First initial
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Chem. 1B Midterm 1
Version B
February 3, 2017
Name:__________________________________________
Print Neatly. You will lose 1 point if I cannot read your name or perm number.
Perm Number:___________________________________
All work must be shown on the exam for partial credit. Points will be taken off for
incorrect or missing units. Calculators are allowed. Cell phones may not be used as
calculators. On fundamental and challenge 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.
1)
2)
6 pts
6 pts
Consider the following exothermic reaction at 25°C and 1 atm.
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
Circle the correct answer for each of the following quantities: w, ΔH, ΔE, ΔS, and
ΔG.
w<0
w>0
w=0
ΔH<0
ΔH>0
ΔH=0
ΔE<0
ΔE>0
ΔE=0
ΔS<0
ΔS>0
ΔS=0
ΔG<0
ΔG>0
ΔG=0
Consider the following equilibrium:
2NO2(g) ⇌ N2O4(g)
ΔG˚=-5.4 kJ (at 881˚C)
Now suppose a reaction vessel is filled with 0.594 atm of dinitrogen tetroxide
(N2O4) at 881.˚C. Answer the following questions about this system:
Under these conditions, will the pressure of N2O4 tend to rise
of fall?
☐Rise
☒Fall
Is it possible to revese this tendency by adding NO2?
In other words, if you said the pressure of N2O4 will tend to
☒Yes
rise, can that be changed to a tendency to fall by adding NO2?
☐No
Similarly , if you said the pressure of N2O4 will tend to fall, can
that be changed to a tendency to rise by adding NO2?
If you said the tendency can be reversed in the second
question, calculate the minimum presure of NO2 needed to
_____atm
reverse it.
Round your answer to 2 significant digits
∆𝐺 = ∆G° + RTln(𝑄) The tendency will reverse when ΔG=0
𝑃𝑁 𝑂
∆G° = −RTln ( 22 4 )
𝑃𝑁𝑂2
𝑇 = 881℃ + 273.15℃ = 1,154𝐾
0.594
𝑘𝐽
−5.4 kJ = −(0.0083145 𝑚𝑜𝑙∙𝐾
)(1,154𝐾) ln (𝑃2 )
𝑁𝑂2
𝑃𝑁𝑂2 = 0.58 𝑎𝑡𝑚
Therefore, when the pressure of NO2 is above 0.058 atm the pressure of N2O4 will
fall
3)
6 pts
Determine the mass of CO2 produced by burning enough C3H8 to produce
1.00×102 kJ of heat.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
ΔHrxn=-2,217 kJ
3 𝑚𝑜𝑙 𝐶𝑂
44.009 𝑔 𝐶𝑂2
)
1 𝑚𝑜𝑙 𝐶𝑂2
−1.00 × 102 𝑘𝐽 ( −2,217 𝑘𝐽2 ) (
= 5.96 𝑔 𝐶𝑂2
2
4)
6 pts
Calculate the standard enthalpy of formation of CCl4(g) using the
thermochemical equations.
CCl4(g) + 4HCl(g)  CH4(g) + 4Cl2(g)
ΔH1 = 397.0 kJ
½H2(g) + ½Cl2(g)  HCl(g)
ΔH2 = -92.3 kJ
C(graphite) + 2H2(g)  CH4(g)
ΔH3 = -74.81 kJ
Standard enthalpy of formation reaction
C(graphite) + 2Cl2(g)CCl4(g) ΔH°f
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g)
-ΔH1 = -397.0 kJ
4HCl(g)  2H2(g) + 2Cl2(g)
-4ΔH2 = 369.2 kJ
C(graphite) + 2H2(g)  CH4(g)
ΔH3 = -74.81 kJ
C(graphite) + 2Cl2(g)CCl4(g)
ΔH°f = -397.0 kJ +369.2 kJ -74.81 kJ =-102.6 kJ
5)
6 pts
What is the heat capacity of the metal if 60.0 g of metal at 95.0˚C is put into a
coffee cup calorimeter that contains 75.0 g of water at 25.0˚C. The final
temperature of the system is 35.0˚C The heat capacity of the calorimeter is
10.0 °𝐶𝐽
𝑞𝑚𝑒𝑡𝑎𝑙 = −(𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑐𝑎𝑙 )
𝑞𝑚𝑒𝑡𝑎𝑙 = 𝑚𝑚𝑒𝑡𝑎𝑙 𝐶𝑚𝑒𝑡𝑎𝑙 (𝑇𝑓 − 𝑇𝑖(𝑚𝑒𝑡𝑎𝑙) )
𝑞𝑚𝑒𝑡𝑎𝑙 = (60.0𝑔)𝐶𝑚𝑒𝑡𝑎𝑙 (35.0℃ − 95.0℃) = −(3,600 𝑔 ∙ ℃)𝐶𝑚𝑒𝑡𝑎𝑙
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑤𝑎𝑡𝑒𝑟 𝐶𝑤𝑎𝑡𝑒𝑟 (𝑇𝑓 − 𝑇𝑖(𝑤𝑎𝑡𝑒𝑟) )
𝐽
𝑞𝑤𝑎𝑡𝑒𝑟 = (75.0 𝑔) (4.184 𝑔∙℃) (35.0℃ − 25.0℃) = 3,135 𝐽
𝑞𝑐𝑎𝑙 = 𝐶𝑐𝑎𝑙 (𝑇𝑓 − 𝑇𝑖(𝑤𝑎𝑡𝑒𝑟) )
𝐽
𝑞𝑐𝑎𝑙 = (10.0 ℃) (35.0℃ − 25.0℃) = 100 𝐽
−(3,600 𝑔 ∙ ℃)𝐶𝑚𝑒𝑡𝑎𝑙 = −(3,135 𝐽 + 100 𝐽)
𝐽
𝐶𝑚𝑒𝑡𝑎𝑙 = 0.899 𝑔∙℃
6)
6 pts
What is the maximum amount of work that can be obtained when 0.100 mol
Ar(g) in a volume of 1.00 L at 25˚C is allowed to expand to 2.00 L?
