Download Ans_PS08b_full_121 F16

Chemistry 121
Mines, Fall 2016
Answer Key, Problem Set 8b – (full)
1. WO1; 2. WO2; 3. MP*; 4. MP (6.11); 5. MP*; 6. MP*; 7. MP (6.48); 8. MP*; 9. MP (6.66); 10. MP*; 11. WO3; 12. WO4; 13. WO5;
14. MP*; 15. MP*; 16. MP (6.60); 17. MP (6.93); 18. MP*; 19. MP (6.99); 20. MP (6.75); 21. MP*; 22. MP*; 23. MP (6.39); 24. MP
(6.42); 25. MP (6.43); 26. MP (6.78); 27. MP (6.82); 28. WO6; 29. WO7; 30. MP*; 31. MP (6.88(a,d)); 32. MP (6.85)
------------------General Early Ideas and Definitions (KE, PE, sys vs surroundings, heat, E, sign conventions)
1. WO1. Define each of the following and give the symbol associated with the last two:
(a) kinetic energy (KE)—energy of motion
(b) potential energy (PE)—energy of position (and often considered “stored” energy)
(c) internal energy (of a system)—the sum total of the kinetic energy and potential energy of all the
nanoscopic entities in the system. Symbol is E.
(d) heat—energy that transfers (or “flows”) from a hotter substance to a cooler one. Symbol is q.
(Note: KE, PE, and E are properties of a substance or system, but “heat” is not. Heat only “exists” during the transfer; once it
“gets there” it is no longer considered to be “heat”—it merely contributes to the internal energy getting larger.)
2. WO2. (a) When you change the temperature of a substance or system, what property always changes with it, no matter
what?
The average kinetic energy (per particle) in the system must change as T changes. The average
kinetic energy depends only on temperature (from kinetic molecular theory).
(b) For a sample of a pure substance that is not undergoing any physical change, what is the qualitative and quantitative
relationship between heat and temperature?
Qualitative: As heat flows into a pure substance, its temperature must increase (if it is not undergoing some
physical change). As heat flows out of a pure substance, its temperature must go down.
Quantitative: q = Cs x m x T This is consistent with the qualitative answer above, because it shows
that if q is positive (heat transfers in), then T is positive (T goes up), and if q is negative
(heat transfers out), then T is negative (T goes down).
(c) Explain how it can be that water is being “heated” and yet its temperature is not increasing.
What happens to the “heat
energy” that flows into the water as the water is boiling?
Short answer. In this scenario, all the energy that transfers into the system as heat ends up
as increased potential energy (of the now separated molecules) rather than kinetic energy.
Explanation. Since average kinetic energy and T are proportional, if the average kinetic energy
does not go up, T will not go up (or more properly worded, if T stays the same, the average kinetic
energy must stay the same). Since energy is conserved, if energy transfers into a system (net), it
must end up “somewhere”. Since the internal energy of the system is the sum of KE and PE, if the
energy doesn’t end up increasing KE, it must increase PE (because those are the only two
choices).
“It takes energy to separate things that are attracted to one another” is the idea in PS8a that I referred you to. This is
basically just another way of saying that potential energy increases when two things that attract one another are
separated. In this case, the molecules in the liquid are attracted to one another, so energy is required to actually “do”
the vaporization process. During boiling, this energy transfers into the system as “heat” during heating, and it ends up
raising the potential energy of the system without raising the average kinetic energy of the particles. That is how T can
remain the same even though the system is still being “heated”. Once all the liquid is vaporized, if energy continues to
enter the system, the temperature of the system (now a gas) will go up, as average kinetic energy will now increase
again.
3. MP Solution to this problem not in key at this time.
PS8b-1
Answer Key, Problem Set 8b
4. MP (6.11).
If internal energy of the products of a reaction is higher than that of the reactants, what is the sign of E for the
reaction? In which direction does energy flow?
Answer: E is positive and energy flows from the surroundings to the system. One could draw an energy
diagram to show this as follows:
Energy
Products
E = Eproducts – Ereactants > 0
Reactants
Thermal Energy Transfer, (Specific) Heat Capacity, and “No Process” Calorimetry Situations
5. MP Solution to this problem not in key at this time.
6. MP Solution to this problem not in key at this time.
7. MP (6.48).
How much heat is required to warm 1.50 kg of sand from 25.0 C to 100.0 C?
Answer: 95,000 J = 95 kJ
Reasoning / Work:
The specific heat capacity of a substance represents the amount of energy needed to raise 1.0 g of
it by 1 degree Celsius. From Table 6.4, the specific heat of sand is 0.84 J/(gC). 1.50 kg = 1500
g, so the energy needed would be 1500 times as great to raise the sample by 1 degree. Going
from 25 to 100 degrees represents a change in T of +75 degrees, so multiply by 75 times more. In
equation form, we have that heat energy needed (q) equals: q = Cs x m x T
qsand 
0.84 J 
1000 g 
o
x  1.50 kg x
 x (100.0 - 25.0) C  94500 J  95000 J or 95 kJ
o
kg 
g C

8. MP Solution to this problem not in key at this time.
9. MP (6.66).
A 32.5-g iron rod, initially at 22.7 C is submerged into an unknown mass of water at 63.2 C, in an insulated
container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 C. What is the mass of the water?
