CHEM 35
General Chemistry
EXAM #3
November 15, 2000
Name: Largo, Key
SSN:
Lab T.A.:
INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the
questions. For questions involving calculations, show all of your work -- HOW
you arrived at a particular answer is MORE
important than the answer itself! Circle your final answer to numerical
questions.
The entire exam is worth a total of 150 points.
Attached are a periodic table and a formula sheet jam-packed with useful stuff.
Good Luck!
Page |
Possible Points |
Points Earned |
2 |
30 |
30 |
3 |
30 |
30 |
4 |
15 |
15 |
5 |
30 |
30 |
6 |
20 |
20 |
7 |
25 |
25 |
TOTAL: |
150 |
150 |
1. For the
Hydrogen atom:
a. (10 pts) Calculate
the energy change (in Joules) associated with an n=2 to n=4 electronic
transition.
DE = RH(1/(ni)2
- 1/(nf)2)
= (2.1798741 x 10-18
J)(1/(2)2 – 1/(4)2)
= (2.1798741 x 10-18
J)(1/4 – 1/16)
= (2.1798741 x 10-18
J)(0.1875)
= 4.0872639 x 10-19
J
= 4.0873 x 10-19
J
b. (5 pts) Is a
photon emitted or absorbed as a result of this transition?
EMITTED ABSORBED circle one
c. (10 pts) Calculate
the wavelength (nm) of the photon having the same energy as the energy change
associated with this transition.
E = hc/l ® l
= hc/E
l
= (6.62606876 x 10-34 J-s)(2.9979 x 108 m/s)
4.0872639 x 10-19
J
= 4.860046 x 10-7
m
= 4.860046 x 10-7
m x 109 nm
m
= 4.860 x 102
nm or 486.0 nm
d. (5 pts) Is the
photon in the visible portion of the electromagnetic spectrum?
YES NO circle one
2. (5 pts
each) Write the ground state electron configurations for the
following atoms and ions (use noble gas abbreviations for core electrons where
appropriate):
a. Na: [Ne]3s1
b. Co: [Ar]4s23d7
c. Cu3+:
[Ar]3d8
d. Au: [Xe]4f146s15d10
3. (10 pts) Give the
values of the 4 quantum numbers (n, l, ml, and ms) for
each of calcium’s two valence electrons.
Ca: [Ar]4s2 Þ n =
4, l = 0, ml = 0, ms = +½
n = 4, l = 0, ml = 0, ms = -½
4. (5 pts
each) Circle the number next to the appropriate response for
each of the following:
a. That no
two electrons in an atom can possess the same set of 4 quantum numbers is a
consequence of:
1. The
Aufbau Principle
2. The
Heisenberg Uncertainty Principle
3. Hund’s
Rule
4. Pauli
Exclusion Principle
5. Robert’s
Rules of Order
b. Einstein’s
explanation of the photoelectric effect:
1. illustrated
the wave properties of matter
2. inspired
Thomas Edison to invent the light bulb
3. applied
Planck’s quantum theory of electromagnetic radiation
4. utilized
his newly developed theory of relativity
5. was
inspired by his observations of how a violin string vibrates
c. The Bohr
model of the atom:
1. was first
proposed by Balmer more than 30 years before Bohr
2. accurately
predicts the line emission spectrum for the hydrogen atom
3. is based
on the wave properties of the electron in a hydrogen atom
4. was
originally made out of balsa wood using simple hand tools in Bohr’s garage
5. is also
known as the “plum pudding” model of the atom
5. (5 pts
each) For the following, circle the species in each row with the
desired property:
a. largest
atomic radius Na Al S Ar K
b. smallest
radius Al3+ O2- F- Ne Na+
c. greatest
electron affinity F Cl Br I At
d. smallest
1st ionization energy
B Al C Si P
6. (10 pts) Recall
the thermite reaction:
2 Al(s)
+ Fe2O3(s) ® Al2O3(s) + 2 Fe(s)
This
highly exothermic reaction is used for welding massive objects, such as
propellers for large ships. Using the standard molar enthalpies of formation
given below, calculate DHo
(kJ) for this reaction.
DHof (Fe2O3(s))
= -822.16 kJ
DHof
(Al2O3(s)) = -1669.8 kJ
DHo
= [(-1669.8 kJ) + 2(0)] – [(-822.16 kJ) + 2(0)]
= -847.64 kJ
= -847.6 kJ
7. (10 pts) The
specific heat of copper metal is 0.385 J/g-K. How many Joules of heat are
necessary to raise the temperature of a 1.42-kg block of copper from 25.0 oC
to 88.5 oC?
m
= 1.42 kg x 1000 g = 1.42 x 103 kg
kg
DT
= 88.5 – 25.0 oC = 63.5 oC = 63.5 K
cs
= 0.385 J/g-K
q
= mcsDT = (1.42 x 103 kg)(0.385
J/g-K)(63.5 K)
= 3.4715 x 104 J
=
3.47 x 104 J
8. (10 pts) Given the
following data:
N2(g)
+ O2(g) ® 2 NO(g)
DH = +180.7
kJ
2 NO(g)
+ O2(g) ® 2 NO2(g) DH = -113.1 kJ
2 N2O(g)
® 2 N2(g)
+ O2(g) DH =
-163.2 kJ
use
Hess’s law to calculate DH (kJ)
for the following reaction:
N2O(g) + NO2(g)
® 3 NO (g)
½
Rxn 3: N2O(g)
®
N2(g) + ½O2(g) DH
= ½(-163.2) = - 81.60 kJ
-½
Rxn 2: NO2(g) ®
NO(g) + ½O2(g)
DH = -½(-113.1) = + 56.55 kJ
Rxn 1: N2(g) + O2(g)
® 2 NO(g) DH
= +180.7 = + 180.7 kJ
N2O(g) + NO2(g)
® 3 NO (g) DH
= +155.65 kJ
= +155.6 kJ
9. For N2:
a. (5 pts) Write the ground-state
electron configuration, using the appropriate building-up sequence.
7 x 2 = 14 e-
N2: (s1s)2(s*1s)2(s2s)2(s*2s)2(p2p)4(s2p)2
b. (10 pts) Show the orbital populations
on a molecular orbital energy diagram.
For the
p-electrons:
s*2p
p*2p
2p
2p
¯ s2p
¯
¯
p2p
N N2 N
c. (5 pts) What is
its bond order?
bo = ½(6 – 0) = 3
d. (5 pts) Is the
system paramagnetic? Why/why not?
All
of the electrons are paired, so the system is
not paramagnetic.