The maximum work that can be done by a system occurs when the
process is carried out reversibly.
𝑉𝑓
𝑤𝑟𝑒𝑣 = −𝑛𝑅𝑇𝑙𝑛 ( )
𝑉𝑖
2.00 𝐿
𝐽
= −(0.100 𝑚𝑜𝑙) (8.3145 𝑚𝑜𝑙∙𝐾 ) (298𝐾)𝑙𝑛 (
)
1.00 𝐿
𝑤𝑟𝑒𝑣 = −172 𝐽
3
Challenge Problems
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)
8 pts
In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an
initial temperature of 25.00˚C. After dissolution of the salt, the final
temperature of the calorimeter contents is 23.34˚C. Assuming the solution has
𝐽
a heat capacity of 4.18 ℃∙𝑔
and assuming no heat loss to the calorimeter,
𝑘𝐽
calculate the enthalpy change for the solution of NH4NO3 in units of 𝑚𝑜𝑙
The question wants you to calculate ΔHrxn. Reaction of interest is:
NH4NO3(s)  NH4+(aq) + NO3-(aq)
You need to calculate ΔHrxn. Since, it is a coffee cup calorimeter, pressure is
constant and ΔHrxn=qrxn. The question also tells you that qcal=0. The heat from the
reaction will go into the solution therefore qrxn=-qsol and H rxn  q sol
.
∆𝐻𝑟𝑥𝑛 = −𝑚𝑠𝑜𝑙 𝐶𝑠𝑜𝑙 ∆𝑇𝑠𝑜𝑙
𝑚𝑠𝑜𝑙 = 1.60 𝑔 + 75.0 𝑔 = 76.5 𝑔
∆𝑇𝑠𝑜𝑙 = 23.34℃ − 25.00℃ = −1.66℃
𝐽
∆𝐻𝑟𝑥𝑛 = −(76.6 𝑔) (4.18 ℃∙𝑔) (−1.66℃) = 532 𝐽
Calculate ΔH per mol
1 𝑚𝑜𝑙 𝑁𝐻 𝑁𝑂
1.60 𝑔 𝑁𝐻4 𝑁𝑂3 (80.06 𝑔 𝑁𝐻4 𝑁𝑂3 ) = 0.0200 𝑚𝑜𝑙 𝑁𝐻4 𝑁𝑂3
4
3
Since there is only one mole of NH4NO3 in the equation, divide by 0.0200 mol.
