# C kJ ° mol - mol kJ K mol J •

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```Note:
The actual exam may not
resemble this exam, but solving
this exam will help prepare you.
CHM 1220/1225
Sample Exam V – Version 1 – Solution
1.
Suppose 0.562 g of graphite is placed in a calorimeter with an excess of oxygen at 25.00C and 1 atm pressure. The
excess oxygen ensures that all of the carbon burns to form CO2. After the graphite is ignited, the calorimeter temperature
rises to 25.89C. The heat capacity of the calorimeter and its contents is 20.7 kJ/C. Write the thermochemical equation
(including the enthalpy of reaction).
C(graphite) + O2(g)  CO2(g)
qcal = CT = 20.7
kJ
(25.89 – 23.00)C = 18.4 kJ (note: this includes 1 extra digit; it is limited by T to 2sf)
C
 qcal
Hreaction =
=
= -3.9 x 102 kJ
- 18.4 kJ
mol
 1 mol 

0.562 g
 12.01 g 
2.
a.
Find H° for
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

kJ
Hf° values in
mol
CH4(g): -74.87
CO2(g): -393.5
H2O(g): -241.8
H° = [1 mol(-393.5 kJ/mol) + 2 mol(-241.8 kJ/mol)] – 1mol(-74.87 kJ/mol)
H° = -877.1 kJ + 74.87 kJ
H° = -802.2 kJ
b.
3.
The reaction is (circle one)
Find S° for
ENDOTHERMIC
EXOTHERMIC
NEITHER
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
J
Sf° values in
mol  K
CH4(g): 186.1
O2(g): 105
CO2(g): 213.7
H2O(g): 188.7
S° = [1 mol(213.7 J/(mol · K) + 2 mol(188.7 J/(mol · K)] – [1 mol(186.1 J/(mol · K) + 2mol(105 J/(mol · K)]
S° = 591.1 J/K – 396.1 J/K
S° = 195.0 J/K
4.
a.
Using your values for H° and S° from the previous two problems, find G for the same reaction at 1000°C
CH4(g) + O2(g)  CO2(g) + 2H2O(g)
G° = H° - TS° = -802.2 kJ – (1273 K)(0.1950 kJ/K)
G° = -802.2 kJ – 248.2 kJ
G° = -1050.4 kJ
b.
5.
The reaction is (circle one)
SPONTANEOUS NONSPONTANEOUS
Find the value of K for the reaction at 298K and 1 atm
Hf° values in kJ
NH3(g): -45.90
AT EQUILIBRIUM
2NH3(g)  N2(g) + 3H2(g)
mol
Gf° values in kJ
NH3(g): -16.40
mol
Sf° values in J
mol  K
NH3(g): 192.7
G° = -RT ln K
2 mol(16.40 x 103 J/mol) = -8.3145
ln K= -13.2380
K = e--13.2380 = 1.78 x 10-6
N2(g): 130.6
H2(g): 191.6
J (298 K) ln K
mol  K
CHM 1220/1225 - Exam V
A
Page 2 of 3
6.
A gas is enclosed in a cylinder with a movable piston. The gas is compressed with 5.40 J of work and releases 1.50 J
heat to the surroundings. Identify the values of q, w and U. Be sure to include units!
q = -1.50 J
w = +5.40 J
U = q + w = +3.90 J
7.
Given:
N2H4(l) + N2O4(l)  3N2(g) + 4H2O(g); H = -1049 kJ
What is the enthalpy change when 5.00 g N2H4 react with 15.00 g N2O4?
N2O4: 92.0126 g;
N2: 28.0134 g;
Molar masses:
N2H4: 32.0452 g;
H2O: 18.0153 g
Find limiting reactant:
5.00 g N2H4 1 mol = 0.1560 mol (3sf , keep 1 extra digit)
32.0452 g
= 0.16302 mol (4 sf; keep1 extra digit)
15.00 g N2O4 1 mol
92.0126 g
N2H4 is limiting:
5.00 g N2H4 1 mol  - 1049 kJ = -164 kJ; 164 kJ are released
1 mol
32.0452 g
8.
Determine H° for the coal gasification process, 2C(s) + 2H2O(g)  CH4(g) + CO2(g), using the following reactions.
CO(g) + H2(g)  C(s) + H2O(g) ;
H° = -131.3 kJ
CO(g) + H2O(g)  CO2(g) + H2(g) ;
H° = -41.2 kJ
CO(g) + 3H2(g)  CH4(g) + H2O(g);
H° = -206.1 kJ
First, get C(s) as a reactant and CH4(g) and CO2(g) as a products ; H2O can be addressed later
Reverse #1; multiply by 2
2C(s) + 2H2O(g)  2CO(g) + 2H2(g);
H° = +262.6 kJ
#3
CO(g) + 3H2(g)  CH4(g) + H2O(g);
H° = -206.1 kJ
#2
CO(g) + H2O(g)  CO2(g) + H2(g) ;
H° = -41.2 kJ
See what we have now:
2C(s) + 2H2O(g) + 2H2(g) + 2CO(g)  CH4(g) + CO2(g) + 2CO(g) + H2O(g) + 2H2(g);
H° = 15.3 kJ
After removing compounds that are on both sides of the reaction:
2C(s) + H2O(g)  CH4(g) + CO2(g);
H° = 15.3 kJ
9.
For a reaction system that is at equilibrium, which of the following must be true? Circle your answer.
a. G = 0
b. H = 0
c. U =0
d. S = 0
e. q = 0
10. A reaction must be spontaneous at all temperatures when
a. G is negative
b. H is positive and S is positive
c. H is positive and S is negative
d. H is negative and S is positive
e. H is negative and S is negative
11. What is the First Law of Thermodynamics. You may answer with an equation or in words.
The change in internal energy of a system is the change in heat plus the work done on the system.
U = q + w
CHM 1220/1225 - Exam V
A
Page 3 of 3
12. Predict whether entropy increases or decreases for each of the following:
H2(g) + C2H6(g) ⇄ 2 CH4(g) ................................................................................. INCREASES .............. DECREASES
2SO2(g) + O2(g) ⇄ 2SO3(g) .................................................................................... INCREASES ............... DECREASES
Br2(l) ) ⇄ Br2(g) .................................................................................................... INCREASES .............. DECREASES
2NO2(g) ⇄ 2NO(g) + O2(g) .................................................................................... INCREASES .............. DECREASES
Pb(NO3)2(aq) + Na2SO4(aq) ⇄ PbSO4(s) + 2NaNO3(aq) ...................................... INCREASES ............... DECREASES
Physical Constants (Note: these are not exact numbers)
Avogadro’s number .......................... NA =
Molar gas constant............................ R =
.........................
.........................
6.0221367 x 1023
mol
0.08205783 L  atm
mol  K
J
8.314510
mol  K
8.314510 cal
mol  K
Note:
The actual exam may not
resemble this exam, but solving
this exam will help prepare you.
CHM 1220/1225
Sample Exam V – Version 2 – Solution
1. How much energy is liberated at constant pressure if 1.065 g of hydrochloric acid reacts with excess calcium
carbonate?
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g); Hº = -15.2 kJ
1.065 g HCl 
1 mol HCl  15.2 x 10 3 J = -222 J; 222 J are liberated (3 sf due to Hº)

