Chem 107 - Hughbanks
Final Exam, May 11, 2011
Name (Print)
UIN #
Section 503
Exam 3, Version # A
On the last page of this exam, you’ve been given a periodic table and some physical constants. You’ll
probably want to tear that page off the to use during the exam – you don’t need to turn it in with the rest
of the exam.
The exam contains 13 problems, with 9 numbered pages. You have the full 75 minutes to complete the
exam. Please show ALL your work as clearly as possible – this will help us award you partial credit
if appropriate. Even correct answers without supporting work may not receive credit. You may use an
approved calculator for the exam, one without extensive programmable capabilities or the ability to store
alphanumeric information. Print your name above, provide your UIN number, and sign the honor code
statement below.
On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam.
SIGNATURE:
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Solutions
(For grading)
Version A
Scores
(1) (10 pts) Consider the elements in the second, third, and fourth rows of the
periodic table (lithium through krypton, elements #3 through #36). Answer the
following questions by putting the symbol(s) of the element in the blanks
provided. (2 pts. each)
__Ne__
(a) Which one element has the greatest ionization energy?
__ Zn__
(b) Which one element has a 3d104s2 valence electronic configuration?
__ K__
(c) Which one element has the lowest electronegativity?
Mg, Ca
(d) List two of these elements (called ‘X’) that might form an ionic
chloride compound with the formula XCl2?
C, Si
(e) List two of these elements (called ‘X’) that might form a covalent
compound with hydrogen with the formula XH4? (Assume that the
Lewis diagram for XH4 obeys the ‘octet rule’.)
(2) (10 pts) Write ‘T’ or ‘F’ in the blanks to indicate whether the statements are true
or false. (2 pts each.)
1
2
3
4
5
6
7
8
9
10
11
12
13
Tot.
/10
/10
/8
/6
/8
/8
/7
/8
/8
/15
/20
/26
/16
/150
___F__
(a) Stronger bonds tend to be longer bonds.
___F__
(b) Exothermic reactions are always spontaneous.
___T__
(c) In the reaction MgCO3(s) → MgO(s) + CO2(g), ΔS > 0.
___F__
(d) According to the Second Law of Thermodynamics, any reaction for which ΔS < 0 can not
occur.
___T__
(e) In a reaction where no gases are consumed or formed, ΔE and ΔH are nearly equal.
(3) (8 pts) By the use of Lewis diagrams and considering hybridization at the central atom, decide which
of the following molecules or ions are bent.
(I)
NO2+ (the nitrogen atom is in the middle)
–
(II) NO2 (the nitrogen atom is in the middle)
(III) HCN
(IV) SO2 (the sulfur atom is in the middle)
(V) HCO+ (the carbon atom is in the middle)
(A) I and V
(B) II and V
(C) III and IV
(D) II and IV
(E) I, II, and IV
Ans. 3_____D_____
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Solutions
Version A
(4) (6 pts) Arrange the following sets of anions in order of increasing ionic radii:
Cl–
(a) S2–, P3–, Cl– :
O2–
(b) S2–, O2–, Se2– :
Cl–
(c) Br–, I–, Cl– :
S2–
<
S2–
<
<
Br–
P3–
<
.
Se2–.
<
I–
<
.
(5) (8 pts) Classify the following compounds as acids or bases, weak or strong, in accord with their
behavior in aqueous solution (circle the correct answer in each case).
(A) HClO4 :
weak acid
weak base
strong acid
strong base
(B) CsOH :
weak acid
weak base
strong acid
strong base
(C) H2CO3 :
weak acid
weak base
strong acid
strong base
(D) CH3CH2NH2 :
weak acid
weak base
strong acid
strong base
(6) (8 pts) Given the following data for the reaction A + B → C, which of the expressions below is a
fully correct statement of the rate law?
(A) rate =
Expt.
Initial [A]
Initial [B]
Initial Rate of
Formation of C
1
2
3
0.20 M
0.30 M
0.30 M
0.10 M
0.10 M
0.20 M
5.0 × 10–6 M/s
1.13 × 10–5 M/s
1.13 × 10–5 M/s
∆[C]
= k[A]2
∆t
(C) rate = −
∆[B]
= k[A][B]2
∆t
(E) rate = −
∆[C]
= k[B]2
∆t
(B) rate = −
(D) rate =
∆[A]
= k[A]2 [B]
∆t
∆[C]
= k[B]
∆t
Ans. 6_____A_____
2
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Solutions
Version A
(7) (7 pts) An atom in its ground state absorbs a photon with a 200 nm wavelength. The atom then
absorbs a second photon with a wavelength of 250 nm. The atom then returns to its ground state
with emission of a photon. What is the wavelength of the emitted photon?