532 𝐽
𝐽
𝑘𝐽
∆𝐻𝑟𝑥𝑛 =
= 2.66 × 104 𝑚𝑜𝑙 = 26.6 𝑚𝑜𝑙
0.0200 𝑚𝑜𝑙
1b)
8 pts
In the same calorimeter 200. g of ice (Ti(ice) = 0˚C) was added to
440.g water (Ti(water)=80.0˚C). When the system reached thermal equilibrium the
𝑘𝐽
𝐽
temperature of the system was 30.1˚C. What is ΔHfus in 𝑚𝑜𝑙
? 𝐶𝐻2 𝑂(𝑙) 4.18 𝑔∙℃
−𝑞𝐻2 𝑂 = 𝑞𝑖𝑐𝑒
𝑞𝑖𝑐𝑒 = 𝑞𝑓𝑢𝑠 + 𝑚𝑖𝑐𝑒 𝐶𝐻2 𝑂 ∆𝑇0℃→𝑇𝑓
𝑞𝑓𝑢𝑠 = 𝑛𝑖𝑐𝑒 ∆𝐻𝑓𝑢𝑠
−𝑚water 𝐶𝐻2 𝑂 ∆𝑇𝑇𝑖(𝑤𝑎𝑡𝑒𝑟) →𝑇𝑓 = 𝑛𝑖𝑐𝑒 ∆H𝑓𝑢𝑠 + 𝑚𝑖𝑐𝑒 𝐶𝐻2 𝑂 ∆𝑇0℃→𝑇𝑓
−𝑚water 𝐶𝐻2 𝑂 ∆𝑇𝑇𝑖(𝑤𝑎𝑡𝑒𝑟)→𝑇𝑓 − 𝑚𝑖𝑐𝑒 𝐶𝐻2 𝑂 ∆𝑇0℃→𝑇𝑓
∆H𝑓𝑢𝑠 =
𝑛𝑖𝑐𝑒
Calculate the moles of ice
1 𝑚𝑜𝑙 𝐻 𝑂
200. 𝑔 𝐻2 𝑂 (18.02 𝑔 𝐻2 𝑂) = 11.1 𝑚𝑜𝑙
2
∆H𝑓𝑢𝑠 =
𝐽
𝐽
−(440. 𝑔) (4.18 𝑔∙℃
) (30.1℃ − 80.0℃) − (200. 𝑔) (4.18 𝑔∙℃
) (30.1℃ − 0.0℃)
𝑘𝐽
= 6.00 𝑚𝑜𝑙
11.1 𝑚𝑜𝑙
4
2a)
8 pts
Using thermodynamic data from your constant sheet, calculate ΔG˚ at 25˚C for
the process
2SO2(g) + O2(g)  2SO3(g)
where all gases are at 1.00 atm pressure. Also calculate ΔG at 25˚C for this
same reaction but with all gases at 10.0 atm pressure.
Compound
SO2(g)
O2(g)
SO3(g)
𝑘𝐽
ΔGf˚(𝑚𝑜𝑙
)
-300
0
-371
°
∆𝐺𝑟𝑥𝑛
= ∑ ∆𝐺𝑓° (𝑝𝑟𝑜𝑑) − ∆𝐺𝑓° (𝑟𝑒𝑎𝑐)
°
∆𝐺𝑟𝑥𝑛
= 𝑛𝑆𝑂3 ∆𝐺𝑓° (𝑆𝑂3 ) − 𝑛𝑆𝑂2 ∆𝐺𝑓° (𝑆𝑂2 ) − 𝑛𝑂2 ∆𝐺𝑓° (𝑂2 )
𝑘𝐽
𝑘𝐽
𝑘𝐽
) − (1) (0 𝑚𝑜𝑙)
𝑚𝑜𝑙
𝑃𝑆𝑂3 2
𝑅𝑇𝑙𝑛 (
)
𝑃𝑆𝑂2 2 𝑃𝑂2
°
∆𝐺𝑟𝑥𝑛
= (2) (−371 𝑚𝑜𝑙) − (2) (−300.
°
°
∆𝐺𝑟𝑥𝑛 = ∆𝐺𝑟𝑥𝑛
+ 𝑅𝑇𝑙𝑛(𝑄) = ∆𝐺𝑟𝑥𝑛
+
𝑘𝐽
𝑘𝐽
∆𝐺𝑟𝑥𝑛 = −142 𝑚𝑜𝑙 + (0.0083145 𝑚𝑜𝑙∙𝐾) (298. 𝐾)𝑙𝑛 (
𝑘𝐽
= −142 𝑚𝑜𝑙
10.02
)
(10.02 )(10.0)
𝑘𝐽
= −148 𝑚𝑜𝑙
2b)
10 pts
𝑘𝐽
What is ΔGrxn (in 𝑚𝑜𝑙
) when the temperature is 87˚C and the pressures of all the
gases are 10.0 atm. Assume that ΔH and ΔS are temperature independent.