Answer: 35 g
Strategy:
1) Recognize this as a “heat transfer” kind of problem. Thus you will have a qsys = -qsurr type of
setup.
2) Further recognize that this problem involves NO CHEMICAL OR PHYSICAL CHANGES—i.e.,
no process other than heat flow from a hotter object or sample to a cooler object or sample.
This means that EACH sample (the “system” and the “surroundings”) changes temperature
only as a result of a heat flow. Thus you will have TWO specific heat capacities, TWO “delta
T’s” and TWO masses. In other words qsys = -qsurr reduces to
Cs,sys x msys x Tsys =  (Cs,surr x msurr x Tsurr)
3) The above reduces further in this problem to:
Cs,iron x miron x Tiron =  (Cs, water x mwater x Twater)
PS8b-2
Answer Key, Problem Set 8b
4) Noting that T ≡ Tf Ti, and Cs,water= 4.18 J/g·°C and Cs,iron = 0.449 J/g·°C (Table 6.4), all of the
unknowns are given in the problem except for mwater, which can be determined algebraically.
Execution of Strategy:
0.449 J/g·°C x 32.5 g x (59.5°C  22.7°C) =  (4.18 J/g·°C x mwater x (59.5°C  63.2°C))
537.0 J = -(-15.466) J/g x mwater
mwater 
537.0 J
 34.72 g = 35 g
15.466 J/g
Enthalpy Changes, General (meaning of, endothermic vs exothermic, sign of, prediction of endo/exo)
10. MP Solution to this problem not in key at this time.
11. WO3. 6.58 (plus added parts (d) and (e)).
Is each process exothermic or endothermic? Also indicate the sign of H.
Strategy: Assess whether (only) separation of attracted particles is occurring, or whether (only) “coming
together” of attracted particles is occurring. If it is (only) separation, which requires energy, then
the process is endothermic, and thus H > 0. If it is (only) “coming together”, which releases
energy, the the process is exothermie, and thus H < 0.
NOTE: If a process involves some separation and some “coming together” (bond breaking and bond
making), then you cannot make a prediction without additional information (see WO4).
(a) dry ice evaporating
endothermic, H > 0
CO2(s) → CO2(g); Molecules (which attract one another) are separating from one another. This requires
energy (raises PE)
(b) a sparkler burning
exothermic, H < 0
Light and heat are being emitted from the system. Exothermic means “heat transfers out” from system.
(c) the reaction that occurs in a chemical cold pack used to ice athletic injuries
endothermic, H > 0
System getting cold as a result of a process occurring means energy is being absorbed (energy will then
transfer in from the surroundings because of the temperature difference.)
(d) O3(g) → O2(g) + O(g)
endothermic, H > 0
This particular chemical change involves only the breaking of a bond (no new bonds are made). Pulling
apart bonded atoms must require energy. (Note that one molecule splits into a smaller one plus a free
atom.)
(e) NO2(g) + NO2(g) → O2NNO2(g)
exothermic, H < 0
This particular chemical change involves only the making of a bond (an N-N bond) (no bonds are
broken). The coming together of two atoms that attract one another to form a bond must release
energy. (Note: Two molecules combine into one larger one.)
12. WO4. (follow up to WO3)
Describe specifically why you cannot predict whether the following process is
endothermic or exothermic without additional information.
H2(g) + Br2(g)  2 HBr(g)
Shortest Answer: Because unlike the chemical changes shown in WO3 ((d) and (e)), this process
involves both bond breaking and bond making. As such, you can’t know without further
information whether the amount of energy to break the bonds that are broken in the reactants will
be more or less than the amount of energy released upon making the bonds in the products.
PS8b-3
Answer Key, Problem Set 8b
More Detailed Answer: Unlike the problems in WO3, this process does not involve only the separation or
only the coming together of particles that attract one another. Visualize the reaction in terms of
actual atoms and molecules rather than just seeing the symbols in the equation. Think of H2 as
being H-H, where two H atoms are bonded, Br2 as Br-Br where two Br atoms are bonded
together, and HBr as H-Br, where an atom of H is bonded to an atom of Br. So imagine the
smallest amount of this chemical reaction that could occur (“one equation unit’s worth of reaction”
in my made-up jargon): one molecule of H2 reacts with one molecule of Br2 to form two
molecules of HBr. In order to rearrange the atoms in the reactants to make the products in this
reaction, you can imagine that you’d have to break one H-H bond and one Br-Br bond in the
reactants, but when the products are formed, you’d make two H-Br bonds. The breaking of the
H-H and Br-Br bonds will require energy, but the making of the H-Br bonds will release energy.
So without knowing which bonds are stronger (i.e., how much energy it would require to break HH bonds, Br-Br bonds, and H-Br bonds), you cannot say whether the overall process will require
energy (be endothermic) or release it (be exothermic).
13. WO5. 6.22 & 6.23 in Tro.
6.22. From a molecular viewpoint, where does the energy emitted in an exothermic chemical reaction come from?
In short, it comes from the potential energy of the atoms/ions in the system. In an exothermic process, the
rearrangement of atoms/ions (or even molecules, in the case of a physical change) results in a lowering of
potential energy. That energy is converted into KE in the system, which ultimately transfers out into the
surroundings as “heat”. (Technically, the conversion of PE into KE initially makes the system hotter, and
then the energy would be transferred from the system to the surroundings as heat (heat transfers from
hotter matter to colder matter).
