Chemistry 114
Second Hour Exam
Name:____________
1. (10 points) When X-rays of 1.54 Å were diffracted off a crystal of a substance, a first order
diffraction peak was seen at 18.32o. What is the interplanar spacing in this crystal.
n8 =2d sin 2
1(1.54Å) = 2d sin(18.32o)
1.54/2 = d(.3143)
1.54/(2 ×.3143) = d
d=2.45Å
2. (10 points) Compare and contrast the type of bonding used un making a diamond with the bonding
used in making a lead metal. What physical properties of these solids are explained by this bonding.
Note the use of lead metal NOT pencil lead (graphite)
Diamond - atomic network solid - uses direction covalent bonds to hold atoms in place. Strong
standard covalent bond lock all atoms and electrons into the lattice make this a very strong material.
The direction covalent bonds to not conduct electricity and do not allow the material to be bent or
pulled into a wire.
Lead(metal) - Metallic atomic solid - uses nondirectional covalent bonds to hold atoms in
lattice. These non-directional bonds allow electrons to flow freely within the lattice to make this
material a good conductor of electricity. The bonds may also be distorted and reformed easily, so the
material may be bent or pulled into a wire.
3. (10 points) Diethyl ether has a vapor pressure of 300 torr at 10o C, and 480 torr at 20oC. What is
the )H vaporization of this substance?
ln
VP2 ∆ H  1 1 
=
 − 
VP1
R  T1 T2 
 300 torr 
∆H
1
1 

ln
 =
−


 480 torr  8.3145 J / K ⋅ mol  273 + 20 273 + 10 
∆H
( .003413− .003534 )
8.3145
− 0.47(8.3145 J / K ⋅ mol ) = ∆ H ( − .000121K −1 )
ln(.625) =
0.47(8.3145)/.000121 = ∆ H vap
∆ Hvap = 3.23x104 J / mol or 32.3kJ / mol
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4. (10 points) Below is a phase diagram of a substance.
What form is this substance in at 0.1 atm
pressure and 300o K ?
__gas_________
What form is this substance in at 1 atm
pressure and 200o K ?
___solid______
What form is this substance in at 10 atm
pressure and 200o K?
_______solid__
Where is this material’s triple point?
About 300K, 3atm
What is this material’s critical temperature
and pressure?
About 400K, 9atm
5. ( 10 points) I am going to make a solution with 50 ml of water (density 1.00 g/ml) and 10 ml of
methanol (CH3OH, density .791) Assume the final solution has a volume of 59.5 ml. Calculate the
Molarity of the solution
Molarity = mole solute/liter solution
moles solute?? g solute 10 ml ×.791g/ml =7.91 g
MW solute = 12+4+16 = 32
Mole solute = 7.91g/32 g/mol
=.247 mole
Molarity = .247 mol/.0595 L = 4.15 M
Molality of the solution
Molality = mole solute/kg solvent
= .247 mole/.05 kg
=4.94m
Mole fraction of methanol in the solution
OmeOH = mole MeOH/total moles
moles H2O = 50 ml × 1 g/ml × 1/18g/mol
=2.78 mol
OmeOH = .247/(.247+2.778) = .0816
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6. The vapor pressure of water at room temperature is 25 torr, while the vapor pressure of methanol is
19 torr. What is the vapor pressure of the solution made in Problem 5 at room temperature. (Assuming
that it is an ideal solution)
VPsoln = VP 1 + VP 2
= O1VP01 + O2 VP02
2 component system so 1= O1 + O2
O1 = Methanol (last problem) = .0816
O2 = 1-.0816 = .9184
=.0816(19 torr) + .9184(25 torr)
= 24.5 torr
When I mix this solution I find it gets slightly warm. Why would this occur? What does this tell
you about the vapor pressure you calculated in part A of this problem
If the solution gets warm when I mix it, this implies a negative )H of solution, which implies that
a favorable interaction occurs between the solute and the solvent. This interaction means the these
molecules will be attracted to each other, and won’t be as likely to move into the vapor where they
can’t interact. This should result in a vapor pressure that is LOWER than expected, and a NEGATIVE
deviation from Raoult’s law.
7. Benzene has a boiling point of 80 oC and a freezing point of 5.5 oC. It has a Kb of 2.53 oC@kg/mol
and a Kf of 5.12 oC@kg/mol. If I dissolve a material in benzene that gives the solution a freezing point of
2.5oC, what will the boiling point of this solution be? (Assume 1 atm of pressure)
)TFP = Kfm; )TBP = KB m
Get m from freezing point depression and plug into equation for boiling point elevation
)TFP = 5.5-2.5 = 3.0oC = 5.12(m)
m = 3.0/5.12 = .586 mol/kg solute
)TBP = 2.53 oC@ kg/mol × .586 mol/kg
= 1.5 oC
Boiling point gets elevated so new boiling point is
80+1.5 = 81.5oC
8. What is a van’t Hoff factor? If I dissolved iron(III) chloride in water, what would you expect its
van’t Hoff factor to be? What are van’t Hoff factors frequently lower than theoretical values?
The van’t Hoff factor is a number that tells you how many ions (or particles) are made when 1
mole of salt is dissolved in a solution. For FeCl3 this would be 4
van’t Hoff factors are frequently lower than expected because some ion pairing occurs in solution. This
ion pairing keeps ions from totally dissociating and acting as totally independent particles.
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9. Use the method of initial rates to determine the rate constant and order parameters for the following
reaction:
X+Y 6Z
Experiment
1
2
3
Initial [X]
.05
.1
.1
Initial [Y]
.05
.05
.1
measured rate (mol/L@s)
.002112
.004224
.009055
For component X
.004224
k (.1) a (.05) b
.1a
 .1 
=

a
b =
a = 
 .05
.002112 k (.05) (.05)
.05
a
2 = 2a ; a = 1
For component Y
.009055
k (.1)1 (.1) y
.1y
 .1 
=
=
=
 
.004224 k (.1) 1 (.05) y .05 y  .05
y
2.144 = 2 y
log( 2.144 ) = y log( 2)
log( 2.144 ) / log(2) = y
.3312/.301 = y = 11
.
k
.002112 = k (.05)1(.05)1.1
.002112 = k(.05).0371
k = .002112/(.05×.0371) = 1.14 l/mol@sec
10. I have a first order reaction with a half-life of 7 minutes. That is, 50% of the reactant is used up
after 7 minutes of reaction time. If I start with 0.5 M reactant, and let the reaction run for exactly 15
minutes, how much of the reactant is used up?
For first order reaction T1/2 = .693/k
7 min = .693/k
k = .693/7 min
k = .099 min-1
Using this k in the integrated rate equation
ln [At] = -kt + ln[A0]
ln[X] = -.099(15) + ln(.5)
= -1.485 + -.693
= -2.178
-2.178
[X] = e
=.113 M remains )TBP
.5-.113 = .387 moles of reactant are used up
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