(A) 50 nm
(B) 300 nm
(C) 111 nm
(D) 450 nm
(E) 222 nm
Ans. 7_____C_____
(8) (8 pts) Put the following molecules in order of increasing boiling point and clearly explain the
reasons that justify your answer: H2O, H2S, H2Se, H2Te.
lowest boiling point
H2 S
<
highest boiling point
H2Se
<
H2Te
<
H2 O
.
Explanation:
Hydrogen bonding in H2O causes water to have the strongest intermolecular forces, hence its
relatively high boiling point. The other three increase as the strength of the dispersion forces for the
chalcogenide increases, from sulfur to tellurium.
(9) (8 pts) Two resonance structures are shown for the cation below. How many hydrogen atoms for
this cation must lie in the “plane of the paper”?
(a) 1
(b) 3
(c) 4
(d) 5
(e) 7
Ans. 9_____C_____
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Solutions
Version A
Questions 10-12 – show all work (61 pts)
(10) (15 pts) Nitrous oxide (N2O) decomposes according to the reaction shown below. The rate of this
reaction is exceedingly slow in pure N2O, but chlorine gas accelerates it significantly.
2 N2O(g) → 2 N2(g) + O2(g)
The rate law depends on the pressures of both N2O and Cl2 and can be written as
rate = k [P(N2O)]m[P(Cl2)]n
where the partial pressures P(N2O) and P(Cl2) are used instead of concentrations.
Initial rate date were collected on this reaction at T = 800 K:
Expt.
P(N2O) (torr)
P(Cl2) (torr)
initial rate (torr min–1)
1
300
40
3.0
2
150
40
1.5
3
300
10
1.5
(a) (3 pts) What is the specific term used to describe the role of chlorine, Cl2, in this reaction?
Cl2 is a catalyst
(b) (7 pts) Determine the order of this reaction with respect to N2O and Cl2.
m =
1
n=
½
(c) (5 pts) Give the value of the rate constant, k, with the correct units.
k = (rate)/{[P(N2O)] [P(Cl2)]1/2} = (3.0 torr min–1) /{[300 torr] [40 torr]1/2} = 0.00158 torr–1/2 min–1
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Solutions
Version A
(11) (20 pts) Consider the reaction between reactants A and B to give products D and E — the energy
diagram below applies. C is an intermediate product that is not present when the reaction is over.
(Note: There are five parts to this question, part (e) is on the top of the next page.)
Step 1
Step 2
A + B ⎯⎯⎯
→ C ⎯⎯⎯
→ D+E
(a) (3 pts) Give the values of ΔE and the energy of activation (Ea) for the reactions indicated.
∆E(A + B → C) = +50 kJ
∆E(A + B → D + E) = –10 kJ
Ea(A + B → C) = +90 kJ
(b) (3 pts) Which step is rate determining (circle the correct answer here and in part c)?
Step 1
Step 2
Both
Neither
Neither
No way to tell
No way to tell
(c) (3 pts) The overall reaction is…
Exothermic
Endothermic
(d) (5 pts) Catalyst P increases the rate of the first step (A + B → C) but does not affect the second step
(C → D + E). Catalyst Q increases the rate of the second step, but does not affect the rate of the
first step. Which catalyst will be most effective in increasing the rate of the overall reaction (A + B
→ D + E)? (You must give a correct brief explanation for your answer to receive credit.)
Since the first step is rate determining, catalyst P will be more effective because that is the step it
affects.
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Solutions
Version A
(10) continued…
(e) (6 pts) The rate constant for the reaction (A + B → D + E) is 0.0072 M-1 s-1 at a temperature of 30.0
°C. Use the Arrhenius equation to calculate the rate constant at a temperature of 45.0 °C?
k = A • exp{–Ea/RT} or ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)
ln(k2/k1) = (Ea/R)(1/T1 – 1/T2) = (90000 J mol-1)/(8.314 J mol-1 K-1) • (1/(303 K) – 1/(318 K))
ln(k2/k1) = 1.685 ⇒ k2/k1 = (0.0072 M-1 s-1)e1.685 = 0.0388 M-1 s-1
(12) (26 pts) The following two reactions are involved in alternative methods of purification of iron
from its ores. Some relevant thermodynamic data is shown below.
