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Utexas Chemistry 302 Mccord
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kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 1
This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.
001 10.0 points
Which of the following will best help deter-mine the direction of heat flow in a system?
1. temperature correct
2. pressure
3. internal energy
4. work
5. enthalpy
Explanation:
Heat flows spontaneously from a hotterbody to a colder body. In other words, heatflows from a body of high temperature to abody of lower temperature.
002 10.0 points
In the manufacture of nitric acid by the oxi-dation of ammonia, the first product is nitricoxide. The nitric oxide is then oxidized tonitrogen dioxide:
2NO(g) + O2(g) −→ 2NO2(g)
Calculate the standard reaction enthalpyfor the reaction above (as written) using thefollowing data:N2(g) + O2(g) −→ 2NO(g)
∆H◦ = 180.5 kJN2(g) + 2O2(g) −→ 2NO2(g)
∆H◦ = 66.4 kJ
1. −100.3 kJ/mol rxn
2. −690.72 kJ/mol rxn
3. −520.2 kJ/mol rxn
4. −128.2 kJ/mol rxn
5. −975.0 kJ/mol rxn
6. −114.1 kJ/mol rxn correct
7. −252.4 kJ/mol rxn
Explanation:
Using Hess’ Law and the given standard re-action enthalpies, the first reaction is reversedand added to the second:
2NO(g) −→ N2(g) + O2(g)
∆H◦ = −180.5 kJ
N2(g) + 2O2(g) −→ 2NO2(g)
∆H◦ = 66.4 kJ
2NO(g) + O2(g) −→ 2NO2(g)
∆H◦ = −114.1 kJ
003 10.0 points
What mass of ethanol (C2H5OH(ℓ)) must beburned to supply 500 kJ of heat? The stan-dard enthalpy of combustion of ethanol at 298K is −1368 kJ ·mol−1.
1. 126 g
2. 29.7 g
3. 16.8 g correct
4. 2.74 g
5. 10.9 g
Explanation:
500 kJ
1368 kJ/mol= 0.365497 mol of ethanol
(0.365497 mol)(46 g/mol)= 16.8129 g ethanol
004 10.0 points
Carbon monoxide reacts with oxygen to formcarbon dioxide by the following reaction:
2CO(g) + O2(g) → 2CO2(g)
∆H for this reaction is −135.28 kcal.
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 2
How much heat would be released if 12.0moles of carbon monoxide reacted with suffi-cient oxygen to produce carbon dioxide? Useonly the information provided in this ques-tion.
1. 811.68 kcal correct
2. 135.28 kcal
3. 270.56 kcal
4. 1623.36 kcal
5. 541.12 kcal
6. 405.84 kcal
Explanation:
∆H = −135.28 kcal nCO = 12.0 mol(
−135.28 kcal
1 mol rxn
)(1 mol rxn
2 mol CO
)
×(12 mol CO) = −811.68 kcal
so 811.68 kcal were released.
005 10.0 points
Burning 1 mol of methane in oxygen to formCO2(g) and H2O(g) produces 803 kJ of en-ergy. How much energy is produced when 3mol of methane is burned?
1. 803 kJ
2. 2,409 kJ correct
3. 1,606 kJ
4. 268 kJ
Explanation:
n1 = 1 mol q1 = 803 kJn2 = 3 mol
If we know howmuch heat is evolved when 1mole of methane is combusted, then we knowhow much heat would be evolved if 3 mol ofmethane were combusted:(
803 kJ
1 mol methane
)
(3 mol methane)
= 2409 kJ
006 10.0 points
Calculate the standard reaction enthalpy forthe reaction.