Why does the reaction mixture undergo an increase in temperature even though energy is emitted?
As noted above, the PE “lost” is initially converted into KE of nanoscopic particles (thermal energy) in the
system, raising the temperature of the system. The energy only “transfers out” after this initial T increase.
When you feel a reaction vessel and its “hot”, you are really sensing heat energy transferring from the
system into your fingers (the surroundings).
6.23. From a molecular viewpoint, where does the energy absorbed in an endothermic chemical reaction go?
It goes into raising the PE of the system. The atoms/ions (or molecules) in the products (after
rearrangement) have a greater PE than they did when they were arranged however they were before in the
reactants.
Why does the reaction mixture undergo a decrease in temperature even though energy is absorbed?
Initially, when the process occurs (think of it as occurring rapidly), the system converts its own thermal
energy (KE of particles) into potential energy as the process occurs, lowering the T of the system. Energy
from the surroundings then transfers into the system because of the (created) T difference.
Meaning of a Thermochemical Equation / Stoichiometry with energy
14. MP Solution to this problem not in key at this time.
15. MP Solution to this problem not in key at this time.
16. MP (6.60).
What mass of natural gas (CH4) must burn to emit 267 kJ of heat?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Hrxn = -802.3 kJ
NOTE: I do not like the subscript “rxn” here, although that is the conventional way of representing the H
for a thermochemical equation. I think it should be symbolized Heqn because it only reflects the H
for an amount of reaction indicated by the coefficients of the balanced equation, interpreted as
moles. Since I like to distinguish any “actual reaction” (with actual amounts) from the equation, the
subscript “rxn” can be confusing or misleading. Please make sure that you realize that the “H of
PS8b-4
Answer Key, Problem Set 8b
reaction” here does not mean the H of the “actual reaction” (with actual amounts). Please go back
to your Exp 14 lab report form and note the difference between the Hsys (for your actual dissolution
or reaction) and the H for the thermochemical equation (which Tro would call Hrxn)
Answer: 5.34 g CH4
Strategy:
1) 267 kJ of heat is to be generated by the chemical reaction represented by the thermochemical
equation shown, which has a stated Hrxn = -802.3 kJ. Since, 267 kJ of heat is to be generated, H
for the actual reaction will be 267 kJ. The question is asking about the amount of CH4 that would
need to be burned (excess O2 is assumed), so interpret the enthalpy change for the thermochemical
equation in terms of energy and amount of CH4. Namely, that “802.3 kJ of energy is released for
every one mole of CH4 that is burned.” It is per one mole only because the coefficient of CH4 in this
balanced equation is 1.
802.3 kJ produced
1 mol CH4 reacted
, and use it to calculate moles of CH4 reacted. (i.e., do the stoichiometry). Note that since kJ of
1 mol CH4 reacted
(so
energy is “known”, you must use the conversion factor in its reciprocal form:
802.3 kJ produced
really, you could just go directly to this from the balanced equation—I just directly “translated” the H
from the equation into kJ/mol of CH4 because I’m used to thinking about it this way.)
2) Create the corresponding “energy-mole” conversion factor (with Hrxn and CH4):
3) Calculate grams of CH4 from moles using the molar mass of CH4: 12.01 + 4(1.008) = 16.04 g/mol
Execution:
This is reasonable, because this is less
than a mole of CH4, and 267 kJ is less
than 802 kJ. (Actually, you could go further
1 mol CH4 reacted 16.04 g
267 kJ produced x
x
 5.338..  5.34 g CH4
802.3 kJ produced mol CH4
and say that this is about a third of a mole,
which makes sense since 267 is about a third of
802)
17. MP (6.93) Determine the mass of CO2 produced by burning enough of each of the following fuels to
produce 1.00 x 102 kJ of heat. Which fuel contributes least to global warming per kJ of heat produced?
Note: I am leaving off state designations for all gases. Also, I will leave off the “rxn” subscript from Tro, since if the H comes after a
chemical equation, you should know it refers to “1 mole’s worth of reaction”. Lastly, I am leaving off the degree superscript
here because standard states are not involved in this problem and so it is unnecessary to designate the H as such.
Answers: (a) 5.49 g CO2 (for CH4) ; (b) 5.96 g CO2 (for C3H8); (c) 6.94 g CO2 (for C8H18)
Thus, CH4 contributes the least to global warming per kJ of heat produced, since it
produced the least amount of CO2 per 100. kJ heat produced.
