Download Practice test 3

Name (please print):___________________________
Exam 3
th
Chemistry 142, Spring 2005
April 27 , 2005
Part 1. Answer 7 of the following 8 multiple choice questions. If you answer more than 8 cross
out the one you wish not to be graded, otherwise only the first 7 will be graded. 4 points each.
1.
14
Complete the nuclear reaction: 6 C →
14
7
N+?
0
-1
A) e
0
B) 1e
4
C) 2He
1
D) 0n
0
E) 0γ
2.
In a voltaic cell, electrons flow from the __________ to the __________.
A) salt bridge, anode
B) anode, salt bridge
C) cathode, anode
D) salt bridge, cathode
E) anode, cathode
3.
184
Calculate the nuclear binding energy for the isotope 72 Hf (183.9510 amu).
-8
A) 1.11x10 J
B) 7.51x10
-19
C) 2.37x10
-10
predicted mass = 185.4941 amu w/o electrons
J
mass defect = 1.5826 amu w/o electrons
J (2.31x10
-10
J without correcting for mass of electrons)
14
D) 1.36x10 J
15
E) 6.85x10 J
4.
The reduction half reaction occurring in the standard hydrogen electrode is __________.
A) H2 (g, 1atm) → 2H (aq, 1M) + 2e−
+
B) 2H (aq) + 2OH− → 2H2O(l)
+
C) O2(g) + 4H (aq) + 4e− → 2H2O(l)
+
D) 2H (aq, 1M) + 2e− → H2(g, 1atm)
+
+
E) 2H (aq, 1M) + Cl2(aq) → 2HCl(aq)
Name (please print):___________________________
5.
Which one of the following is a redox reaction?
A) NaOH + HCl → NaCl + H2O
2+
B) Pb
+ 2Cl− → PbCl2
C) AgNO3 + HCl → HNO3 + AgCl
D) 3NO2 + H2O → 2HNO3 + NO
E) None of the above is a redox reaction
6.
What can be said about a chemical system that has reached a minimum in free energy?
A) The temperature is OK.
B) The system entropy is zero.
C) The system has achieved equilibrium.
D) The reaction is complete.
E) The reaction is very fast.
7.
As a result of __________, we are able to measure absolute values of the entropy of a sample
and are not forced to define relative values as we were when we defined enthalpies of formation,
∆H°f.
A) the first law of thermodynamics
B) the second law of thermodynamics
C) the third law of thermodynamics
D) Hess’s law
E) reversibility
8.
If ∆G° for a reaction is greater than zero, then __________.
A) Keq = 0
B) Keq = 1
C) Keq > 1
D) Keq < 1
E) More information is needed
Name (please print):___________________________
Part 2. Answer 4 of the following 5 short answer questions. If you answer more than 4 cross out
the one you wish not to be graded, otherwise only the first 4 will be graded. 6 points each.
9.
Using your table of standard reduction potentials, from each of the following pairs of substances,
choose the one that is the stronger oxidizing agent.
(a) F2(g), Cl2(g)
10.
(b) O2(g) in acidic solution, O2(g) in basic solution
Under each of the following conditions, predict the sign of ∆G.
(a) An exothermic reaction with a positive entropy change:______negative________________
(b) An endothermic reaction with a negative entropy change:_____positive________________
(c) An exothermic reaction with a negative entropy change at high temperature:_______positive_
11.
2+
The standard cell voltage for the cell Zn(s) + Cu
2+
→ Zn
+ Cu(s) is 1.10 V. Calculate ∆G°°.
∆G°° = -nFE°° = –(2)(96485 C/mol)(1.10 J/C) = -212 kJ
12.
In the following reaction, indicate the oxidizing agent and the reducing agent:
3+
2+
+
3+
Tl (aq) + 2Cr (aq) → Tl (aq) + 2Cr (aq)
Tl
13.
3+
= oxidizing agent; Cr
2+
= reducing agent
How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of
molten AlCl3 with an electrical current of 12.0 A?
4.00 g Al ×
1mol Al 3 mol e 96485 C 1 sec
×
×
×
= 3.57 × 10 3 seconds
26.98 g 1 mol Al
mol e
12.0 C
Name (please print):___________________________
Part 3. Answer all of the following questions. 16 points each.
14.