CH4(g) + H2O(g) → CO(g) + 3H2(g)
given2H2(g) + CO(g) → CH3OH(ℓ)
∆H◦ = −128.3 kJ ·mol−1
2CH4(g) + O2(g) → 2CH3OH(ℓ)∆H◦ = −328.1 kJ ·mol−1
2H2(g) + O2(g) → 2H2O(g)∆H◦ = −483.6 kJ ·mol−1
1. +206.1 kJ ·mol−1 correct
2. +216 kJ ·mol−1
3. +42.0 kJ ·mol−1
4. +155.5 kJ ·mol−1
5. +412.1 kJ ·mol−1
Explanation:
We need to reverse the first reaction, halvethe second, halve and reverse the third andadd the results:
CH3OH(ℓ) → 2H2(g) + CO(g)∆H = 128.3 kJ/mol
CH4(g) + 0.5O2(g) → CH3OH(ℓ)∆H = −164.05 kJ/mol
H2O(g) → H2(g) + 0.5O2(g)
∆H = +241.8 kJ/mol
CH4(g) + H2O(g) → CO(g) + 3H2(g)
∆H = 206.05 kJ/mol
007 10.0 points
The value of ∆H for the reaction
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ)
is −2220 kJ/mol rxn. How much heat isgiven off when 11.0 g of propane gas (C3H8)is burned at constant pressure?
1. 25.96 kJ
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 3
2. 1665.0 kJ
3. 2220.0 kJ
4. 22420.0 kJ
5. 555.0 kJ correct
6. 6660.0 kJ
7. 50.5 kJ
Explanation:
∆H = −2220 kJ/mol mC3H8= 11 g
q =(
−2220 kJ
mol rxn
)( 1 mol rxn
1 mol C3H8
)
×
(1 mol C3H8
44 g C3H8
)
(11 g C3H8)
= −555 kJ or 555 kJ released
008 10.0 points
Calculate the standard enthalpy change forthe reaction
2HCl(g) + F2(g) → 2HF(ℓ) + Cl2(g)
given
4HCl(g) + O2(g) → 2H2O(ℓ) + 2Cl2(g)
∆H0 = −202.4 kJ/mol rxn
1
2H2(g) +
1
2F2(g) → HF(ℓ)
∆H0 = −600.0 kJ/mol rxn
H2(g) +1
2O2(g) → H2O(ℓ)
∆H0 = −285.8 kJ/mol rxn
1. ∆H0 = −1015.4 kJ/mol rxn correct
2. ∆H0 = −516.6 kJ/mol rxn
3. ∆H0 = +1587.2 kJ/mol rxn
4. ∆H0 = −1088.2 kJ/mol rxn
5. ∆H0 = +1116.6 kJ/mol rxn
6. ∆H0 = +516.6 kJ/mol rxn
7. ∆H0 = −1116.6 kJ/mol rxn
8. ∆H0 = +1088.2 kJ/mol rxn
9. ∆H0 = +1015.4 kJ/mol rxn
10. ∆H0 = −1587.2 kJ/mol rxn
Explanation:
The first equation needs to be multiplied by1
2in order to get the equation we’re interested
in. Thus its ∆H0 is multiplied by1
2as well.
The second equation needs to be multipliedby two in order to get the equation we’reinterested in. We also multiply its ∆H0 bytwo.The third equation needs to be reversed, so
the sign of its ∆H0 should change.Then we add the equations to get the equa-
tion we’re interested in. We also add the ad-justed ∆H0 values to get the answer, −1015.4kJ/mol rxn.
009 10.0 points
Calculate the standard enthalpy of formationof bicyclo[1.1.0]butane
H
H
C
C
CH2H2C
given the standard enthalpies of formation of717 kJ ·mol−1 for C(g) and 218 kJ ·mol−1 forH(g) and the average bond enthalpies of 412kJ ·mol−1 for C H and 348 kJ ·mol−1 forC C.
