NAME __________________
CHEMISTRY 162 Test#4w15
Score ____/63
PART I Multiple Choice (5 pts)
Multiple Choice(5 pts) B – D – A – B – B
PART II Qualitative (13 pts)
1. In each pair complete the following 3 parts; i) identify ALL the intermolecular forces present
ii) circle the substance in each pair with the higher boiling point iii) explain your choice for “ii”
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a) CH3Br or CH3F
Dispersion – Dipole
b) CH3CH2CH2OH or CH3CH2OCH3
Dispersion – Dipole – Hydrogen
Heavier mass
More surface area
2. a) a & b - 4
b) a & c - 5
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CH3C-OH
O
3. a) Sublimation b) Melting
3
c) Vaporization
SOLID
Heavier mass
c) b & c - 5
CH3C-OH
O
HO
FCH2CHF
CH3OCH2NH2
CH3OCH2NH2
HO
FCH2CHF
d) Freezing
e) Deposition
LIQUID
B-MELTING
D-FREEZING
VAPORIATION-C
P
CONDENSATION-F
SUBLIMATION-A
DEPOSITION-E
Temp
GAS
c) C2H6 or C3H8
Dispersion
f) Condensation
PART III Calculations ( pts)
1. The vapor pressure of carbon tetrachloride is 0.132 atm at 23.0oC and 0.526 atm at 58.0oC.
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a) Calculate the Hvap in this temperature range
4
R = 8.21 J/mol-K
ln
P1 = 0.132 atm
P2 - H vap  1 1 

 - 
P1
R  T2 T1 
P2 = 0.526
ln
1.38 
T1 = 273.15 + 23.0 = 296.15 K T2 = 331.15 K
0.526 - H vap  1
1 



0.132
8.31  331.15 296.15 
- H vap
0.000357
8.31
- H vap 
(1.38  8.31)
 3.2 *`10 4 J/mol
0.000357
b) using your answer in “a” and the vapor pressure at 23.0oC, calculate the expected vapor pressure at 38.0oC.
4
R = 8.21 J/mol-K
P1 = 0.132 atm
- H vap  1 1 

ln

- 
T T 
P1
R
 2 1
P2 = X
P
3.2 *10 4  1
1 
ln X 


P1
8.31  311.15 296.15 
P2
PX
 1.886
0.132
5
2. C2H5Cl = 64.5 g/mol
3.
5
T1 = 296.15 K T2 = 311.15 K
e
 PX

 ln  0.6344
 P1

PX  (1.886) * (0.132)  0.249 atm
n = 10.0g/64.5 = 0.155 mols
Hvap = 26.4 kJ/mol
q = n*Hvap = (0.155 mol)*(26.4 kJ/mol) = 4.09 kJ
OR
= (0.155 mol)*(2.64*104 J/mol)*(1 kJ/1000 J) = 4.09 kJ
T1 = 34.9 + 273 = 308 K
T2 = 76.7 + 273 = 350 K
R = 8.314 J/mol·K
-Hvap = 40500 J/mol
Ln P2 – (-1.88849) = 1.9152855
P2 - H vap  1 1 
 - 

P1
R  T2 T1 
P2
- 40500  1
1 
Ln



0.1513
8.31  350 308 
Ln P2 - Ln 0.1513 
Ln
(-4873.6)0.002857 - 0.00325
4. Solid phase: 125 to -114 = 11 T
q = m*Csolid*T
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q = (25.0 g)(0.97 J/gK)*(11 K)
= 266.75 J
solid/liquid phase: no T
(25.0 g)*(mol/46.0g) = 0.543 mol
q = n(Hfus) endothermic
q = (0.543 mol*(5.02 kJ/mol)
= 2.75 kJ
TOTAL E = 266.8 + 2750 + 9717.5 = 12734.3 J
Ln P2 = 1.9152855 - 1.88849
Ln P2 = 0.0267955
P2 = exp^0.0267955 = 1.027
P2 = 1.02 atm
Liquid phase: 144 to 55 = 169 T
q = m*Cliquid*T
q = (25.0 g)(2.3 J/gK)*(169 K)
= 9717.5 J
(12.73 kJ)
5. In the heating curve diagram: 1) on the vertical axis, LABEL the location for the boiling & freezing points
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2) indicate the phase or condition of a substance and briefly explain what is happening to that substance along
the path of the line in each numbered region in the area under the diagram.
BOILING PT
o
T
FREEZING PT
1
2
3
4
5
T I M E
List descriptions of: 1- intermolecular attraction that occurring 2-behavior of atoms/ions/molecules in various phases
3-KE of molecules 4-temperature effects at various phases 5-etc
Region 1: SOLID
Region 2: SOLID-LIQUID
Region 4: LIQUID-GAS
Region 3: LIQUID
Region 5: GAS
PART IV Review Equilibrium (9 pts)
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[HBr]2
1. a) K eq 
[H 2 ][Br2 ]
[HBr]2
1
b) K eq 
[H 2 ]
[0.20]2
 0.020
2. a) K eq 
[2.00]
c) K eq 
[H 2 ]
N2O4(g) - 2 NO2(g)
b)
I.
2.00
+1.00
C. -x
E. 3.00-x
c) Kp = Kc(RT)x
x=2–1=1
0.20
+2x
0.20+2x
T = 50 + 273 = 323
K eq : 0.020 
[0.20  2x] 2
[3.00 - x]
(0.20 + 2x)2 = 0.020(3.00 – x)
4x2 + 0.82x – 0.020 = 0
use quadratic formula, solve for “x”
values = -0.227 & 0.022 x= 0.022
at equilibrium:
N2O4: 3.00 – 0.022 = 2.98 M
NO2: 0.20 + 2*0.022 = 0.244 M
Kp = 0.020(0.0821* 323)1 = 0.020*2.71 = 0.54