Chem. 1B Midterm 2
Version B
Feb. 24, 2016
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of last name
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All work must be shown on the exam for partial credit. Points will be taken off for
incorrect or no units. Calculators are allowed. Cell phones may not be used for calculators.
On fundamental and short answer problems you must show your work in order to receive
credit for the problem. If your cell phone goes off during the exam you will have your
exam removed from you.
Fundamentals
(of 36 possible)
(
)+(
)=
Problem 1
(of 16 possible)
Problem 2
(of 18 possible)
Multiple Choice
(of 30 possible)
Midterm Total
(of 100 possible)
1
Fundamental Questions
Each of these fundamental chemistry questions is worth 6 points. You must show work to
get credit. Little to no partial credit will be awarded. Make sure to include the correct
units on your answers.
1a) 4 pts
A chemistry graduate student is studying the
rate of this reaction:
2H3PO4(aq) P2O5(aq) + 3H2O(aq)
He fills a reaction vessel with H3PO4 and
measures its concentration as the reaction
proceeds:
Use this data to answer the following
questions. Write the rate law for this reaction.
(You must show your work to get credit)
rate = k______________
Time (m)
0
10
20
Time
(minutes)
0
10.
20.
30.
40.
[H3PO4]
0.0800 M
0.0141 M
0.00773 M
0.00533 M
0.00406 M
1
[H3PO4] ln[H3PO4] [𝐻3 𝑃𝑂4 ]
0.0800 -0.0659-2.526 -1.736 12.5 58.4
0.0141
-4.262
70.9
58.5
-0.600
-0.00637
0.00773 -4.862
129.4
The graph of t vs [𝐻 1𝑃𝑂 ] is a straight line therefore, rate=k[H3PO4]2
3
1b) 2 pts.
Calculate the value of the rate constant k. Round your answers to 2
significant digits. Also be sure your answer has the correct unit symbol.
1
[𝐴]
2)
6 pts
4
1
= 𝑘𝑡 + [𝐴] therefore the slope equal k
0
1
1
𝑟𝑖𝑠𝑒 0.00406 𝑀 − 0.0800 𝑀
𝐿
𝑠𝑙𝑜𝑝𝑒 =
=
= 5.8𝑚𝑜𝑙∙𝑚
𝑟𝑢𝑛
40. 𝑚 − 0 𝑚
𝐿
Or 0.097𝑚𝑜𝑙∙𝑠
What mass of I2 from KI can be produced in 1.0 h with a current of 15 A?
60 𝑚 60 𝑠
𝐼𝑡 (15 𝐴) (1.0 ℎ( 1 ℎ )( 1 𝑚 ))
𝑛= =
= 0.56 𝑚𝑜𝑙 𝑒 −
𝐶
𝐹
96,486𝑚𝑜𝑙
2I- I2 + 2e1 𝑚𝑜𝑙 𝐼
253.80 𝑔 𝐼
0.56 𝑚𝑜𝑙 𝑒 − (2 𝑚𝑜𝑙 𝑒2−) ( 1 𝑚𝑜𝑙 𝐼 2 ) = 71 𝑔 𝐼2
2
3)
6 pts
Balance N2H4(aq) + CO32-(aq) N2(g) + CO(g) in basic conditions
N2H4  N2 + 4H+ + 4e-
CO32- + 4H+ + 2e- CO + 2H2O
N2H4  N2 + 4H+ + 4e2(CO32- + 4H+ + 2e- CO + 2H2O)
N2H4 +2CO32- + 4H+ N2 + 2CO + 4H2O
N2H4 +2CO32- + 4H+ + 4OH- N2 + 2CO + 4H2O + 4OHN2H4(aq) + 2CO32-(aq) N2(g) + 2CO(g) + 4OH-(aq)
2
4)
6 pts
Which of the reaction profiles represents the effect of a catalyst on the rate
of reaction?
A is the catalysized reaction.