1. +175 kJ ·mol−1
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 4
2. −472 kJ ·mol−1
3. +312 kJ ·mol−1
4. −124 kJ ·mol−1
5. −36 kJ ·mol−1 correct
Explanation:
We can write an equation in which we com-pletely decompose bicyclo [1,1,0] butane:C4H6(bicyclobutane, g) → 4C(g) + 6H(g)
∆H◦
rxn = 5 (BEC−C) + 6 (BEC−H)(The comment on the right comes from
dissecting the structure given in the questionand noting how many of each kind of bond ispresent). To find ∆H for this reaction we canuse Hess’ Law with formation enthalpies:
∆H◦
rxn =[
4∆H◦
f C(g) + 6∆H◦
f H(g)
]
−
[
1∆H◦
f C4H6(g)
]
= [4 (717 kJ/mol) + 6 (218 kJ/mol)]
−∆H◦
f C4H6(g)
= 4176 kJ/mol−∆H◦
f C4H6(g)
Now we use the comment on which bondswere broken:
∆H◦
rxn = 6BEC−C + 6BEC−H
= 5 (348 kJ/mol) + 6 (412 kJ/mol)
= 4212 kJ/mol
We can set the two sides of the equationequal since they represent the same reaction:
4212 kJ/mol = 4176 kJ/mol−∆H◦
f C4H6(g)
∆H◦
f C4H6(g)= −36 kJ/mol .
010 10.0 points
Calculate the enthalpy change for the reaction
2 SO2(g) + O2(g) → 2 SO3(g)
∆Hf for SO2(g) = −16.9 kJ/mol;∆Hf for SO3(g) = −21.9 kJ/mol.
1. +5.0 kJ/mol rxn
2. +10.0 kJ/mol rxn
3. −77.6 kJ/mol rxn
4. −5.0 kJ/mol rxn
5. −10.0 kJ/mol rxn correct
Explanation:
Reactants:∆Hf SO2(g) = −16.9 kJ/mol∆Hf O2(g) = 0 kJ/molProducts:
∆Hf SO3(g) = −21.9 kJ/mol
∆Hrxn =∑
n∆Hf products
−
∑
n∆Hf reactants
= (2 mol)(−21.9 kJ/mol)
− (2 mol)(−16.9 kJ/mol)
− (1 mol)(0 kJ/mol)
= −10.0 kJ/mol rxn
011 10.0 points
Consider the following substances:
HCl(g) F2(g) HCl(aq) Na(s)
Which response includes ALL of the sub-stances listed that have ∆H0
f = 0?
1. HCl(g), Na(s) and F2(g)
2. HCl(g), Na(s), HCl(aq) and F2(g)
3. Na(s)
4. HCl(g) and Na(s)
5. Na(s) and F2(g) correct
Explanation:
∆H0f = 0 for elements in their standard
states. This would be true for both Na(s) andF2(g).
012 10.0 points
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 5
If the products of a reaction have a higherenergy than the reactants, then the reaction
1. is not spontaneous.
2. must be spontaneous.
3. is endothermic. correct
4. is exothermic.
Explanation:
Energy must flow into the system for theproducts to have a higher energy than the re-actants. ∆H will be positive and the reactionwill be endothermic.
013 10.0 points
Calculate the average S F bond energy inSF6, using the following ∆Hf values:
SF6(g) = −1209 kJ/molS(g) = 279 kJ/molF(g) = 79 kJ/mol
1. 981 kJ/mol bonds
2. 1209 kJ/ mol bonds
3. 1962 kJ/mol bonds
4. 289.0 kJ/mol bonds
5. 582 kJ/mol bonds
6. 196 kJ/mol bonds
7. 1565 kJ/mol bonds
8. 416 kJ/mol bonds
9. 202 kJ/mol bonds
10. 327 kJ/mol bonds correct
Explanation:
The S F bond energy (in SF6) is definedas the ∆H for the reaction in which ONEMOLE of S F bonds in SF6 are broken togive gaseous atoms. Each molecule of SF6
has SIX S F bonds, so each mole of SF6 has
SIX moles of S F bonds. Thus we need to
break the S F bonds in only1
6mol of SF6.
So we need the ∆H for the reaction
1
6SF6(g) → F(g) +
1
6S(g)
Use Hess’s Law and the given values of ∆Hf
to calculate ∆H for this reaction.
014 10.0 points
Using the bond energy data provided, cal-culate ∆H for the following reaction.
