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#3. Chapter 5 HW. 5) Strategy=. #9. (a) W= Force applied over a distance (b) W=FD. 15. (a) q is negative because the system loses heat and w is negative because the system does work. Δ E = q + w = -113kJ + (-39kJ) = -152kJ. The process is exothermic - PowerPoint PPT Presentation
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Chapter 5 HW
• 5) • Strategy=
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#9
• (a) W= Force applied over a distance
• (b) W=FD
15
• (a) q is negative because the system loses heat and w is negative because the system does work.
• ΔE = q + w = -113kJ + (-39kJ) = -152kJ. The process is exothermic
• (b) ΔE = q + w = +1.62kJ - .847kJ=.75kJ. The process is endothermic.
• (c) q is positive because system gains heat and w is negative because the system does work.
• ΔE = q + w = +77.5 kJ - 63.5kJ=14kJ. The process is endothermic
17
• (a) ΔE = q + w since q is neg and w is positive, ΔE depends on the relative magnitudes of each.
• (b) since q is positive and w is positive, ΔE is positive
• (c) since q is positive and w is negative, ΔE depends on the relative magnitudes of each.
19
• (a) A state function is a property of a system that depends only on the physical state of the system, not on the route used by the system to get to the current state
• (b) Internal energy and Enthalpy are state functions. Work is not a state function
• (c) Temp is a state function, regardless of how hot or cold the samples has been, temp only depends on its present condition.
See back of book for 23
25
• Since ΔH is negative , the reactants 2Cl(g) have a higher enthalpy.
Question 27
• A.) Exothermic (H is negative)
2.4g Mg 1 mol Mg
24.305 g Mg -1204 kJ
2 mol Mg 59kJ
96.0 kJ 2 mol MgO
1204 kJ
40.30 g MgO
1 mol MgO6.43gMgO
C.)
B.)
5.27 D
• 2 MgO(s) 2Mg(s) + O2(g)
• Reversed reaction so sign on H is reversed
7.50 g 1 mol MgO
40.30 g MgO
1204 kJ
2 mol MgO122kJ absorbed
Question 29(a)
.200 mol AgCl 65.5kJ
1mol AgCl 13.1kJ
Question 29 (b)
2.50g AgCl 1 mol AgCl
143.3g AgCl
65.5 kJ
1 mol Agcl 1.14 kJ
Question 29 (c)
22.9J+ =0.0229kJ+ AgCl mol 1
kJ 65.5
mmol 1
mol 101AgCl mmol 0.350
-3
31
• At constant P, ΔE= ΔH-P ΔV. In order to calc ΔE, mor einfo about the condition must be known. For an ideal gas at constant P and T, P ΔV= ΔnRT must be known to calc ΔE from ΔH.
35
• (a) CO2 (g) + 2H2O (l) → CH3OH (l) + 3/2 O2 (g) ΔH=726.5kJ
• (b) 2CH3OH (l) + 3 O2 (g) → 2CO2 (g) + 4H2O (l) ΔH=2(726.5kJ) = -1453kJ
• (c) The exothermic reaction is more likely to be favored thermodynamically
• (d) Vaporization is endothermic. If the products were gas, less heat energy would be available to release to the surroundings since ti takes energy to convert liquid to gas. It is about 88kJ per 2 moles water. This is not enough to make the overall reaction endothermic.
Question 39
JCCCg
Jkg 41047.3)0.255.88(
385.
kg 1
1000gCu 42.1
41
44.4kJ/mmol 1
NaOH 40.00g
NaOH 6.50g
7.219kJ
7.22kJ1000j
1kJC16.2º
Cg•º
4.184Jnsol' 106.5g
Question 45 (a)
C total = 2.500 g glucose 15.57 kJ
1g glucose
1
2.70C 14.42 = 14.4kJ/ C
Question 45 (b)
C H2 0 = 2.700 kg H2 0 4.184 kj
1 kg C = 11.30 kJ/ C
C empty calorimeter =14.4 kJ
1 C -
11.30 kJ
1C = 3.12 = 3.1 kJ/ C
Question 45 (c)
q = 2.500 g glucose 15.57 kJ
1 g glucose 38.93 kJ produced
C H 2 0 = 2.700 kg H 2O 4.184 kJ
1 kg C = 8.368 kJ/ C
C total = 8.368 kJ
1C +
3.12kJ
1 C = 11.49 = 11.5 kJ/ C
38.98 kJ = 11.49 kJ
C T; T = 3.39 C
#47• If a reaction can be described as a series of
steps, ΔH for the reaction is the sum of the enthalpy changes for the each step. As long as we can describe a route where we have the ΔH for each step is known, then we can calculate the overall ΔH .