Strategy: Analogous to the prior problem, except here one must interpret the H of the thermochemical
equation in terms of moles of CO2 produced rather than moles of fuel reacted. For example, for
2217 kJ produced
. As such, in the 3rd step, you will calculate grams of CO2 from
(b):
3 mol CO 2 produced
moles, and thus the molar mass of CO2 must be used: 12.01 + 2(16.00) = 44.01 g/mol
Execution:
(a) CH4 + O2 → CO2 + 2 H2O; H = -802.2 kJ
5
1.00 x 10 2 kJ produced x
1 mol CO 2 produced
44.01 g
x
 5.4854..  .49 g CO 2
802.3 kJ produced mol CO 2
PS8b-5
Answer Key, Problem Set 8b
(b) C3H8 + 5 O2 → 3 CO2 + 4 H2O; H = -2217 kJ
(c) C8H18 +
25
2
5
1.00 x 10 2 kJ produced x
3 mol CO 2 produced
44.01 g
x
 5.9553..  .96 g CO 2
2217 kJ produced
mol CO 2
O2 → 8 CO2 + 9 H2O; H = -5074.1 kJ
1.00 x 10 2 kJ produced x
8 mol CO 2 produced
44.01 g
x
 6.938..  6.94 g CO 2
5074.1 kJ produced mol CO 2
Calorimetry When a Process Does Occur in the System (“bringing multiple ideas together”)
18. MP Solution to this problem not in key at this time.*
*However, the way to approach this problem is similar to the way to approach each of the
next two problems. As such, it might be worth looking at the first three steps of the
strategy listed for Problem 20. After that, recognize that unlike in Problem 20, here it is
the final temperature that is asked for. Also recognize that here (as in Problem 19), the
H “per mole” is given (it is for the dissolution of one mole of CaCl2(s), so it is
effectively like giving you the H for the thermochemical equation). So this problem is a
true blend of problems 19 and 20, and also relates to your Part B of Experiment 14!
These three problems are likely the most difficult problems on this set for many
students because they bring together so many ideas from this unit. I hope you will take
the time needed to “put it all together”.
19. MP (6.99)
H2O(l) → H2O (g) ; H = +44.01 kJ
Estimate the mass of water that must evaporate from the skin to cool the body by 0.50 °C. Assume a body mass of 95 kg and a
specific heat for the body of 4.0 J/g·°C.
Answer: 78 g of H2O
Strategy:
(Note: I have formulated this part of the key in reference to Problems 16 and 17. That said, you may
also wish to read steps 1 through 4 of the strategy for the next problem in this key (Problem 20) since
that strategy applies equally well to this problem. It just approaches it from a more “general”
perspective.)
You are given a thermochemical equation and are asked about an amount of a reactant that must
“react” (here it is technically physical change rather than chemical change). Note that this is the same
kind of question as in Problems 16 and 17 above. However, what is different here is that you are not
given the amount of heat (to absorb)—you must calculate it based on the description of the
“surroundings’” temperature change (here, the body is considered the surroundings). So you must do
this kind of problem in two steps. Without a formal equation, what you do first is use the specific heat
capacity, mass, and the magnitude of the temperature change for the body to calculate qbody
(magnitude). Then you use this as the “energy needed” to be used up by the water evaporation
reaction. More specifically:
1) In equation terms, one can represent the first step I described above as a calculation of the H of the
actual reaction (Hsys) from the q of the surroundings (qbody = qsurr) using:
Hsys = -qsurr = -qbody = -(Cs, body x mbody x Tbody)
Note that an opposite sign that appears here formally since the qbody described above was considered as a positive quantity
(magnitude only) when technically it is a negative quantity (heat transfers out of the body and so qbody is negative).
PS8b-6
Answer Key, Problem Set 8b
2) Now you can do with this Hsys what you did in Problems 16 and 17. That is, use it, along with an


44.01 kJ
 to
“energy-moles” conversion factor from the thermochemical equation 
1
mol
H
O
that
evaporates
2


figure out the amount of reactant that reacts.
**Be careful, however, to make sure that you convert one of the energy units to be consistent with
the other! I will convert J to kJ rather than kJ to J, although either one is obviously “correct”.**
3) Convert moles to grams using the molar mass of H2O to finish the problem.
Execution of Strategy:
Hsys   (4.0 J/g·°C x 95,000 g x (-0.50°C)) = +190,000 J
+190,000 J x
190 kJ x
(T is negative because you are “cooling”)
1 kJ
 190 kJ
1000 J
1 mol H2 O evaporates
18.02 g
x
 77.79.. 78 g H2O evaporates
44.01 kJ
1 mol H2 O
NOTE: This problem is very similar to problem #4 on PS8a! It might be worth taking a look at the problem
now and comparing the two. Although the wording is different (and the PS8a problem asks for moles
instead of grams, and deals with an exothermic chemical reaction rather than an endothermic physical
change), the thought process and sequence of steps is effectively the same.
20. MP (6.75)
Zn(s) + 2 HCl(aq) → ZnCl2(aq)
+ H2(g) ; H = ???
When 0.103 g of Zn is combined with enough HCl to make 50.0 mL of solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 22.5 °C to 23.7 °C. Find H for the thermochemical equation. (Assume the solution
has d = 1.0 g/mL and Cs  4.18 J/g·°C)
Answer: H  -160,000 J  -160 kJ  -1.6 x 102 kJ (per mole of “reaction”)
Strategy:
(NOTE: This problem is very similar to Parts B and C of Experiment 14! Perhaps take a look at
that now while working this problem!)
1) Recognize this as a “heat transfer” kind of problem. That means there will be a qsys = qsurr
type of setup.
2) Unlike problem 9 on this set, this problem does involve a process (here, it is a chemical
change)—i.e., it does NOT involve the flow of heat from an initially hotter object or sample to a
cooler object or sample. This means a couple of things: First, we choose the atoms that
undergo rearrangement to be the system. Second, it makes more sense to view the pertinent
“equality” from (1) to be:
Hsys = qsurr
Writing Hsys instead of “qsys” helps prevent you from (incorrectly) thinking that you need a specific heat capacity for
the system, and trying to set the problem up like Problems 9 - 11 of this set!