We need to find ΔGrxn
°
∆𝐺𝑟𝑥𝑛 = ∆𝐺𝑟𝑥𝑛
+ 𝑅𝑇𝑙𝑛(𝑄)
First find ΔG˚rxn at 87˚C. The tables on the constant sheet list ΔG˚f at 25˚C,
therefore, you need to convert to 87˚C. Since ΔH and ΔS are temperature
independent you can use ΔG=ΔH-TΔS to change the temperature of ΔG˚rxn
°
∆𝐻𝑟𝑥𝑛
= ∑ ∆𝐻𝑓° (𝑝𝑟𝑜𝑑) − ∆𝐻𝑓° (𝑟𝑒𝑎𝑐)
°
∆𝐻𝑟𝑥𝑛
= 𝑛𝑆𝑂3 ∆𝐻𝑓° (𝑆𝑂3 ) − 𝑛𝑆𝑂2 ∆𝐻𝑓° (𝑆𝑂2 ) − 𝑛𝑂2 ∆𝐻𝑓° (𝑂2 )
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
°
∆𝐻𝑟𝑥𝑛
= (2) (−396 𝑚𝑜𝑙) − (2) (−297 𝑚𝑜𝑙) − (1) (0 𝑚𝑜𝑙) = −198 𝑚𝑜𝑙
°
°
°
∆𝑆𝑟𝑥𝑛
= ∑ 𝑆𝑝𝑟𝑜𝑑
− 𝑆𝑟𝑒𝑎𝑐
°
°
°
∆𝑆𝑟𝑥𝑛
= 𝑛𝑆𝑂3 𝑆𝑆𝑂
− 𝑛𝑆𝑂2 𝑆𝑆𝑂
− 𝑛𝑂2 𝑆𝑂° 2
3
2
𝐽
𝐽
𝐽
𝐽
°
∆𝑆𝑟𝑥𝑛
= (2) (257 𝑚𝑜𝑙∙𝐾) − (2)(248 𝑚𝑜𝑙∙𝐾
) − (1) (205 𝑚𝑜𝑙∙𝐾) = −187 𝑚𝑜𝑙∙𝐾
Determine ΔG˚rxn at 87˚C
𝑘𝐽
𝑘𝐽
°
°
°
∆𝐺𝑟𝑥𝑛
= ∆𝐻𝑟𝑥𝑛
− 𝑇∆𝑆𝑟𝑥𝑛
= −198 𝑚𝑜𝑙 − (360. 𝐾)(−0.187 𝑚𝑜𝑙∙𝐾
)
𝑘𝐽
= −131 𝑚𝑜𝑙
Determine ΔGrxn when the gases are at 10.0 atm
∆𝐺𝑟𝑥𝑛 =
°
∆𝐺𝑟𝑥𝑛
+ 𝑅𝑇𝑙𝑛(𝑄) =
𝑘𝐽
°
∆𝐺𝑟𝑥𝑛
+ 𝑅𝑇𝑙𝑛 (
𝑘𝐽
𝑃𝑆𝑂3 2
𝑃𝑆𝑂2 2 𝑃𝑂2
∆𝐺𝑟𝑥𝑛 = −131 𝑚𝑜𝑙 + (0.0083145 𝑚𝑜𝑙∙𝐾 ) (360. 𝐾)𝑙𝑛 (
)
10.02
)
(10.02 )(10.0)
𝑘𝐽
= −138 𝑚𝑜𝑙
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 form 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.
1. What is ΔS for 88.0 g of CO2 undergoing the following reaction at constant pressure?
CO2(s, 150.K)  CO2(g, 195. K)
Helpful Information: Tsub = 195K, ΔHsub = 25.2 , CCO2(s) = 1.07
A) 283
B) 233
C) 154
D) 24.9
E) None of the above
2. For which of the following reaction(s) is the enthalpy change for the reaction not equal to
ΔH°f of the product?
I. 2H(g)  H2(g)
II. H2(g) + O2(g)  H2O2(l)
III. H2O(l) + O(g)  H2O2(l)
A) II only
B) II and III
C) I only
D) III only
E) I and III
3. A 5.000 g sample of methanol, CH3OH, was combusted in the presence of excess oxygen
in a bomb calorimeter. The temperature of the water increased from 24.0000°C to
29.765°C. The heat capacity of the calorimeter is 19,390 . Calculate ΔE for the reaction
in .
A) -111.8
B) 1,512
C) -22.36
D) 716
E) None of the above
6
4. For the vaporization of a liquid at a given pressure,
A) ΔG is positive at all temperatures.
B) ΔG is negative at low temperatures but positive at high temperatures (and zero at
some temperature).
C) ΔG is negative at all temperatures.
D) ΔG is positive at low temperatures but negative at high temperatures (and zero at
some temperature).
Consider a process carried out on 1.00 mol of a monatomic ideal gas by the following two
different pathways. The first pathway is A (3.00 atm, 20.0 L) to C (1.00 atm, 20.0 L) to D
(1.00 atm, 50.0 L); and the second pathway is A (3.00 atm, 20.0 L) to B (3.00 atm, 50.0 L)
to D (1.00 atm, 50.0 L). In each case, the gas is taken from state A to state D.
5. Calculate ΔHACD.
A) –175 L•atm
B) 25 L•atm
C) 175 L•atm
D) –25 L•atm
E) None of the above
6. Which statement is true of a process in which 1 mol of a gas is expanded from state A to
state B?
A) It is not possible to have more than one path for a change of state.
B) The final volume of the gas will depend on the path taken.
C) When the gas expands from state A to state B, the surroundings are doing work
on the system.
D) The amount of work done in the process must be the same, regardless of the
path.
E) The amount of heat released in the process will depend on the path taken.
A, E, E, D, D, E
7