36.46 g
2 mol HCl
2. Given:
SO3(g) + H2O(l)  H2SO4(aq); Hº = -130. kJ
Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l); Hº = -509.2 kJ
Find Hº for the following reaction:
Pb(s) + PbO2(s) + 2SO3 )  2PbSO4(s)
Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l); Hº = -509.2 kJ
2SO3(g) + 2H2O(l)  2H2SO4(aq); Hº = -260. kJ
Pb(s) + PbO2(s) + 2SO3 )  2PbSO4(s); Hº = -769 kJ (no dp due to 1st reaction)
3.
At its normal boiling point of 77ºC, the standard enthalpy change of vaporization is 25.21 kJ/mol. What is the
standard entropy change of vaporization of Freon-10 at its normal boiling point?
H
;
S 
T
S 
kJ
J
mol = 0.0720 kJ = 72.0
350.K
mol  K
mol  K
25.21
4. What is Sº at 298 K for the following reaction?
2O3(g)  3O2(g)
Substance
O3(g)
O2(g)
Sº (J/(mol K)) at 298 K
238.8
205.0
S = 3 mol(205.0) J/(mol·K) - 2 mol(238.80)J/(mol·K) = +137.4 J/K
S =3 mol(0 kJ/mol) – 2 mol(142.7 kJ/mol) = -285.4 kJ
G = H - TS = -285.4 kJ + (298K)(-0.1374 kJ/K) = -244.5 kJ
5. What is Kth at 298 K for the following reaction?
CO(g) + 2H2(g)  CH3OH(g); Gº= -103.8 kJ
Gº= -RT ln Kth
6.
Kth = e
 G
RT
=e
103.8x10 3 J
J 

 8.3145
 (298K)
molK 

= e0.04189 = 1.04
Find the Gº at 298 K for the following reaction:
2HBr(g)  H2(g) + Br2(g); Hº = +545.0 kJ; Sº = -14.1 J/(mol K) at 298 K
Gº = Hº - TSº = 545.0 kJ – (298K)(-0.0141 kJ/K) = 549.2 kJ
Note:
The actual exam may not
resemble this exam, but solving
this exam will help prepare you.
CHM 1220/1225
Sample Exam V – Version 3 – Solution
1.
How much energy, in J, is liberated at constant pressure if 0.894 g of hydrochloric acid reacts with excess calcium
carbonate?
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g); ∆Hº = -15.2 kJ
0.894 g HCl 
1 mol HCl  15.2 x 10 3 J = -186 J; 186 J are liberated