Consider the following reaction:
Ag+(aq) + Cl−(aq) → AgCl(s)
Given the following table of thermodynamic data at 298 K:
∆Hf°
(kJ/mol)
105.90
-167.2
-127.2
Substance
+
Ag (aq)
Cl−(aq)
AgCl(s)
S°°
(J/K•mol)
73.93
56.5
96.11
Calculate ∆G° and the value of Keq for the reaction at 298 K.
Use the thermodynamic data to calculate ∆H°° and ∆S°° and use Gibbs equation to calculate ∆G°°.
∆H°° = [products] – [reactants] = -65.9 kJ
∆S°° = [products] – [reactants] = -34.32 J K
-1
mol-1
Then use ∆G°° = -RT lnKeq to find the equilibrium constant.
∆G°° = −55.7 kJ/mol
-∆G°/RT
Keq = e
∆G°° = – RT ln Keq
3
-1
-1
-1
= e-55.7x10 J mol /(8.314 J K mol 298 K)
Keq = 5.74 × 109
15.
The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is
+0.76 V:
+
2+
Zn(s) + 2H (aq) → Zn (aq) + H2(g)
Under nonstandard conditions, with a partial pressure of H2 = 1.0 atm and [Zn2+] = 1.0 M, the cell
potential is +0.66 V. Calculate the pH in the cathode compartment.
+
Use the Nernst equation to find [H ], then calculate pH.
E = Eo −
0.05916 V
log Q (at 298 K)
n
0.66 V = 0.76 V −
0.05916 V
log
2
[ Zn 2+ ] PH2
[ H + ]2
(1)( 1)
= 0.10 V (2) / 0.05916V = 3.381
[ H + ]2
log
+
1/[H
]2 = 103.381 = 2.565x105
[H+] = 0.02040 M
pH = 1.69
Name (please print):___________________________
16. The radioactivity of equal amounts of CO2 from different samples of carbon is measured with the
same counter. Sample A, which is from the burning of freshly grown wood, gives 121.2 counts per
minute. Sample B, which is from the burning of charcoal from an ancient campfire, gives 42.9 counts per
14
minute. Calculate the age of sample B. (The half-life of C is 5715. yr)
k = 0.693/ t½ = 1.213x10-4 yr-1
N
ln   = – k t
No
 42.9 
ln 
 = – (1.213x10-4 yr-1) t
121.2


-1.0386 = –
(1.213x10-4 yr-1) t
3
t = 8562. yr = 8.56x10 yr
17.
Suppose that the following reactions were to be used in a voltaic cell:
Cu2+(aq) + 2e− → Cu(s)
+
Ag (aq) + e− → Ag(s)
E°°red = 0.34 V
E°°red = 0.80 V
Sketch the voltaic cell that yields a spontaneous reaction under standard conditions, clearly
labeling all of the following in your sketch:
A. anode (and the species in the anode half-cell)
B. cathode (and the species in the cathode half-cell)
C. direction of anion and cation movement through the salt bridge
D. reading on a voltmeter (standard cell potential)
E. direction of electron movement
2+
A. Anode is Cu(s) (species in anode half-cell is Cu )
B. Cathode is Ag(s) (species in cathode half-cell is Ag+)
C. Anions move toward anode through salt-bridge; cations move toward cathode through saltbridge
D. E°°cell = 0.80 – 0.34 = +0.46 V
E. Electrons move from the anode to the cathode
eCu
–
+0.46V
+
+
-
←NO3 K+ →
Cu2+
Ag+
←NO3
Ag
Name (please print):___________________________
Name (please print):___________________________
Formulas and Constants
R = 8.314 J mol-1⋅K-1
1 V = 1 J/C
mp = 1.00728 amu
F = 96,485. C/mol 1 W = 1 J/s
1 J = 1 VC
c = 2.9979x108 m/s
mn = 1.00866 amu
∆S =
1 A = 1 C/s
NA = 6.022x1023 mol-1
q rev
T
∆G = ∆G o + RT lnQ
∆G = −nFE
E = Eo −
RT
lnQ
nF
E = Eo −
0.05916 V
log Q (at 298 K)
n
Eo =
0.025693 V
0.05916 V
ln K eq =
log K eq
n
n
N
ln   = – k t
No
t½ = ln 2/k = 0.693/k
(at 298 K)