5)
6 pts
Draw the following galvanic cell:
Pb(s) + Cl2(g)  Pb2+(aq) + 2Cl-(aq)
Calculate E˚, show the direction of electron flow and the direction of ion
migration through the salt bridge, and identify the cathode and anode.
Pb Pb2+ + 2eE˚ = 0.13 V
Cl2 + 2e- 2ClE˚=1.36 V
2+
Cl2(g) + Pb(s)  Pb (aq) + 2Cl-(aq)
6)
6 pts
E˚=1.49 V
In the first-order reaction A  2B, when the initial concentration of A was
0.0335 M it took 95 s for the concentration to drop to 0.0275 M. How
much more time would be needed for the concentration to drop by an
additional 0.0035 M.
1st Order
𝑙𝑛[𝐴] = −𝑘𝑡 + 𝑙𝑛[𝐴]𝑜
Determine k
𝑙𝑛(0.0275 𝑀) = −𝑘(95 𝑠) + 𝑙𝑛(0.0335 𝑀)
1
𝑘 = 0.0021𝑠
Determine time
𝑙𝑛(0.0240 𝑀) = −(0.00211𝑠)𝑡 + 𝑙𝑛(0.0275 𝑀)
𝑡 = 65𝑠
3
Short Answer Questions
Each of the following short answer questions are worth the noted points. Partial credit will
be given. You must show your work to get credit. Make sure to include proper units on
your answer.
1a) 6 pts
Calculate the voltage of the following cell at 298 K
Zn(s)|Zn2+(aq) 1.7 M||Ag+(aq) 2.4 M|Ag(s)
Reaction of Interest
Zn Zn2+ + 2eE˚ = 0.76 V
2(Ag+ + e-  Ag)
E˚ = 0.80 V
+
2+
Zn(s) + 2Ag (aq)  Zn (aq) + 2Ag(s)
E˚ = 1.56V
2+ ]
[𝑍𝑛
𝑅𝑇
𝑅𝑇
𝐸 = 𝐸° −
𝑙𝑛(𝑄) = 𝐸 ° −
𝑙𝑛 (
)
[𝐴𝑔+ ]2
𝑛𝐹
𝑛𝐹
𝐽
(8.3145 𝑚𝑜𝑙∙𝐾
)(298𝐾)
1.7
𝐸 = 1.56𝑉 −
𝑙𝑛 ( 2 ) = 1.58 𝑉
𝐶
2.4
(2)(96,485𝑚𝑜𝑙)
1b) 10 pts What is the voltage for the same cell at 800 K
Zn(s)|Zn2+(aq) 1.7 M||Ag+(aq) 2.4 M|Ag(s)
Hint: Assume ΔH and ΔS are temperature independent.
°
𝐽
𝐽
𝐽
𝑘𝐽
°
°
𝑆𝑍𝑛
= 42 𝑚𝑜𝑙∙𝐾
, 𝑆𝐴𝑔
= 43 𝑚𝑜𝑙∙𝐾
, 𝑆𝐴𝑔
, ΔH˚f(Ag+) = 105𝑚𝑜𝑙
+ = 73
𝑚𝑜𝑙∙𝐾
Find ΔG at 800 K
∆𝐺° = ∆𝐻° − 𝑇∆𝑆°
