Review Session Chem202 - Illinois Wesleyan Universitygenchem/reviewsession1_key.pdfReview Session...
Transcript of Review Session Chem202 - Illinois Wesleyan Universitygenchem/reviewsession1_key.pdfReview Session...
Review Session Chem202Gases, solids, liquids
Rank the following pairs of compounds withincreasing boiling points.
Give the dominant intermolecular force.
CH3-CH2-O-H CH3-O-CH3
-25 °C
Dipole-Dipole interactionNo hydrogen bonding
78 °C
Hydrogen bonding
Rank the following pairs of compounds withincreasing boiling points.
Give the dominant intermolecular force.
H2C
H2C
H2C CH2
CH2O
H2C
H2C CH2
CH2
cyclopentane tetrahydrofuran
49 °C
London Dispersion
65 °C
Dipole-Dipole
Rank the following compounds with increasingboiling points.
Give the dominant intermolecular force.
NH3C CH3
CH3
NH3CH2C H
CH3
NH3CH2CH2C H
H
37 °C
Onehydrogenbond
49 °C
Two hydrogenbonds
4 °C
No hydrogenbonds
Rank the following compounds with increasingboiling points.
Give the dominant intermolecular force.
H3C
H2C
CH2
H2C
CH3H3C
CHCH2
CH3H3C C CH3
CH3 CH3
CH3
10 °C
spherical
36 °C
linear
28 °C
branched
London-Dispersion forces increase with increasing surface area
Estimate the pressure (inatmospheres) inside a container,
given that its volume is 5.0 L, T = 23°C and it contains 0.010 mg
Nitrogen gas.
!
P =nRT
V=1.0•10
"5g
28.02gmol"1
•(8.206•10
"2LatmK
"1mol
"1)•298K
5.0Lit=1.7•10
"6atm
The composition of gases in container (V= 0.75 L) is 0.10g Neon and 0.20 g Xe. Calculate their partial pressuresand the total pressure (in atmospheres) at 40 °C.
!
n(Ne) =0.10g
20.180g /mol= 0.0050mol
n(Xe) =0.20g
131.2g /mol= 0.0015mol
ntot = 0.0050mol + 0.0015mol = 0.0065mol
Ptot =nRT
V=
0.0065mol•0.08206Latm
molK• 313.15K
0.75L= 0.22atm
PNe = Ptot • XNe = 0.22atm •0.0050mol
0.0065mol= 0.17atm
PXe = 0.052atm
Potassium superoxide (KO2) reacts with carbondioxide and releases oxygen:
4 KO2 + 2 CO2 2 K2CO3 + 3 O2
Calculate the mass of KO2 needed to react with 50 Lit ofcarbon dioxide at 25 °C and 1.0 atm.
!
nCO2 =PV
RT=
1atm " 50Lit
0.08206Lit " atm
molK•298.15K
= 2.04mol
2mol CO2
= 4mol K2O
nK2O = 4.08mol
mK2O= 4.08mol• 71.1g /mol = 290g
At 177 °C and 200 Torr, a sample ofdiallyldisulfide vapor has a density of 1.04 g/Lit.
What is the molar mass of diallyldislufide?
!
P =200Torr
760Torr•1atm = 0.263atm
PV = nRT
V
n=RT
P=
0.08206Lit atm
molK• 450K
0.263atm=140
L
mol
M =140Lit
mol•1.04
g
Lit=146
g
mol
Which compound has the higherboiling point?
Cl
Cl
Cl
Cl
180 °C 174 °C (dipolemoments cancel)
Which compound has thehigher viscosity?
Br Br
Br
Br
Greater intermolecular forces (permanent dipolemoment). Greater intermolecular forces increase theviscosity.
Phase Diagram of UF6
0 275 °C13570 200
1 psi
10
100
1000
90 psi
• What is the physical stateof UF6 at 35 °C and 100psi? Solid•Describe the changes in thephysical state, if UF6 isheated from 0 to 70 °C at 1psi. Sublimation• Describe the changes inthe physical state, if UF6 isheated from 0 to 275 °C at90 psi.Melting, then boiling•At 90 °C the pressure of aUF6 sample is increasedfrom 1 psi to 100 psi. Doesthe physical state change?Yes: Deposition•What can you say about thepressure-dependence of themelting point? independent
A sample consists of 8.00 kg of gaseousnitrogen and fills a 100-Lit flask at 300 °C.What is the pressure of the gas, using the vander Waals equation? What pressure would bepredicted by the ideal gas equation?
a = 1.3909 atm·Lit2mol-2b = 0.03913 Lit·mol-1
!
(P + an2
V2)(V " nb) = nRT
(P + an2
V2) =
nRT
(V " nb)
P =nRT
V " nb" a
n2
V2
M(N2) = 28gmol"1
n(N 2) = 8.00 #103g1molN2
28gN2
$
% &
'
( ) = 286mol
T = 573K
P =
286mol0.08206Lit atm
molK573K
100Lit " 286mol 0.03913Lit
mol
"1.390atmL
2
mol2
(286mol)2
(100Lit)2
=140atm
Ideal Gas :P =nRT
V=
286mol0.08206Lit atm
molK573K
100Lit=134 atm
1. Rearrange van der Waals
2. Calculate numbers of mols
3. Calculate absolute temperature
4. Plug values invan der Waals equation
5. Apply Ideal Gas law