3) Since no “process” occurs in the surroundings (which is the solution here), its T change
depends only on the heat transfer, Cs, and m according to: qsol’n = Cs,sol’n x msol’n x Tsol’n.
Thus:
Hsys = -(Cs,sol’n x msol’n x Tsol’n)
4) Since all the variables on the right side are “known”*, you can calculate Hsys corresponding to
the actual (amount of) reaction that occurred, which involved 0.103 g of Zn.
* 50.0 mL x 1.0 g/mL = 50.0 g solution; Cs,sol’n is said to be 4.18 J/g·°C, Ti and Tf are given.
5) The H for the thermochemical equation must correspond to the reaction of 1 mol of Zn (since
the coefficient of Zn is 1 in the balanced equation. Thus, you must either set up a proportion,
or simply divide “kJ” by “mol Zn” to get “kJ/mol of Zn reacted.
PS8b-7
Answer Key, Problem Set 8b
Execution of Strategy:
Hsys   (4.18 J/g·°C x 50.0 g x (23.7°C  22.5°C)) = -250.8 J (Note: T = 1.2 °C  2 SF)
mol Zn 
0.103 g Zn
= 0.001575…mol Zn
65.38 g/mol Zn
J per mol of Zn reacted 
-250.8 J
 -159197  -160,000 J  -160 kJ
0.001575 mol Zn
Work vs Heat, Sign Conventions, and Relation to Change in Internal Energy (E, q, w relationships)
21. MP Solution to this problem not in key at this time. But see General Reasoning/Strategy in #23
below (6.39), which applies here as well.
22. MP Solution to this problem not in key at this time. But see General Reasoning/Strategy in #23
below (6.39), which applies here as well.
23. MP (6.39).
Identify each energy exchange as primarily heat or work and determine whether the sign of E is positive or
negative for the system.
Answers: (a) heat, E is positive; (b) work, E is negative; (c) heat, E is positive
General Reasoning/Strategy:
Heat or work?
1) If a hotter sample of matter is put next to a cooler one, heat transfers from the hotter one to the colder
one (this is the definition of heat!). So if there is a scenario in which two samples of matter at
different temperatures are put next to each other (and nothing else happens process-wise), the
energy exchange is clearly heat.
2) In order for there to be work, something needs to move on a macroscopic scale (work is “force acting
over a distance). So if there is a volume change (in the absence of a physical or chemical change) or
a change in position of a macroscopic object, then the energy exchange is “work” rather than heat.
3) When neither of the above is the case, most likely there is a physical or chemical change occurring
(in what we would define as the system). In general, whenever there is a physical or chemical
change occurring in the system, the potential energy changes associated with the nanoscopic
“rearrangement” or “repartnering” lead to changes in KE energy, which ultimately leads to heat
transfer either into or out of the system [because the system would initially get hotter or colder as PE
was converted to KE or vice versa] (see Problem #13 WO5 above). So these situations involve
(mostly) heat. (Even if there is some work, it will generally be small compared to the value of the
enthalpy change (heat at constant P) when there is a physical or chemical change happening).
Positive or negative?
Heat: If heat transfers from system to surroundings, heat is negative.
If heat transfers from surroundings to system, heat is positive.
Work: If work is done by the system (on the surroundings), work is negative (chemists’ convention) and
internal energy (E) decreases (if heat is zero).
If work is done on the system (by the surroundings), work is positive and internal energy (E)
increases (if heat is zero).
NOTE: Expansion  work done by system (on surroundings)  wsys is negative.
Compression  work done on system (by surroundings)  wsys is positive.
Strategy Applied:
(a) It is heat and not work because the process occurring in the system is a physical change (turning
liquid water into a gas) and there is minimal “macrocscopic moving” of anything. It is positive
because evaporation requires energy (it involves separating molecules from one another—see
PS8b-8
Answer Key, Problem Set 8b
Problems WO3-WO4 on this set (above) or Problem #5 on PS8a). So heat transfers into the sweat
from the body, qsys (and H) is positive, and thus E is positive (assume work is minimal).
(NOTE: Although a bit of gas is produced, and thus a bit of work is done on the surroundings, that pales in comparison to
the change in enthalpy associated with the nanoscopic rearrangements.)
(b) It is work and not heat because a gas expands against a pressure, where there is no physical or
chemical change (the contents of the balloon remain a gas throughout). It is negative because the
contents of the balloon push the atmosphere away, thus doing work on it (and using its own energy
to do that work).
(c) It is heat and not work because a hot sample was put next to a colder one. Heat transferred from
the flame (surroundings) to the solution (system), so qsys is positive.
24. and 25. MP (6.42) & MP (6.43).
6.42. A system absorbs 196 kJ of heat and the surroundings do 117 kJ of work on the system. What is the change in internal
energy of the system?
6.43. The gas in a piston (defined as the system) warms and absorbs 655 J of heat. The expansion performs 344 J of work
on the surroundings. What is the change in internal energy for the system?
