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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