Exam 2 Key - ars

Chemistry 101 - Exam II
12 November 2014
Name ______________________
Show all work for credit. State any assumptions made to solve a problem. Give all
numerical answers with the correct number of significant figures. All answers in
scientific notation must be in correct scientific notation (i.e., 6.0221023 not
6.022E23 or 6.022e23). Each instance of incorrect scientific notation will result in
the loss of 3 points.
1. (18 points) Textbook Problem 7.89 - The iodine molecule can be photodissociated into
iodine atoms in the gas phase with light of wavelengths shorter than about 792 nm. A
100.0 mL glass tube contains 55.7 mtorr of gaseous iodine at 25.0 C. What minimum
amount of light energy must be absorbed by the iodine in the tube to dissociate 15.0% of
the molecules? By the way, we did this one in class last week.
The energy of each photon is given by the wavelength.
6.626 × 10 −34 J s 2.998 × 10 8 m s−1
hc
= =
= 2.50818 × 10 −19 J
Ephoton
λ
 10 −9 m 
792 nm 

 1 nm 
We need 1 photon for each I2 molecule. The number of I2 molecules is:
10−3 torr
10−3 L
6.022 × 1023 I2 molecules 15 I2 molecules dissociated
PV ( 55.7 mtorr ) 1 mtorr (100.0 mL ) 1 mL
NI2 =
=−1 −1
×
×
RT
1 mol I2
100 I2 molecules
( 62.364 L torr mol K ) (298.2 K )
(
)(
(
)
)
(
)
= 2.7054 × 1016
The total energy is:
NI2 × E photon = 2.7054 × 1016 × 2.50818 × 10 −19 J = 6.79 × 10 −3 J
2. (23 points) Decanoic acid, also known as capric acid because it smells like goats, (C10H20O2)
has an enthalpy of combustion of is -1453.0 kcal mol-1. What is the enthalpy of formation,
ΔHf, of capric acid in kJ mol-1?
C10H20O2 + 14 O2  10 CO2 + 10 H2O H = -1453.0 kcal
=
∆H
∑
products
n∆H f −
∑
reactants
n∆H f
  14 mol O2   0.00 kJ  
x kJ
 4.184 kJ   10 mol CO2   −393.5 kJ   10 mol H2 O   −285.8 kJ    1 mol C8H16 O2  
−1453.0 kcal mol −1 
 =  mol rxn   mol CO  +  mol rxn   mol H O   −  mol rxn   mol C H O  +  mol rxn   mol O  
1
kcal

 






2 
2
8 16 2 
2 
  










14
mol
O
10
mol
CO
10
mol
H
O
−
−
0.00
kJ
4.184
kJ
393.5
kJ
285.8
kJ








2
2
2
x =−  −1453.0 kcal mol −1 
+
 − 

 − 




 1 kcal   mol rxn   mol CO2   mol rxn   mol H2 O    mol rxn   mol O2  

= −713.6 kJ (mol C10H20 O2 )
−1
3. (21 points) Write the electron configuration and orbital diagram for the following elements
using the noble gas shorthand.
a. molybdenum
[Kr] 5s1 4d5
[Kr]  
5s
4d
b. protactinium
[Rn] 7s2 5f2 6d1
[Rn]   
7s
5f
6d
c. tellurium
[Kr] 5s2 4d10 5p4
[Kr]   
5s
4d
5p
4. (20 points) Balance the following reaction and estimate its enthalpy of reaction (in kJ) using
bond energies.
H
H
S
H
C
H
C
C
C
H
H
O
N
H
4
Broken:
8 N-H (386 kJ)
4 C-N (305 kJ)
8 C-C (346 kJ)
16 C-H (411 kJ)
4 C=C (614 kJ)
4 C-S (289 kJ)
4 C=O (745 kJ)
4 S-H (364 kJ)
25 O=O (494 kJ)
+ 25 O2  16 CO2 + 14 H2O + 2 N2 + 4 SO2
Formed:
32 C=O (799 kJ)
28 H-O (459 kJ)
2 NN (942 kJ)
2 S-O (265 kJ)
2 S=O (532 kJ)
∆H ≈ ( 8 )( 391 kJ) + ( 4 )( 286 kJ) + ( 8 )( 346 kJ) + (16 )( 414 kJ) + ( 4 )( 614 kJ) + ( 4 )( 289 kJ) + ( 4 )( 745 kJ) + ( 4 )( 364 kJ) + ( 25 )( 498 kJ) 
− ( 32 )( 799 kJ) + ( 28 )( 463 kJ) + ( 2 )( 942 kJ) + ( 4 )( 265 kJ) + ( 4 )( 532 kJ) 
=
−9 448 kJ ( or − 10 516 kJ if you did 2 double bonds in SO2 )
5. (22 points) In a calorimetric experiment, 0.6654 g of liquid o-xylene, C8H10, was burned
completely to make carbon dioxide gas and liquid water. The temperature of the water rose
from 26.38C to 28.66C. The heat capacity of the calorimeter and its contents is 12.53 kJ/C.
What is E and H for the combustion process?
C8H10 + 21/2 O2 8 CO2 + 5 H2O
qcal = C∆T = (12.53
kJ

) ( 28.66 C − 26.38 C ) = 28.56840 kJ

C

qrxn = −qcal
q
−28.56840 kJ
∆E = rxn =
=−4558.106 kJ
=−4.56 × 103 kJ
mol
mol
1
mol
C
H
1
mol
rxn
nrxn
8 10
×
0.6654 g C8H10 ×
106.1650 g C8H10 1 mol C8H10
∆H = ∆E + ∆nRT
=
−4558.106 kJ  −2.5 mol gas   8.3145 × 10 −3 kJ 
+
−4564.304 =
−4.56 × 10 3 kJ
 ( 298.15 K ) =
  mol gas K
mol
mol rxn
mol
rxn



6. (22 points) For each of the following ions, draw the electron dot structure with the optimum
formal charges, state the electron group and molecular geometries, determine the polarity,
state the hybridization on the central atom and the number of  and  bonds.
a. I3-
b. IF4+
-
+
I
I
I
EGG = trigonal bipyramidal
MG = linear
non-polar
sp3d
2 , 0 
F
F
I
F
F
EGG = trigonal bipyramidal
MG = see-saw
polar
sp3d
4 , 0 
7. (24 points) The hypophosphite ion is 47.66% P, 3.10% H, and 49.24% O. The molar mass of
ion is 64.989 g/mol. Find the molecular formula of the ion and draw its Lewis structure if all
of the hydrogens are attached directly to the phosphorus.
1 mol P
×
=
47.66 g P=
1.5388 mol P/1.5388
mol 1.000
30.973 g P
1 mol H
=
3.10 g H ×
3.0756 mol H/1.5388 mol =
1.9987 =
2
1.00794 g H
1 mol O
=
49.24 g O ×
3.0776 mol O/1.5388 mol =
2.0000 =
2
15.9994 g O
The empirical formula is PH2O2 which has an empirical mass of 64.988 g mol-1.
Molar mass
64.989 g mol−1
=
= 1.0000
= 1
Empirical mass 64.988 g mol−1
So the molecular formula is the same as the empirical formula, PH2O2. It has the Lewis structure:
H
P
O
O
H
The charge has to be 1- because of the way the formal charges add together.

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