Answers: 313 kJ and 311 J
Reasoning:
In both of these problems, you need to realize that the internal energy of a system can only be changed
two ways: via “heating” (or removing heat) or via work (being done either by the system or on the system).
Namely:
Esys  q + w
(chemists’ convention)
where q is “heat” and w is “work”. A positive q means heat flows into the system (raising the system’s
energy) and a negative q means heat flows out of the system. A positive w (in a chemistry context) means
work is done on the system (raising its internal energy), and a negative w means work is done by the
system (lowering its internal energy).
In 6.42 then, q = +196 kJ and w = +117, so Esys  +196 + (+117) = +313 kJ
In 6.43, q = +655 J and w = -344 J, so Esys  +655 + (-344) = +311 J
Hess’s Law
26. MP (6.78).
NOTE: This is not really a “Hess’s Law” problem as much as an “Understanding what a thermochemical
equation means” problem. This was covered in the PS8a handout and problem number 2. This
problem was put here because you need to use these ideas when you “manipulate” the given
equations in a Hess’s Law type problem en route to finding a set of equations that will sum up to the
target equation (see my handout with a General Procdure to Hess’s Law type problems).
Answers:
(a) 465 kJ ( 3 x 155 kJ). Process is exactly three times as much “reaction” as original.
(b) -155 kJ. Process is exact reverse of original so sign is opposite.
(c) -77.5 kJ ( -½ x 155 kJ). Process is exactly one-half as much “reaction” as the reverse of the original.
PS8b-9
Answer Key, Problem Set 8b
27. MP (6.82).
Calculate Hrxn (which means the H for the “target” thermochemical equation, or what I often call Hoverall in a
Hess’s Law problem such as this) for the reaction (equation):
CH4 + 4 Cl2 → CCl4 + 4 HCl
(all gases)
given:
C(s) + 2 H2 → CH4;
H1 = -74.6 kJ
C(s) + 2 Cl2 → CCl4;
H2 = -95.7 kJ
H2 + Cl2 → 2 HCl;
H3 = -92.3 kJ
Answer: Hoverall  -205.7 kJ
Explanation: Following the approach described in my handout (and in that pain-stakingly made video lecture
that will post soon, if not already), go substance by substance in the target equation and check to make
sure that a substance appears in only one place in the set of equations given. Then make the position
(reactant or product) and “amount needed” (i.e., coefficient) match by either reversing the equation or
multiplying the equation through by a number (or both). In this case, you must be careful to “skip” Cl2
(the second reactant) during this process since it appears in more than one of the given equations.
Execution:
1) CH4 is only in equation 1, but it is a product there so the equation must be flipped: H1’ = +74.6 kJ
2) Cl2 is in equations (2) and (3), so SKIP IT!
3) CCl4 is only in equation 2, and it is on the correct side and with correct coefficient, so it needs no
modification at all: H2’ = -95.7 kJ
4) HCl is only in equation 3, but it has a coefficient of 2 there, and we “need” a 4. So multiply the equation
by two: H3’  2 x H3  2 x -92.3 kJ  -184.6 kJ
(You should really check to make sure these add up to the target, but I will omit that here for now.)
Hoverall  +74.6 + (-95.7) + (-184.6)  -205.7 kJ
28. WO6.
Calculate the standard enthalpy change, ∆Ho, for the formation of 1 mol of strontium carbonate (the material
that gives the red color in fireworks) from its elements.
Sr (s)

C(graphite)

O2 (g)
3
2

SrCO3 (s)
The information available is
(1) 2 Sr (s) 
O2 (g)
(2) SrO (s)  CO2 (g)
(3) C(graphite)
 2 SrO (s)

 O2 (g)
H  -1184 kJ
SrCO3 (s)

CO2 (g)
H  -234 kJ
H  -394 kJ
Answer: -1220 kJ
Strategy: Same as last problem. But note that here it is easiest to skip the 3rd substance (O2) since it
is present in more than one of the given equations.
Execution:
1) Sr(s) only in (1), but there’s a 2 there and need a 1  take ½ x (1): H1°’ = ½ x (-1184 kJ) = -592 kJ
2) C(graphite) only in (3). Take equation “as is” b/c correct coefficient and side: H3°’ = -394 kJ
3) **SKIP O2 BECAUSE IT IS IN MULTIPLE EQUATIONS**
PS8b-10
Answer Key, Problem Set 8b
4) SrCO3 only in (2). Take equation “as is”. H2°’ = -234 kJ
Hoverall  -592 + (-394) + (-234)  -1220 kJ
Standard Enthalpies of Formation and Calculating Hrxn from them
29. WO7. (i) What is the standard enthalpy of formation of a substance?
(ii) Do 6.82(b,d only) in Tro and add parts (e) and (f): (e) F2(g) ; (f) F(g)
(iii) Why is the standard enthalpy of formation of any element (in its standard state) zero?
(iv) Why is the answer in iii(f) above positive?
Answers, Some With Reasoning Within:
(i) The standard enthalpy of formation of a substance is the enthalpy change corresponding to the
formation of one mole of a substance (in a specific state) from its elements in their standard states.
(ii) Write an equation for the formation of [one mole of] each compound from its elements in their
standard states, and find Hof for each from Appendix IIB.