Reaction # 1
Reaction # 2
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
∆S˚ = +15.2 J/K
2 Fe2O3(s) + 3 C(s, graphite) → 4 Fe(s) + 3 CO2(g)
∆S˚ = +557.98 J/K
Substance
∆Hf˚ (kJ mol-1)
∆Gf˚ (kJ mol-1)
S˚ (J K-1 mol-1)
Fe2O3(s)
–824.2
??
87.40
CO(g)
–110.5
–137.2
197.6
C(s, graphite)
5.740
Fe(s)
??
CO2(g)
–393.5
–394.4
213.6
NOTE: You do not need to fill in all the blanks in the table - just answer the questions below!
(a) (5 pts) Calculate ∆H˚ for the reaction #1.
∆H˚ = 2(0) + 3(–393.5) – 3(–110.5) – (–824.2) = – 24.8 kJ
Continued on next page…
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Solutions
Version A
(b) (5 pts) Calculate S˚ for solid iron.
∆S˚ = +15.2 = 2(S˚Fe) + 3(+213.6) – 3(197.6) – 87.40
= 2S˚Fe + 3(+213.6) – 3(197.6) – 87.40 = 2S˚Fe –39.4 J K-1
2S˚Fe = 39.4 + 15.2 = 54.6 J K-1 mol-1
(c) (8 pts) Calculate ∆Gf˚ for Fe2O3.
For reaction #1,
∆H˚ = – 24.8 kJ and ∆S˚ = +15.2 J/K
⇒ ∆G˚ = – 24.8 kJ – (298 K)(+.0152 kJ) = –20.27 kJ
∴ –20.27 kJ = 2(0) + 3(–394.4) – 3(–137.2) – ∆Gf˚(Fe2O3) = – 771.6 kJ – ∆Gf˚(Fe2O3)
∆Gf˚(Fe2O3) = – 771.6 kJ + 20.27 kJ = – 751.23 kJ mol-1
(d) (8 pts) At what temperatures is reaction #2 spontaneous?
∆H˚ = 4(0) + 3(–393.5) – 2(0) – 2(–824.2) = +467.9 kJ
set ∆G = 0 = ∆H – T∆S ⇒ T = ∆H/∆S = (+467900 J)/(557.98 J/K) = 839 K
Above 839 K, the process should be spontaneous.
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Solutions
Version A
(13) (16 pts) On the next page is a temperature-pressure phase diagram for uranium hexafluoride (UF6).
UF6 is an important compound used by nuclear chemists and engineers in uranium isotope
separation (of 235UF6 from 238UF6). Note that the pressure axis is logarithmic and expressed in SI
units (1.0 kPa = 1.0 kiloPascal = 1000 Pa = 1000 N m–2 = 0.01 bar = 0.009869 atm). Read the
directions below carefully before proceeding with this problem.
(a) (12 pts) There are twelve (12) missing labels on this diagram. From the following list of
words, supply the correct labels. Put the LETTER before each word (A, B, C, … Q) into the
correct box in each case. The three boxes with black borders are labels for the entire region in
which they sit. Six of the labels refer to the processes indicated by arrows next to the labels that
cross phase boundaries. There are three more labels: one of them refers to the dashed horizontal
line in the diagram, one refers to the point where the dashed line intersects one of the phase
boundaries, and the third refers to the point where the three regions of the diagram meet. Note:
Five labels do not belong on the diagram – just skip those!
(A) Condensation (to a liquid)
(B) Gas
(C) Liquefaction (melting)
(D) Normal boiling point
(E) Plasma
(F) Vaporization (boiling)
(G) Solid
(H) Desublimation (opposite of sublimation)
(I) Normal melting point
(J) Liquid
(K) Equivalence point
(L) Triple Point
(M) Sublimation (solid vaporization)
(N) Normal sublimation point
(O) Atmospheric pressure
(P) Equilibrium point
(Q) Solidification (freezing)
(b) (4 pts) At 10 atm. pressure, give approximate values for the temperatures (in ˚C) at which UF6
melts and boils.
Melts:
60 ˚C
Boils:
141 ˚C
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Solutions
Version A
Q
F
C
J
A
F
G
L
Q
N
C
M
J
O
H
B
B
9