Homework5 Answer (Due Friday, 02/23/2011) - Wetalldid – · PDF file ·...

Homework5 Answer (Due Friday, 02/23/2011)

Q4.1 P4.1, P4.2, P4.3, P4.5 P4.6 P4.8, P4.18, P4.19, P4.21, P4.29, P4.31, P4.32

Q4.1) Under what conditions are H and U for a reaction involving gases and/or liquids or solids

identical?

H = U + (p V). H ≈ U for reactions involving liquids and solids, since, in a good

approximation, the volume does not change in a chemical reaction. On the other hand, for chemical

reactions involving gases, (p V) = n R T, and H ≈ U if the number of moles of reactants and

products is identical.

P4.1) Calculate H reaction

and Ureaction

at 298.15 K for the following reactions:

a. 4NH3(g) + 6NO(g) 5N2(g) + 6H2O(g)

b. 2NO(g) + O2(g) 2NO2(g)

c. TiCl4(l) + 2H2O(l) TiO2(s) + 4HCl(g)

d. 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l). Assume complete dissociation of

NaOH, H2SO4, and Na2SO4.

e. CH4(g) + H2O(g) CO(g) + 3H2(g)

f. CH3OH(g) + CO(g) CH3COOH(l) Table 4.1 4.2 Appendix B

a) gNO,ΔH 6g,NHΔH 4gO,HΔH 6g,NΔH 5ΔH f3f2f2freaction

1-

-1-1-1-1reaction

mol kJ 1815.0-

mol kJ 1.396mol kJ 5.944mol kJ 41.826mol kJ 05ΔH

1-

-1-1-1reactionreaction

mol kJ 1817.5

K 298.15mol K J 8.3144721mol kJ 1815.0T Rn ΔHΔU

b) gNO,ΔH 2g,OΔH-g,NOΔH 2ΔH f2f2freaction

1-

-1-1-1reaction

mol kJ 116.2

mol kJ 1.392mol kJ 0mol kJ 3.232ΔH

1-


mol kJ 113.7


c) O,HΔH 2,TiClΔHgHCl,ΔH 4s,TiOΔHΔH 2f4ff2freaction

-1 -1 -1 -1reaction

-1

ΔH 944.0 kJ mol 4 92.3 kJ mol 804.2 kJ mol 2 285.8 kJ mol

62.6 kJ mol

-1 -1 -1reaction reaction

-1

ΔU ΔH n R T 60.8 kJ mol 4 8.314472 J K mol 298.15 K

52.7 kJ mol

d) The overall reaction is:

OH 2 aq H 2 aq OH 2 2

aq,OHΔH 2aq,HΔH 2O,HΔH 2ΔH ff2freaction

1-

-1-1-1reaction

mol kJ 111.6

mol kJ 230.02mol kJ 02mol kJ 85.822ΔH

1-

-1reactionreaction

mol kJ 111.6

0mol kJ 111.6T Rn ΔHΔU

e) CH4(g) + H2O(g) CO(g) + 3H2(g)

gg O,HΔH,CHΔHg,HΔH 3gCO,ΔHΔH 2f4f2ffreaction

1-

-1-1-1-1reaction

mol kJ 05.92

mol kJ .8142mol kJ 4.67mol kJ 03mol kJ 5.101ΔH

1-


mol kJ 00.92


f) CH3OH(g) + CO(g) CH3COOH(l)

gCO,ΔHgOH,CHΔHCOOH,CHΔHΔH f3f3freaction

1-

-1-1-1reaction

mol kJ 172.8

mol kJ 10.51mol kJ 0.201mol kJ 3.484ΔH

1-


mol kJ 167.8



and Ureaction

for the oxidation of benzene. Also calculate

Hreaction Ureaction

Hreaction

The chemical equation for the oxidation of benzene is:

lOH 3gCO 6gO 7lHC 22221

66

The standard enthalpy for this reaction is:

1111

if,ii mol kJ 3268 mol kJ 49.1mol kJ 285.83mol kJ 393.56H νΔH

U is calculated as:

11113 mol kJ 3264 K 298.15K mol J 8.3144721.5-mol J 103268

T Rn ΔHΔU

And:

00122.0

mol J 103268

mol J 103264 mol J 103268

ΔH

ΔUΔH13

1313

P4.3) Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of

formation of CO2(g) and H2O(l) to determine H f for benzene.

HC g O g CO 6 OH 3 662215

22 ,HCΔH- 66combustion

g CO 6 g O 6 C 6 22 s g,COΔH 6 2f

OH 3 g O g H 3 2223

2 O,HΔH 3 2f

HC s C 6 g H 3 662 O,HΔH 3g,COΔH 6,HCΔH- 2f2f66combustion

-1-1-1-166f mol kJ 9.64 mol kJ 285.8-3mol kJ 393.5-6mol kJ 2683,HCΔH

P4.5) Several reactions and their standard reaction enthalpies at 25°C are given here:

H reaction kJ mol1

CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g) –127.9

Ca(s) + 1/2 O2(g) CaO(s) –635.1

CaO(s) + H2O(l) Ca(OH)2(s) –65.2

The standard enthalpies of combustion of graphite and C2H2(g) are –393.51 and –1299.58 kJ

mol–1, respectively. Calculate the standard enthalpy of formation of CaC2(s) at 25°C.