Find ΔH
∆𝐻°𝑟𝑥𝑛 = ∆𝐻°𝑓 (𝑍𝑛2+ ) + 2∆𝐻°𝑓 (𝐴𝑔)
−∆𝐻°𝑓 (𝑍𝑛) − 2∆𝐻°𝑓 (𝐴𝑔+ )
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
∆𝐻°𝑟𝑥𝑛 = −153𝑚𝑜𝑙
+ 2(0𝑚𝑜𝑙
− 2(105𝑚𝑜𝑙
) − 0𝑚𝑜𝑙
) = −363𝑚𝑜𝑙
Find ΔS
∆𝑆°𝑟𝑥𝑛 = 𝑆°𝑍𝑛2+ + 2𝑆°𝐴𝑔 − 𝑆°𝑍𝑛 − 2𝑆°𝐴𝑔+
𝐽
𝐽
∆𝑆°𝑟𝑥𝑛 = −112𝑚𝑜𝑙∙𝐾
+ 2(43𝑚𝑜𝑙∙𝐾
)
𝐽
𝐽
𝐽
−42 𝑚𝑜𝑙∙𝐾 − 2 (73𝑚𝑜𝑙∙𝐾) = −214𝑚𝑜𝑙∙𝐾
𝐽
𝐽
∆𝐺° = ∆𝐻° − 𝑇∆𝑆° = −363,000𝑚𝑜𝑙 − (800𝐾) (−214𝑚𝑜𝑙∙∙𝐾)
𝐽
∆𝐺° = −192,000 𝑚𝑜𝑙
∆𝐺° = −𝑛𝐹𝐸°
𝐽
−192,000𝑚𝑜𝑙
∆𝐺°
𝐸° = −
=−
= 0.995 𝑉
𝐶
𝑛𝐹
(2)(96,485𝑚𝑜𝑙
)
𝐸 = 0.995𝑉 −
𝐽
(8.3145 𝑚𝑜𝑙∙𝐾
)(800𝐾)
𝐶
(2)(96,485𝑚𝑜𝑙
)
𝑙𝑛 (
1.7
) = 1.04 𝑉
2.42
4
2a) 8 pts
The reaction 5Br-(aq) + BrO3-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
Is expected to obey the mechanism
𝑘1
+
BrO3 (aq)+H (aq) ⇌ HBrO3(aq)
fast equilibrium
𝑘−1
𝑘2
+
HBrO3(aq)+H (aq) ⇌ H2BrO3+(aq)
fast equilibrium
𝑘−2
𝑘3
Br-(aq)+H2BrO3+(aq)→ (Br-BrO2)(aq)+H2O(l)
(Br-BrO2)(aq)+4H+(aq)+4Br-(aq)→products
Write the rate law for this reaction.
slow
fast
𝑟𝑎𝑡𝑒 = 𝑘3 [𝐵𝑟 − ][𝐻2 𝐵𝑟𝑂3+ ]
H2BrO3+ is an intermediate therefore, needs to be eliminated from the rate
law. Use equilibrium to eliminate the intermediate.
𝑘2 [𝐻𝐵𝑟𝑂3 ][𝐻 + ] = 𝑘−2 [𝐻2 𝐵𝑟𝑂3+ ]
𝑘2 [𝐻𝐵𝑟𝑂3 ][𝐻 + ]
+
[𝐻2 𝐵𝑟𝑂3 ] =
𝑘−2
𝑘3 𝑘2 [𝐵𝑟 − ][𝐻𝐵𝑟𝑂3 ][𝐻 + ]
𝑟𝑎𝑡𝑒 =
𝑘−2
HBrO3 is an intermediate therefore, needs to be eliminated from the rate
law. Use equilibrium to eliminate the intermediate.
𝑘1 [𝐵𝑟𝑂3− ][𝐻 + ] = 𝑘−1 [𝐻𝐵𝑟𝑂3 ]
𝑘1 [𝐵𝑟𝑂3− ][𝐻 + ]
[𝐻𝐵𝑟𝑂3 ] =
𝑘−1
−
𝑘3 𝑘2 𝑘1 [𝐵𝑟 ][𝐵𝑟𝑂3− ][𝐻 + ]2
𝑟𝑎𝑡𝑒 =
= 𝑘[𝐵𝑟 − ][𝐵𝑟𝑂3− ][𝐻 + ]2
𝑘−1 𝑘−2
2b) 10 pts Does this mechanism fit with the following experimental data? You must
find the experimental rate law to get credit.
