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CHAPTER 17 REGULAR II HOMEWORK:


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Question 1

Calculate ΔGº (in kJ) at 298 K for the following reaction:

2H2O(g) +2Cl2(g) → 4HCl(g) + O2(g) 

Use the data in Appendix II (Tro); Appendix 4 (Zumdahl)

Correct

 

Explanation

ΔG = [4(-95.3) + 0] - [2(0)+2(-228.6)]

=(-381.2)-(-457.2)

=76

 

Question 2

Is the above reaction under standard conditions spontaneous in the forward direction?

Select one:
 Correct

 

    

 

Question 3

Calculate ΔHº (in kJ) for the following reaction at 298 K:

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g) 

Use the data in Appendix II (Tro); Appendix 4 (Zumdahl)

Correct

 

Explanation

ΔH = [(2(-238.6) + 3(0)] - [2(-393.5) + 4(-285.8)]

=-477.2-(-787 + (-1143.2))

=1453

 

Question 4

Calculate ΔSº (in J/K) at 298 K for :

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g) 

Use the data in Appendix II (Tro); Appendix 4 (Zumdahl)

Correct

 

Explanation

ΔS = [2(126.8)+3(205.2)] -[2(213.8) + 4(70)]

(253.6 + 615.6)- (427.6 +280)

869.2-707.6

161.6

 

Question 5

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for :

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)

Correct

 

Explanation

ΔG=ΔH-TΔS

=1453*10^3-298(161.6)

=140,4843 J = 1404.8 kJ

 

Question 6

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for :

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)

Correct

 

Explanation

ΔG=1453e3-1900(161.6)

=1145960 J

=1145.96 kJ

 

Question 7

Is the above reaction spontaneous in the forward direction at either of the temperatures? (Pick 2 answers.)

Select one or more:
 Correct
 Correct

 

    

 

Question 8

What is ΔGrxno (in kJ) at 1447 K for the following reaction?

2POCl3(g) → 2PCl3(g) + O2(g) 

POCl3(g): ΔHfo = -592.7 kJ/mol and Sº = 324.6 J/K mol)
PCl3(g): ΔHfo = -287.0 kJ/mol and Sº = 311.7 J/K mol)
O2(g): ΔHfo = ? kJ/mol and Sº = 205.0 J/K mol)

Hint given in feedback.

Correct

 

Explanation

Find ΔHº and ΔSº. ΔG =ΔHº -TΔSº.

ΔHf = [0 + 2(-287)]-2(-592.7)

=-574 + 1185.4

=611.4 kJ


ΔS = [(205) + 2(311.7)] - 2(324.6)

=828.4-649.2

=179.2 J


ΔG = ΔH - TΔS

=611.4*10^3-1447(179.2)

=352097.6 J = 352.1 kJ

 

Question 9

At what temperature (in K) does the above reaction become spontaneous?

Hint given in feedback.

Correct

 

Explanation

Use ΔG =ΔHº -TΔSº. The temperature at which ΔG = 0 is the key.


0=611.4e3- T(179.2)

T=3411.83

 

Question 10

The above reaction is spontaneous in the forward direction
(Pick 2)

Select one or more:
 Correct
 Correct

 

    

 

Question 11

What is ΔGo (in kJ) at 491 K for the following reaction?

PbO(g) + CO2(g) → PbCO3(s) 

PbO: ΔHfo = -219.0 kJ/mol and So = 66.5 J/K mol)
PbCO3(s): ΔHfo = -699.1 kJ/mol and So = 131.0 J/K mol) 

CO2: ΔHfo = -393.5 kJ/mol and So = 213.6 J/K mol) 

Correct

 

Explanation

ΔH= (-699.1)-[-219.0+(-393.5)]

=-699.1+612.5

=-86.6


ΔS= (131) - [66.5 + 213.6]

=-149.1


ΔG= -86.6e3-491(-149.1)

=-13391.9

=-13.4 kJ

 

Question 12

At what temperature (in K) does the above reaction become spontaneous?

Correct

 

Explanation

0=-86.63e3-T(-149.1)

T=580.82

 

Question 13

The above reaction is spontaneous in the forward direction
(Pick 2)

Select one or more:
 Correct
 Correct

 

    

 

Question 14

What is ΔGo (in kJ) at 21 ºC for the phase change of nitromethane from the liquid to the gaseous state? 

CH3NO2(l) → CH3NO2(g) 

CH3NO2(l): ΔHo = -113.1 kJ/mol and So = 171.8 J/K mol)
CH3NO2(g): ΔHo = -74.7 kJ/mol and So = 274.4 J/K mol)

Hint given in feedback.

Correct

 

Explanation

Find ΔH and ΔS for the phase change. ΔG =ΔH -TΔS. In calculating ΔG it does not matter what kind of process it is--reaction or phase change.


ΔH= (-74.7) - (-113.1)

=38.4 kJ

ΔS= 274.4 - (171.8)

=102.6 J

ΔG=38.4e3-(21+273)(102.6)

=8235.6 J=8.23 kJ

 

Question 15

Which state is more stable for nitromethane at the temperature in the previous question?

Select one:
 Correct

 

    

 

Question 16

What is the boiling point of nitromethane (in ºC)?


Correct

 

Explanation

goes here...


0= 38.4e3-T(102.6)
T=374.27 K - 273 = 101.27 C
 

Question 17

At 298 K, ΔGo = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ↔ 2N2O4(g)

Calculate ΔG (in kJ) at 298 K when PN2O = 3.10 atm, PO2 = 0.0072 atm, and PN2O4= 0.276 atm.