H2(g) + Cl2(g) −→ 2HCl(g)
Bond Energy
Bond
(
kJ
mol
)
H H 436Cl Cl 242H Cl 432
1. 186 kJ ·mol−1
2. −246 kJ ·mol−1
3. 246 kJ ·mol−1
4. −186 kJ ·mol−1 correct
Explanation:
∆H0rxn =
∑
BErct −
∑
BEprod
= (436 kJ/mol + 242 kJ/mol)
− 2 (432 kJ/mol)
= −186 kJ/mol
The negative sign means the reaction isexothermic by 186 kJ/mol.
015 10.0 points
The standard molar enthalpy of formationof NH3(g) is −46.11 kJ/mol. What is thestandard molar internal energy of formationof NH3(g)?
1. −46.11 kJ/mol
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 6
2. −51.07 kJ/mol
3. −41.15 kJ/mol
4. −43.63 kJ/mol correct
5. −48.59 kJ/mol
6. −38.67 kJ/mol
7. −39.91 kJ/mol
Explanation:
∆H = −46.11 kJ/molT = 298.15 K (standard temperature)
1
2N2(g) +
3
2H2(g) → NH3(g)
ni =
(
1
2+
3
2
)
mol = 2 mol
nf = 1 mol∆n = (1− 2) mol = −1 mol
∆E = q + w, q = ∆H and
w = −P ∆V = −∆nRT
= −(−1 mol)(8.31451 J/mol ·K)
×(298.15 K)
(
kJ
1000 J
)
= 2.47897 kJ
Work is per 1 mole of NH3 formed, so
∆E = ∆H + w
= −46.11 kJ/mol + 2.47897 kJ/mol
= −43.631 kJ/mol
016 10.0 points
The internal energy change is 7 kJ when anideal gas expands from 3.50 to 29.2 liters at aconstant external pressure of 3.47 atm. Whatis the heat absorbed by the gas from its sur-roundings?
Correct answer: 16.0361 kJ.
Explanation:
∆E = 7 kJ P = 3.47 atmVi = 3.5 L Vf = 29.2 L
∆V = 29.2 L− 3.5 L = 25.7 L
∆E = q + w = ∆H − P ∆V
∆H = ∆E + P ∆V
= (7 kJ) +
[
3.47 atm · (25.7 L)
×
101.33 J
1 atm · L·
kJ
1000 J
]
= 16.0361 kJ
017 10.0 points
What is the total motional contribution to themolar internal energy of gaseous BF3?
1. 1.5RT
2. 2.5RT
3. 3RT correct
4. 3.5RT
5. RT
Explanation:
The contribution of each mode of motion to
the total molar internal energy is1
2RT . BF3
is a nonlinear molecule so it has three modesof translational motion and three modes ofrotational motion (assuming no contributionfrom vibration). Therefore,
Um = 6
(
1
2RT
)
= 3 RT
018 10.0 points
For a given transfer of energy, a greater changein disorder occurs when the temperature ishigh.
1. False correct
2. True
Explanation:
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 7
From ∆S =q
Tsince T is in the denomina-
tor, ∆S will be larger (more positive) when-ever T is smaller.
019 10.0 points
Entropy is a state function.
1. False
2. True correct
Explanation:
State functions are denoted by capital-ized letters. They are P (ressure), V (olume),T (emperature), S (entropy), G(ibb’s Free en-ergy), H (enthalpy). The change in the valueof a state function is independent of the pathtaken.
020 10.0 points
Place the following in order of increasing en-tropy.
1. solid, liquid, and gas correct
2. gas, liqiud, and solid
3. liqiud, solid, and gas
4. gas, solid, and liqiud
5. solid, gas, and liqiud
Explanation:
Entropy (S) is high for systems with highdegrees of freedom, disorder or randomnessand low for systems with low degrees of free-dom, disorder or randomness.
S(g) > S(ℓ) > S(s) .
021 10.0 points
Which substance has the higher molar en-tropy?
1. KCl(aq) at 298 K and 1.00 atm correct
2. Unable to determine
3. KCl(s) at 298 K and 1.00 atm
4. They are the same
Explanation:
KCl(aq) has ions distributed more ran-domly in solution than ions localized in acrystal lattice, hence a higher molar entropy.
022 10.0 points
Calculate the standard entropy of vaporiza-tion of ethanol at its boiling point 352 K. Thestandard molar enthalpy of vaporization ofethanol at its boiling point is 40.5 kJ ·mol−1.