[Br-]o (M)
[BrO3-]o (M)
[H+]o (M)
Initial Rate( mol
Ls )
1.0
1.0
1.0
12
2.0
1.0
1.0
24
1.0
1.0
2.0
48
1.0
2.0
1.0
36
General Rate Law: 𝑟𝑎𝑡𝑒 = 𝑘[𝐵𝑟 − ]𝑥 [𝐵𝑟𝑂3− ]𝑦 [𝐻 + ] 𝑧
From experiments 1 and 2 when the concentration of Br- is double the rate
goes up by a factor of 2 therefore 1st order with respect to BrFrom experiments 1 and 3 when the concentration of H+ is double the rate
goes up by a factor of 4 therefore, 2nd order with respect to H+
From experiments 1 and 4
12𝑚𝑜𝑙
= 𝑘(1.0 𝑀)(1.0 𝑀)𝑦 (1.0 𝑀)2
𝐿∙𝑠
36𝑚𝑜𝑙
= 𝑘(1.0 𝑀)(2.0 𝑀)𝑦 (1.0 𝑀)2
𝐿∙𝑠
0.333 = (0.50)𝑦
y=1.6
𝑟𝑎𝑡𝑒 = 𝑘[𝐵𝑟 − ][𝐵𝑟𝑂3− ]1.6 [𝐻 + ]2
No the mechanism is not consistent.
5
Multiple Choice Questions
On the ParScore form you need to fill in your answers, perm
number, test version, and name. Failure to do any of these things will
result in the loss of 1 point. Your perm number is placed and
bubbled in under the “ID number.” Do not skip boxes or put in a
hyphen; unused boxes should be left blank. Bubble in your test
version (B) under the “test form.” Note: Your ParScore forms will
not be returned to you, therefore, for your records, you may want to
mark your answers on this sheet. Each multiple choice question is
worth 5 points.
Identify the choice that best completes the statement or answers the question.
Reaction
Na+ + e–  Na
Al3+ + 3e–  Al
Fe2+ + 2e–  Fe
Co2+ + 2e–  Co
Cu2+ + 2e–  Cu
Ag+ + e–  Ag
Cl2 + 2e–  2Cl–
F2 + 2e–  2F–
E° (volts)
–2.71
–1.66
–0.44
–0.28
+0.34
+0.80
+1.36
+2.87
1. Which of the following would be the best reducing agent?
a. Na+
b. Al3+
c. F2
d. Al
e. F–
2. The equilibrium constant for the following general reaction is 10. at 25°C. Calculate E° for
the cell.
X2(s) + Y+(aq)
X2+(aq) + Y(s)
(unbalanced)
a. 0.015 V
b. 0.060 V
c. 0.030 V
d. 0.045 V
e. None of the above
6
3. What time is required to plate a metal tray (24.0 cm x 12.0 cm) with a coating (thickness
0.00200 cm) of silver (density = 10.54
) using a current of 7.65 A? Neglect the amount
of silver required to coat the edges.
Helpful Information: The reaction of interest is Ag+ + e-  Ag and MAg = 107.9
a. 1130 s
b. 708 s
c. 912 s
d. 1420 s
e. None of the above
4. The rate constant for a reaction increases from 10.0
to 100.
when the temperature is
increased from 315 K to 413 K. What is the activation energy for the reaction in
a. 0.0848
?
b. 18.8
c. 11.0
d. 1.95
e. None of the above
5. In which of the following cases must E be equal to zero?
I.
In any cell at equilibrium
II. In a concentration cell
III. E° can never be equal to zero.
a.
b.
c.
d.
e.
II only
III
I and II
I only
None of the above
6. What reaction occurs at the anode and at the cathode when electrolysis of aqueous MgI2
occurs? Assume standard conditions.
Helpful information: I2 + 2e2IE° = 0.54 V
Mg2+ + 2eMg
E° = -2.37 V
a.
b.
c.
d.
e.
Anode/Cathode
Water/Magnesium
Water/Water
Magnesium/Water
Iodine/Magnesium
None of the above
7