36.46 g
2 mol HCl
2. At its normal boiling point of 77ºC, the standard entropy change of vaporization is 85.2 J/(mol K). What is the
standard enthalpy change of vaporization of Freon-10 at its normal boiling point, in kJ/mol?
S 
H
;
T
H  TS = (77+273)K(85.2
J
= 2.98 x 104 J/mol or 29.8 kJ/mol
mol  K
3. Find (a) ∆Hº, (b) ∆Sº and(c) ∆Gº, and (d) Kth at 298 K for the following reaction?
3O2(g) 2O3(g)
O3(g)
Substance
O2(g)
Sfº (J/(mol K)) at 298 K
205.0
238.80
0
142.7
Hfº (kJ/mol) at 298 K
S = 2 mol(238.80)J/(mol·K) – 3mol(205.0) J/(mol·K) = -137.4 J/K
H =2 mol(142.7 kJ/mol) – 3(0 kJ/mol) = 285.4 kJ
G = H - TS = 285.4 kJ – (298K)(-0.1374 kJ/K) = 326.3 kJ
4. Camphor is a white solid that melts at 179.5˚C. It is a good solvent to find the molar mass of organic compounds.
Its freezing point constant is Kf = 40˚C/m. When 1.07 mg of a sample is dissolved in 78.1 mg of camphor the
melting point was 176.0˚C. (a) What is the molality of the solution? (b) What is the molar mass of the sample in
grams?
T = 179.5C – 176.0C = 3.5C
T 3.5C
=
= 0.0875 m
C
Kf
40.
m
mol solute = m·kg solvent = 0.0875 m ∙78.1 x 10-6 kg = 6.834 x 10-6 mol solute
1.07 x 10 -3
mass solute
= 1.6 x 102 g/mol
=
Mm =
mol solute
6.834 x 10 -6
m=
N
Note:
T
The actual exam may not
rresemble this exam, but so
olving
this exam will help prepare
e you.
CHM
M 1220/1225
Sa
ample Exam V – Version 4 – Solution
11.
A 3.510 g sample of ben
nzene, C6H6, was burned with
w excess oxxygen in a bo
omb calorimeter at constan
nt
volume. The
T temperatu
ure of the calo
orimeter chan
nged from 25..00°C to 37.188°C. The heat capacity, C, o
of the
calorimeteer and its con
ntents is 12.05 kJ/°C.
2C6H6 + 15O
1 2  12CO
O2 + 6H2O
Write you
u answer to th
he following question
q
on th
he line to the right of the q
question. In orrder to receiv
ve credit,
you must show your work
w
below.
a. Wha
at is q?
-1
146.8 kJ
b. Wha
at is ΔH for the balanceed reaction??
-6506 k
kJ or -6.506 x 103 kJ
kJ
(37.18
(
 25.000)C = 1466.769 kJ
C
qreactioon - qcalorimeteer = -146.8 kJ
k
78.111 g C 6 H 6
 146.4 kJ
k
H 

 2 mol = -65532 kJ or -6.532 x 103 kJJ
3.510 g C 6 H 6
1 mo
ol C 6 H 6
q caloriimeter  CT  12.05
22.
33.
(4 sf)
(4 sf)
A gas expaands isothermaally; this meanss there is no ch
hange in internaal energy. q is -76 J. Write y
you answer to
o the
following
g question on the line to thee right of the question. In o
order to receiv
ve credit, you
u must show y
your
work belo
ow.
a. Whatt is ΔU? ΔU
U=0
there is no change in innternal energgy
b. Whatt is w?
w = -q = -(--76 J) = + 766 J
w = +76 J
The entha
alpy change when
w
liquid methanol
m
, CH
H3OH, vaporizzes at 25°C is 38.0 kJ/mol.
For each of
o the followin
ng, write you
ur answer on the
t line to thee right of the q
question. In o
order to receiv
ve credit,
show you
ur work.
a. What is the entropy change wheen 1.00 mol of vapor in equ
uilibrium witth liquid cond
denses to liqu
uid at
25°C?? Be careful with
w the sign and
a the sig fig
gs.
S 
b.
H