Help given in feedback.

Correct

 

Explanation

Hint, first find Q.


ΔG = ΔG* + RTlnQ

Q= (0.276)^2/ (3.10)^2(0.0072)^3= 21237.201

ΔG = -6.36e3 + 8.314(298)ln(21237.201)

=18325.31 kJ = 18.33 kJ

 

Question 18

Under the conditions in the previous question, in which direction is the reaction spontaneous?

Select one:
 Correct

 

    

 

Question 19

At 298 K, ΔGo = + 8.68 kJ for the reaction:

ZnF2(s) ↔ Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ΔG means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.)

Select one:
 Correct

 

    

 

Question 20

Calculate ΔG (in kJ) at 298 K for some solid ZnF2, 0.029 M Zn2+ and 0.057 M F-(aq).
Hint given in feedback.

Correct

 

Explanation

Hint, first find Q.

ΔG = ΔG* + RTlnQ

Q= (0.057)^2(0.029)=0.000094221

=8.68e3 + 8.314(298)ln(0.000094221)

=-14286.76 J = -14.29 kJ

 

Question 21

Under the conditions of the previous question what is the spontaneous direction?

Select one:
 Correct

 

    

 

Question 22

Endothermic reaction; decrease in entropy:
Calculate the equilibrium constant at 41 K for a reaction with ΔHo = 10 kJ and ΔSo = -100 J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)

Hint given in feedback.

Correct

 

Explanation

Hint, first calculate ΔGo at 41 K.

ΔG=ΔH - TΔS

=10e3-41(-100)

=14100


ΔG=-RTlnK

14100= -8.314(41)ln(k)

k=1.085e-18

 

Question 23

Calculate the equilibrium constant at 145 K for the thermodynamic data in the previous question. 

Notice that Keq is larger at the larger temperature for an endothermic reaction.

Correct

 

Explanation

ΔG=ΔH - TΔS

=10e3-145(-100)

=24500


ΔG=-RTlnK

24500= -(8.314)(145)lnK

k=1.4922e-9

 

Question 24

Endothermic reaction; increase in entropy
Calculate the equilibrium constant at 35 K for a reaction with ΔHo = 10 kJ and ΔSo = 100 J/K. 

Correct

 

Explanation

ΔG=10e3-35(100)

=6500

ΔG=-RTlnK

6500= -(8.315)(34)lnK

k=-1.99035e-10

 

Question 25

Calculate the equilibrium constant at 132 K for the thermodynamic data in the previous question. 

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.

Correct

 

Explanation

ΔG=10e3-132(100)

=-3200

ΔG=-RTlnK

-3200=-8.314(132)lnK

k=18.46

 

Question 26

Warm-up question Using data from the Appendix, calculate ΔGo (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ↔ N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions)

Correct

 

Explanation

ΔG  = [0= (-237.1)] - [103.7+0]

=-340.8 kJ     appendix will vary dependent on book and edition due to quick technological advances

 

Question 27

Calculate ΔG (in kJ) at 65 oC for the reaction:

N2O(0.0046 atm) + H2(0.28 atm) ↔ N2(352.3 atm) +H2O(l)

Correct

 

Explanation

Q=(352.3)/(0.0046)(0.28)=273524.8447

ΔG = ΔG + RTln Q

=-337.27e3+(8.314)(65+273)ln 273524.8447

=-302089.54 J = -302 kJ

 

Question 28

Assume that the ΔHo and ΔSo of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 61 oC (in atm).

CH3OH (l) ↔ CH3OH(g) . . . ΔHo = 38.0 kJ and ΔSo = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback.

Correct

 

Explanation

Hint,what is the relationship between Keq and the vapor pressure.


ΔG=ΔH-TΔS

=38e3-(61+273)(112.9)

=291.4


ΔG= -RTlnK

291.4 = -(8.314)(61+273)lnK

K=0.90038

 

Question 29

Calculate the vapor pressure of Hg at 72 oC (in atm).

Hg(l) ↔ Hg(g) . . . ΔHo = 61.32 kJ and ΔSo = 98.83 J/K

Correct

 

Explanation

ΔG=ΔH-TΔS

=61.32e3-(72+273)(98.83)

=27223.65


ΔG= -RTlnK

27223.65 = -(8.314)(72+273)lnK

K=7.6e-5

 

Question 30

Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C—C—C bond angle? The actual bond angle is 60º. (c) Consequently, do you expect this bond to be under a lot of stress? (d) What does this imply about the expected C—C bond strength?

Select one or more:
 Correct
 Correct
 Correct
 Correct

 

    


 

Question 31

Calculate the average C—C bond strength in cyclopropane (in kJ/mol). Its combustion and the experimental enthalpy of reaction are: 

C3H6(g) + 4.5O2(g) → 3CO2(g) + 3H2O(g) ΔHºrxn = -1,957.7 kJ/mol

Since all reactants and products are in the gaseous state, bond energies may be used to estimate the enthalpy of reaction. Do not use the standard C—C bond strength, leave the C—C bond strength as an unknown and solve for its value. For review see chapter 9 p. 392 (Tro) or chapter 13 p. 608 (Zumdahl).

Correct

 

Explanation

ΔH= Summation of bonds broken - Summation of bonds formed

-1957.7kJ=[ 6(C-H) + 3(C-C) + 4.5 (O=O) ] + 6(C=O) + 6(O-H)

1957.7= [6(414)+3(C-C)+4.5(498)]+[-6(799)-6(463)]

-1957.7=[4725+3(C-C)]-7572

5614.3=[4725 + 3(C-C)]

C-C=296.43


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