at 27.0oC.
a. Calculate the moles of N2.
|PV = nRT |
|(99.7)(0.200) = (n)(8.31)(300) ∴ n = 0.00800 mol |
b. Calculate the total moles of gas.
|PV = nRT |
|(90.4)(0.750) = (n)(8.31)(300) ∴ n = 0.0272 mol |
c. Calculate the partial pressure of each gas.
|PN2 = XAPtot = (0.00800/0.0272)(90.4) = 26.6 kPa |
|Ptot = PO2 + PN2 |
|90.4 kPa = PO2 + 26.6 kPa ∴ PO2 = 63.8 kPa |
6. Explain why methane gas does not behave as an ideal gas at low temperatures and high pressures.
|Low temperature molecules form clusters, which produce fewer collisions |
|and less pressure than ideal. High pressure molecules occupy a |
|significant amount of volume, which is unaccounted for in an ideal gas. |
7. In addition to dispersion, what type of forces would you expect in the following molecules?
|C2H2 |NO2 |NH3 |CF4 |
|none |dipole |H-bond |none |
8. Rank butane, ethane, methane and propane from lowest boiling point to highest.
|methane < ethane < propane < butane |
9. Consider the following heating profile for C5H12.
|36oC | | | | D | |
| | | | |E | |
| | | | | | |
| | | | | | |
| | | | | | |
|-130oC | | B | | | |
| | |C | | | |
| | | | | | |
| Heat Added (J) |
a. Highlight the correct answer.
|Which is greater? |Cs |Cl |
|Which is greater? |ΔHfus |ΔHvap |
|Which interval is melting? |B-C |D-E |
|Which is the boiling point? |-130oC |36oC |
b. How much heat is needed to vaporize 5.0 g of liquid C5H12 from 20.oC to 36oC?
(Cl = 120. J/mol•K, ΔHvap = 26.2 kJ/mol)
|Q = nClΔT + nΔHvap |
|Q = (5.0/72)[(120)(16) + 26200] = 2.0 x 103 J |
10. 10 g of water is added to a 10.0-L container filled with dry air at 20oC (PH2O = 20 torr). The container is sealed.
a. How many grams of the water will evaporate?
|PV = nRT |
|(20 x 0.133)(10.0) = n(8.31)(20 + 273) |
|n = 0.0109 moles x 18.0 g/1 mole H2O = 0.197 g |
b. Would the amount of water that evaporates increase (↑), remain the same (=) or decrease (↓) for the following changes?
| |↑ |= |↓ |
|Use a 5.0 L container | | |x |
|Use humid air | | |x |
|Raise the temperature to 25oC |x | | |
|Add 20.0 g of water | |x | |
11. Consider the following phase diagram.
[pic]
Answer the following questions.
|What is the normal melting point? |35oC |
|Which state is the densest? |solid |
|What phase change occurs when going from 40oC to |deposition |
|0oC at 0.5 atm pressure? | |
|What phase change occurs when going from 0.50 atm|condensation |
|to 1.0 atm at 60oC? | |
12. Consider the following solids.
a. Rank the solids from highest melting point (1) to lowest.
|CH4 |H2O |MgO |Na |NaCl |SiO2 |
|6 |5 |2 |4 |3 |1 |
b. Justify your relative ranking of CH4 and H2O.
|H2O is higher because its bonding is stronger ( H-bond + dispersion) |
|compared to CH4 (dispersion). |
c. Justify your relative ranking of MgO and NaCl.
|MgO is higher because of higher ionic charges (2+ and 2-) compared to |
|NaCl (1+ and 1-), and smaller ionic radii compared to NaCl's. |
13. Explain the following observations.
a. NH3 boils at 240 K, whereas NF3 boils at 144 K.
NH3 has dispersion forces and hydrogen-bonding. NF3 has dispersion and dipole forces. The higher boiling point for NH3 is due to the greater strength of the hydrogen-bonding between NH3 molecules.
b. At 25°C and 1 atm, F2 is a gas, whereas I2 is a solid.
The intermolecular force for both F2 and I2 is London dispersion. The strength of London forces is related to the number of electrons surrounding the atom. Thus, I2 with more electrons forms stronger bonds and is a solid.
14. A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g of O2(g) at 298 K.
a. Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.
|24.5 g N2 x 1 mol/28.0 g = 0.875 mol N2 |
|28.0 g O2 x 1 mol/32.0 g = 0.875 mol O2 |
|PV = nRT |
|P(5.00) = (1.75)(0.0821)(298) ∴ P = 8.56 atm |
b. Calculate the following for N2(g).
(1) mole fraction
|XN2 = mol N2/total moles = 0.875/1.75 = 0.500 |
(2) partial pressure
|PN2 = XN2Ptot = (0.500)(8.56 atm) = 4.28 atm |
c. If the cylinder develops a pinhole-sized leak and some of the gaseous mixture escapes, would the ratio moles of N2(g) to moles O2(g) in the cylinder increase, decrease, or remain the same? Justify your answer.
|N2 will effuse faster than O2 because N2 has a lower molar mass ∴ the |
|moles of N2 will decrease faster than O2. |
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