1. +115 J·K−1·mol−1 correct
2. − 40.5 kJ·K−1·mol−1
3. +40.5 kJ·K−1·mol−1
4. +513 J·K−1·mol−1
5. − 115 J·K−1·mol−1
Explanation:
∆Hvap = 40500 J ·mol−1 TBP = 352 K
∆Scond =q
T=
∆Hcon
TBP=
∆Hvap
TBP
=40500 J ·mol−1
352 K= +115.057 J ·mol−1
·K−1
023 10.0 points
What is the entropy change for freezing4.64 g of C2H5OH at 158.7 K? ∆H =−4600 J/mol.
Correct answer: −2.91941 J/K.
Explanation:
∆H = −4600 J/mol T = 158.7 Km = 4.64 g
MW = 2 (12.0107 g/mol)
+ 6 (1.00794 g/mol)
+ 8 (15.9994 g/mol)
= 46.0684 g/mol .
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 8
The molar change in entropy is
∆S =q
T=
∆H
T
=−4600 J/mol
158.7 K
= −28.9855J
mol ·K,
and the change in entropy is(
−28.9855J
mol ·K
)(
4.64 g
46.0684 g/mol
)
= −2.91941 J/K .
024 10.0 points
Suppose that 140 g of ethanol at 25◦C is mixedwith 235 g of ethanol at 85 ◦C at constantatmospheric pressure in a thermally insulatedvessel. What is the ∆Ssys for the process?The specific heat capacity for ethanol is 2.42J/g K.
Correct answer: 3.50979 J/K.
Explanation:
mc = 140 g mh = 235 gTc,i = 25◦C = 298.15 KTh,i = 85◦C = 358.15 K
CP = 2.42J
g ·K
We use the specific heat here (J/g·K) althoughthe problem can also be worked with the mo-lar heat capacity also. The change in entropyfor each portion of ethanol is calculated viathe equation:
∆S = mCP,s ln
(
T2T1
)
Note that the “hot” side (85◦C) will cooldown to the final temperature and the “cold”side (25◦C) will warm up to the final tempera-ture. The amount of heat that flows from oneto the other is
qc = −qh
mc Cp(ethanol) (Tf − Tc,i)
= −mh Cp(ethanol) (Tf − Th,i)
Tf =mc Tc,i +mh Th,i
mc +mh
=(140 g) (298.15 K) + (235 g) (358.15 K)
140 g + 235 g= 335.75 K
The change in entropy for the “cold”ethanol initially at 25◦C is
∆Sc = (140 g)
(
2.42J
g ·K
)
× ln
(
335.75 K
298.15 K
)
= 40.2393 J/K
and for the “hot” ethanol initially at 85◦C,
∆Sh = (235 g)
(
2.42J
g ·K
)
× ln
(
335.75 K
358.15 K
)
= −36.7295 J/K
The total change in entropy is that of theentire system (there is no change in entropyof the surroundings as the vessel isolates thesystem from the surroundings):
∆Stot = 40.2393 J/K+ (−36.7295 J/K)
= 3.50979 J/K
025 10.0 points
The temperature of 2.00 mol Ne(g) is in-creased from 25◦C to 200◦C at constant pres-sure. Calculate the change in the entropy ofneon. Assume ideal behavior.
1. − 19.2 J·K−1
2. +7.68 J·K−1
3. +9.60 J·K−1
4. − 7.68 J·K−1
5. +19.2 J·K−1 correct
kumar (kk24268) – HW02 - Thermo 2 – mccord – (51580) 9
Explanation:
T1 = 25◦C+ 273.15 = 298.15 KT2 = 200◦C+ 273.15 = 473.15 K
For a monoatomic ideal gas, Cp,m = 2.5 R
∆S = nCp,m ln
(
T2T1
)
= (2.00 mol) 2.5 (8.314 J ·mol−1·K−1)
× ln
(
473.15 K
298.15 K
)
= +19.1977 J ·K−1
We expect a positive answer since tempera-ture increased.