Strategy:
1) Recognizing that a standard enthalpy of formation is, by definition, the change in enthalpy
associated with the formation of one mole of a substance, write the formula of the
substance on the right side of an arrow (because it is being “formed”, right?) with a
coefficient of one (and only one!).
2) Look at each element symbol in the product substance, and write the formula for the
stable form of each element, including the state designation, on the left side of the
equation, without a coefficient (for now). Note that the most stable forms of H, N, O, F, Cl,
Br, and I are the diatomic molecules, and the most stable form of C is graphite (written as
C(s, graphite). For metals, the most stable forms are generally the monatomic atoms (i.e, the
same as their symbol, without any subscripts). You will not need to memorize any other
elements whose stable form is not their monatomic atom.
3) Remember that the coefficient for the substance on the right MUST REMAIN ONE. As
such, add coefficients the left side, using fractions if needed, to balance the equation
for each atom type.
Answers:
(b) MgCO3(s):
(d)
Mg(s)  C(s,graphite) 
3
2
O 2 ( g )  MgCO 3 ( s ) ; Hof (MgCO 3 (s)) = -1095.8 kJ/mol
NOTE: 3/2 of a molecule of O mathematically yields three O atoms. You must keep the fraction here since the right
side’s coefficient must be “one”.
1
C(s,graphite)  2 H2 ( g )  O 2 ( g )  CH3 OH( l ) ; Hof (CH3 OH( l )) = -238.6 kJ/mol
CH3OH(l):
2
o
NOTE: Physical states are critical in these problems. The H f for CH3OH(g) is greater than -238.6 kJ (it is -201.0
kJ) because the enthalpy (think PE here) of a substance in its gaseous state is always greater than in its
liquid state—it requires energy to turn a liquid into a gas! (Think about that for a moment and ask me if you
do not see why this is so.
(e) F2(g):
F2(g)  F2(g)
; Hof (F2(g)) = 0 kJ/mol
(this “reaction” is “no reaction”—nothing has
rearranged or changed)
When you form an element in its standard state from “its elements in their standard states”, the equation represents
no change of any kind. The reactant is identical to the product.
(f) F(g):
1
2
F2 ( g )  F( g )
; Hof (F(g)) = +79.38 kJ/mol
PS8b-11
Answer Key, Problem Set 8b
(iii) Shortest Answer: Forming a mole of an element “from a mole of its element(s)” (both in the same
[standard] state) is doing nothing at all! See (ii)(e) above! If there is no actual
chemical or physical change, then there can be no enthalpy change!
More Commentary, More Detailed Answer (Partial “Rant”):
I think it is absolutely silly for Tro to write as a separate “Rule” on p. 258 (see “3. Standard Enthalpy of
Formation”) that for an element in its standard state, Hof = 0. You should not have to memorize this
or think of the definition for Hof of an element as being somehow “different” than the definition for
the Hof for any substance. Just write out the equation for the “formation” and if it is for an element in
its standard state, you should see that nothing has happened and thus the Hof is zero!
When you form a compound from elements, you necessarily must make and/or break some bonds,
but when you for an element from an element (same form and same state), there is no change
physically and so there is no change in energy/enthalpy. This should make sense to you. If not,
please reread and/or ask me.
**NOTE: The choice of definition for Hof essentially makes the most stable forms of elements in their
standard states our “reference” state for enthalpy. It arbitrarily “makes” them the “zeros” of
potential energy.
(iv) This Hof is positive because the process it represents involves (only) the breaking of a chemical
bond (compare this to Problems 11 (d) and (e)!), and we know that it takes energy to break bonds
(endothermic, positive H). In fact, the Hof for F(g) is precisely half of the energy needed to break
apart two F atoms that are bonded together.
30. MP Solution to this problem not in key at this time. However, the strategy, in addition to being
described in the video for this problem in Mastering, is described in my answer to the
next problem (#31).
31. MP (6.86(a,d))
Use standard enthalpies of formation to calculate Hrxn° for each reaction (equation).
General Comment/Strategy:
Finding the H° for a balanced chemical equation means finding the H (under standard
conditions) associated with “unforming” the number of moles of each reactant indicated by its
coefficient and “forming” the number of moles of each product indicated by its coefficient. So we
can use H°’s of formation of each substance, abbreviated as Hf°’s, along with Hess’s Law, to find
the overall H° for any equation! It is simply the sum of two “processes”:
1) unforming all the reactants [which is the opposite of forming them], denoted as Hunforming_R’s°, &
2) forming all the products, denoted . Hforming_P’s°

Hoverall° = Hunforming_R’s° + Hforming_P’s°
= -Hforming_R’s° + Hforming_P’s°
Swapping the order of the two terms yields:
= Hforming_P’s° Hforming_R’s°   np Hfo (products)   nr Hfo (reactants)
which is just 6.15 in Tro! It’s just “products” – “reactants” using Hf°’s for each reactant and product
(times their respective mole amounts (coefficients), and being VERY careful with signs!!) You also
need to make sure to get the proper values from the table (with proper signs). And, of course,
make sure not to forget to multiply each substance’s Hf° by its coefficient in the balanced
equation.