49
• A→B ΔH =+30kJ
• B →C ΔH=+60kJ
• Then the overall is 90kJ
51
kJ 0.1300-H 2O
kJ 1.2940-H 5O
kJ 1640.1H 3O
104264
10424
2464
OPOP
OPP
POP
53
kJ 10.492-H 4HF 26
kJ 1.2940-H HF4 2 2
)680(2H 242
3.52-H 2C 2
34242
22
42
242
CFFHC
FH
kJCFFC
kJHHC
Question 59
-847.6kJ= 2(0)- )(-822.16kJ- 2(0) + kJ) (-1669.8=
Al(s) 2- OFe - Fe(s) 2+ (s)OAl = f32ff32f
rxn
rxn
H
H
Question 61 (a)
196.6kK- =0-)2(-296.9kJ- )2(-395.2kJ =
(g)O -(g)SO 2- (g)SO 2 = 2f2f3f rxn
Question 61 (b)
37.1kJ=kJ) (-924.7-kJ) (-285.83 + kJ 601.8- =
(s)Mg(OH) - O(l)H + MgO(s) = 2f2ff rxn
Question 61 (c)
Hrxn=2Hºf Fe2O3 - 4Hºf FeO - Hºf O2
• 2(822.16kJ) - 4(-271.9kJ) - 2(0) =
• -556.7kJ
Question 61 (d)
rxn = f SiO2 (s) + 4 f HCl(g) - f SiCl4(l) - 2 f H2O(l)
= - 910.9kj + 4(-932.30kJ) - (-640.1kJ) - 2(-285.831kJ) = -68.3kJ
Question 63
Hrxn=3Hºf CO2 + 3Hºf H2O - Hºf
C3H6O
• 3(-393.5kJ) + 3(-285.83kJ) - Hºf
C3H6O(l) = -1790kJ
Hºf C3H6O(l)=-248kJ
Question 67 (a & b)
• C8H18 +25/2O2 8CO2 + 9H2O
• 8C + 9H2 C8H18
Question 67cHrxn=8Hºf CO2 (g)+ 9Hºf H2O(g)
- Hºf -25/2 Hºf O2(g)- Hºf C8H18(l)
• -5069kJ=8(-393.5kJ) + 9(-241.82kJ) -C8H18(l) -25/2(0) = Hºf C8H18(l) =-255kJ
Question 73
Calg 7.59kJ 4.184
Cal 1
mol 1
kJ 2812
g 180.2
mol 10.16
Question 75(a)
• C3H4(g)+4O2(g)3CO2(g) + 2H2O(g)
H = 3(-393.5kJ) + 3(-241.82kJ) -(185.4kJ) - 4(0)= -1850kJ/molC3H4
Question 75 a
kg
kj
g443
43
10 X 4.616kg 1
g 1000
065.40
HC mol 1
HC mol 1
1849.5kJ
Question 75b
• C3H4(g)+9/2O2(g) 3CO2(g) + 3H2O(g)
H = 3(-393.5kJ) + 3(-241.82kJ) -9/2(0)- (20.4kJ)=-1926kj/molC3H6
75b
kg
kj
g463
63
10 X 578.4kg 1
g 1000
080.42
HC mol 1
HC mol 1
kJ4.9261
Question 75c
• C3H8(g)+5O2(g3CO2(g) + 4H2O(g)
H = 3(-393.5kJ) + 4(-241.82kJ) -(-103.8kJ) - 5(0) =
• -2044kJ/molC3H8
Question 75c
kg
kj
g483
83
10 X 635.4kg 1
g 1000
096.44
HC mol 1
HC mol 1
kJ0.2044
Question 100(a)
• Mol Cu=M*L=1.00M*0.0500L=0.0500mol
• G=mol *M=0.0500*63.546=3.1773=3.18g Cu
• Given there is one mole of Cu per mole of compound (CuSO4)
100 ( b & c)• CuSO4 + KOH → K2SO4 + Cu(OH) 2
• Cu 2+ SO4 2- +2K 1+ OH 1-
Cu(OH)2 + K 1+ SO4 2-
• Cu2++2OH- Cu(OH)2
100(d)
• The temp rises so exothermic
• Q=(100g)(-6.2ºC)(4.184J/gºC) =-2.6kJ
• The reaction is carried out involves .050 mol so on molar basis:
• -2.6 kJ/.050mol = -52 kJ/mol
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