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