T
kJ
kJ
kJ
mol = -0.1275167
779
·1 mol = --0.1275167779
·
298 K
mol  K
K
-0.128 kJ/K
k
or -1288 J/K
‐ 38.0 The standard
s
entrropy of forma
ation (at 25°C) of this vapo
or is 243 J/(mo
olK).
Wha
at is the standard entropy of
o formation (at
( 25°C) of th
he liquid?
S = Sf – Si = Sg - Sl
J
-128
= 2443
mol  K
Sl = (243+1288)
37
71
J
mol  K
Page 1
J
mol  K
J
mol  K
Sl
CHM 1220/1225 – Exam I
4.
A
Page 2
Find ∆G for the following reaction at 500.°C. ∆Hf°, Sf° values are given on the additional information sheet.
(Note: you can only use ∆Gf° values for standard temperature.)
2HgO(s)  2 Hg(l) + O2(g)
 kJ 
Hf 

 mol 
J


Sf 

 mol  K 
-90.79
0
0
70.27
76.03
205.0
kJ
)
H = +181.58 kJ
mol
J
J
J
S = [(2mol)(76.03
) + (1 mol)(205.0
)] – [(2 mol)(70.27
)]
mol  K
mol  K
mol  K
J
J
J
= (142.06
+ 205.0
) – (140.54 )
(1 dp due to 205.0)
K
K
K
J
J
J
- 140.54
S = 216.5
= 347.06
K
K
K
G = H - TS
kJ
= 181.58 kJ – (773 K)(0.2165
)
K
= 181.58 kJ – 167.4 kJ
(3 sf  0 dp due to T)
= 14.2 kJ
G = 14.2 kJ
H = 0 – (2mol)(-90.79
5.
Find the value of Kth for the following reaction at 298 K. You may use values of ∆Gf°.
2HgO(s)  2 Hg(l) + O2(g)
 kJ 
Gf 

 mol 
-58.49
G = 0 – (2mol)(-58.49
G = -RT lnKth
 G f 
ln Kth =
=
RT
Kth = e-47.2127
Kth = 3.13 x 10-21
0
0
kJ
)
mol
 116.98 x 10 3 J
= -47.2127
J


8.3145
298 K 
 mol  K 
G = +116.98 kJ
(3 sf due to T)
CHM 1220/1225 – Exam I
6.
A
Page 3
Balance the following oxidation reduction-reduction reaction using the half-reaction method in acid
conditions.
NO + Cr2O72-  Cr3+ + NO3-
oxidation #s: N, Cr
a.
+2
+6
+3
+5
Write the balanced oxidation half-reaction (acidic conditions).
2H2O + NO  NO3- + 4H+ + 3eb. Write the balanced reduction half-reaction (acidic conditions).
6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O
c.
Write the balanced overall reaction (acidic conditions).
4H2O + 2NO + 14H+ + Cr2O72-  2NO3- + 8H+ + 2Cr3+ + 7H2O
2NO + 6H+ + Cr2O72-  2NO3- 2Cr3+ + 3H2O
d.
Convert the overall reaction to basic conditions.
6OH- + 6H+ + 2NO Cr2O72-  2NO3- 2Cr3+ + 3H2O + 6OH6H2O + 2NO + Cr2O72-  2NO3- + 2Cr3+ + 3H2O + 6OH3H2O + 2NO + Cr2O72-  2NO3- + 2Cr3+ + 6OH7.
a.
Circle the stronger oxidizing agent: Li+ or F2
Explain.
Oxidizing agent means reduction half-reaction. More positive E is the stronger
oxidizing agent.
For Li+, E = -3.04 V
for F2, E = +2.87 V
3+
2+
b. Circle the stronger reducing agent: Pb or Cu .
Explain.
Reducing agent means oxidation half-reaction. More positive E is the stronger
reducing agent. Note Eoxidation = -Ereduction
For Pb3+, E = +0.13 V
for Cu2+, E = -0.34 V
8.
Consider the following cell:
Zn(s)|Zn2+(aq) || Na+(aq)|Na(s)
a.
What is the value of n?
Oxidation half-reaction (anode) Zn(s)  Zn2+(aq) + 2e—
Reduction half-reaction (cathode) Na+(aq) + e-  Na(s)
b.
Ereduction° = 0.76 V
Ereduction° = -2.71 V
n=2
What is E°cell.
E°cell = E°oxidation + E°reduction = E°cathode - E°anode
E°cell = -0.76 V + -2.71 V = -2.71 V - 0.76 V
E°cell = -3.47 V
c.
What is ∆G°?
∆G° = -nFE°cell = -(2 mol)(96,485
∆G° = -669,605.9 J = -669.6059 kJ
∆G° = -670. kJ
C
J
)(-3.47 )
mol
C
d. What is Kth?
∆G° = -RT ln Kth
 G
ln Kth =
=RT
Kth = e-270
J
mol
= -270.
J


8.3145
298K 
 mol  K 
 670. x 10 3
(3 sf)
Exam V Sample MC
Answer Section
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DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
easy
easy
moderate
difficult
difficult
moderate
moderate
difficult
easy
easy
easy
easy
easy
moderate
easy
moderate
difficult
easy
moderate
easy
easy
easy
```