PS8b-12
Answer Key, Problem Set 8b
(a) 2 H2S(g) + 3 O2(g) → 2 H2O(l) + 2 SO2(g)
(d) N2O4(g) + 4 H2 → N2(g) + 4 H2O(g)
Answers: (a) -1124.0 kJ ( [2(-285.8) + 2(-296.8)] – [2(+-20.6) + 3(0)] )
(d) -976.4 kJ ( [1(0) + 4(-241.8)] – [1(+9.16) + 4(0)] )
32. MP (6.85)
(values come from Appendix)
(values come from Appendix)
Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the equation:
N2H4(l) + N2O4(g) → 2 N2O(g) + 2 H2O(g)
Calculate Hrxn° for this reaction (equation) using standard enthalpies of formation.
Answer: -380.2 kJ
( [2(+81.6) + 2(-241.8)] – [1(+50.6) + 1(9.16)] )
(values come from Appendix)
Strategy: Same as prior problem.
---------------------------- END OF PROBLEM SET----------------------------------Extra Problems for Practice (from prior years’ answer keys)
6.63.
The propane fuel (C3H8) used in gas barbeques burns according to the thermochemical equation:
C3H8 + 5 O2 → 3 CO2 + 4 H2O; H = -2217 kJ
If a pork roast must absorb 1.6 x 103 kJ to fully cook, and if only 10% of the heat produced by the barbeque is actually
absorbed by the roast, what mass of CO2 is emitted into the atmosphere during the grilling of the pork roast?
Answer: 950 g CO2
Reasoning:
1) If 1.6 x 103 kJ is needed, and only 10% (0.10) of the heat produced by the BBQ is actually absorbed,
than you need to generate 1/0.10 = 10 times as much heat as you need, i.e., 1.6 x 104 kJ. (Another
way to look at this is to say that for every 100 kJ generated by the reaction, only 10 kJ (10% of 100) is
absorbed. Thus again, 10x as much heat needs to be generated. Hence, 1.6 x 104 kJ).
2) Create an “energy-mole” conversion factor (with H and CO2) using the balanced equation:
2217 kJ released
, and use it to calculate moles of CO2 produced. (i.e., do the stoichiometry)
3 mol CO2 produced
3) Calculate moles of CO2 to grams using molar mass
Execution:
1.6 x 10 4 kJ x
3 mol CO2 produced
44.01 g
x
 952.8..  950 g
2217 kJ produced
mol CO2
6.69. Two substances, A and B, initially at different temperatures, come into contact and reach thermal
equilibrium. The mass of substance A is 6.15 g and its initial temperature is 20.5 °C. The mass of
substance B is 25.2 g and its initial temperature is 52.7 °C. The final temperature of both substances at
PS8b-13
Answer Key, Problem Set 8b
thermal equilibrium is 46.7 °C. If the specific heat of substance B is 1.17 J/g·°C, what is the specific heat of
substance A?
Answer: 1.1 J/g·°C
Strategy:
1) Like the last problem, this one only involves heat transfer (no “process” in system). Thus:
Cs,A x mA x TA   (Cs,B x mB x TB)
2) Substitute in properly and solve for Cs,A
Execution of Strategy:
Cs,A x 6.15 g x (46.7°C  20.5°C)   (1.17 J/g·°C x 25.2 g x (46.7°C  52.7°C))
C s ,A 
(-176.9 J)
 1.097 J/g·°C  1.1 J/g·°C
161.13 g  C
6.132.
The internal energy of an ideal gas depends only on its temperature. Which statement is true of an isothermal
(constant-temperature) expansion of an ideal gas against a constant external pressure? Explain.
(a) E is positive
(b) w is positive.
(c) q is positive
(d) E is negative
Answers:
(a) False. The problem states that “the internal energy of an ideal gas depends only on its temperature.”
Thus, if the temperature does not change (“isothermal”), Esys  0
(b) False. If the system expands, the system does work on the surroundings (pushes it away)  w < 0
(contributes to a decrease in Esys)
(c) True. Mathematical answer: If Esys  0 and Esys  q + w and w < 0, then q > 0
Conceptual answer: If the system does work on the surroundings, but its internal energy does not
change, the energy to do that work must be coming from an input of heat from the surroundings.
(d) False. Esys  0 (See (a).)
6.136.
Two (samples of) substances A and B, initially at different temperatures, are thermally isolated from their
surroundings and allowed to come into thermal contact. The mass of (the sample of) substance A is twice the mass of
(the sample of) substance B, but the specific heat of substance B is four times the specific heat of substance A. Which
substance will undergo a larger change in temperature?
Answer: Sample A will change its temperature by a greater amount (in absolute value). (Its change
will, in fact, be twice the change of Sample B.)
Explanation: Since q = Cs x m x T, it should be clear that for a given q, the greater the Cs, the smaller
the change in T (a greater specific heat means “harder to change its T”. Also, for a given q, the greater
the m, the smaller the change in T (the greater the mass, the harder it is to change its T). In fact, T is
inversely proportional to both of these variables. So although sample A is 2x the mass of B (making the
T half as great (2x smaller) as B), B’s 4x larger specific heat would make its T one-quarter as great
(4x smaller). So B’s larger specific heat “wins” here, and its T will be half as great as A’s.
Ta CB
m
4
1 2
x B  x  , meaning that the absolute

CA
1 2 1
mA
Tb
value of A’s temperature change is twice that of B’s.
NOTE: You could also show mathematically that 
PS8b-14
Answer Key, Problem Set 8b
PS8b-15