OH 2 s CaC g HC s OHCa 22222 -1mol kJ 127.9

s OHCa OH s CaO 22 -1mol kJ 65.2-

g Os HC OH g CO 3 225

2222 -1mol kJ 1299.58

g CO 2 g O 2 s C 2 22 -1mol 393.51kJ-2

s CaO g O s C 221 -1mol kJ 635.1-

s CaC s Ca s C 2 2 -1f mol kJ 8.59ΔH

P4.6) From the following data at 25°C, calculate the standard enthalpy of formation of FeO(s)

and of Fe2O3(s):

H reaction kJ mol1

Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) 492.6

FeO(s) + C(graphite) Fe(s) + CO(g) 155.8

C(graphite) + O2(g) CO2(g) –393.51

CO(g) + 1/2 O2(g) CO2(g) –282.98

graphite Cs FeO g CO s Fe - -1mol kJ 155.8

g Og CO g CO 221

2 -1mol kJ 282.98

g CO g O graphite C 22 -1mol 393.51kJ-

s FeO g O s Fe 221 -1

f mol kJ 3.266ΔH

graphite C 3s OFe g CO 3 s Fe 2 32 -1mol kJ 492.6-

g Og CO 3 g CO 3 223

2 -1mol kJ 282.983

g CO 3 g O 3 graphite C 3 22 -1mol 393.51kJ-3

s OFe g O s 2Fe 32223 -1

f mol kJ 2.824ΔH


at 650 K for the reaction 4NH3(g) +6NO(g) 5N2(g) + 6H2O(g)

using the temperature dependence of the heat capacities from the data tables.

K

Td

K

TCK 15.298HK 506H

650

K 298.15

pff

1-1-

38888

25555

mp,3mp,2mp,2mp,p

mol K J

)107744.2(6)107706.2(4)103115.1(6)100176.1(5

103326.56102446.44108228.26103968.25

0.02593)(60.0110340.00795)(60.01187)(5

33.58629.29433.80630.815

gNO,C 6g,NHC 4gO,HC 6g,NC 5ΔC

T

T

T

11

144

733

4

22

113724650

15.298

650

15.298

895.330.6585.16736.0444.13

4

15.298650104772.1

3

15.298650100053.2

2

15.29865000441.0)15.298650(21.38

104772.1100053.200441.021.38

kJmolkJmol

Jmol

dTmolJKTTTdTTCKK

p

1-1-1-1-

f

f3f2f2ff

mol kJ 1815mol kJ 3.916mol kJ 5.944mol kJ 8.2416K 15.298H

gNO,H 6g,NHH 4gO,HH 6g,NH 5K 15.298H

-111f mol kJ 1.1811)895.3()1815(K 506H kJmolkJmol

P4.18) Given the data in Table 4.1 (Appendix B, Data Tables) and the following information,

calculate the single bond enthalpies and energies for Si–F, Si–Cl, C–F, N–F, O–F, H–F:

Substance SiF4(g) SiCl4(g) CF4(g) NF3(g) OF2(g) HF(g)

H f

kJ mol1 –1614.9 –657.0 –925 –125 –22 –271

a) SiF4

The average Si-F single bond enthalpy is calculated as follows:

g F 2s Si g SiF 24 -1reaction mol kJ 9.1614ΔH

g F 4 g F 2 2 -1reaction mol kJ 9.474ΔH

g Si s Si -1reaction mol kJ 0.450ΔH

g F 4g Si g SiF4 -1reaction mol kJ 2382ΔH

The average Si-F bond enthalpy is then:

1--1

bondavrg mol kJ 596

4

mol kJ 2382H

And the average Si-F bond free energy:


-1

ΔU ΔH n R T 2382 kJ mol 4 8.314472 J K mol 298.15 K

2372 kJ mol

1--1

bondavrg mol kJ 593

4

mol kJ 2372U

b) SiCl4

g Cl 2s Si g SiCl 24 -1reaction mol kJ 0.657ΔH

g Cl 4 g Cl 2 2 -1reaction mol kJ 21.314ΔH

g Si s Si -1reaction mol kJ 0.450ΔH

g Cl 4g Si g SiCl4 -1reaction mol kJ 1592ΔH

The average Si-Cl bond enthalpy is then:

1--1

bondavrg mol kJ 0.398

4

mol kJ 1592H

And the average Si-Cl bond free energy:


-1


1582.0 kJ mol

1--1


4

mol kJ 0.1582U

c) CF4

g F 2s C g CF 24 -1reaction mol kJ 0.925ΔH

g F 4 g F 2 2 -1reaction mol kJ 9.474ΔH

g C sC -1reaction mol kJ 7.716ΔH

g F 4g C g CF4 -1reaction mol kJ 1959ΔH

The average C-F bond enthalpy is then:

1--1


4

mol kJ 1959H

And the average C-F bond free energy:

1-


mol kJ 0.1949


1--1


4

mol kJ 0.1949U

d) NF3

g F g N g NF 223

221

3 -1reactionΔH 125.0 kJ mol

g F 3 g F 223 -1

reaction mol kJ 9.473ΔH

g N gN 221 -1


g F 3g N g NF3 -1reaction mol kJ 9.835ΔH

The average N-F bond enthalpy is then:

1--1


3

mol kJ 9.538H

And the average N-F bond free energy:

1-


mol kJ 5.828


1--1


3

mol kJ 28.58U

e) OF2

g F g O g OF 2221

2 -1reaction mol kJ 0.22ΔH

g F 2 g F2 -1reaction mol kJ 9.472ΔH

g O gO 221 -1


g F 2 g O g OF2 -1reaction mol kJ 0.430ΔH

The average O-F bond enthalpy is then:

1--1


2

mol kJ 0.430H

And the average O-F bond free energy:

1-


mol kJ 0.425


1--1


2

mol kJ 0.425U

f) HF

g F21 g H g HF 222

1 -1reaction mol kJ 0.271ΔH

g H gH 221 -1


g F gF 221 -1


g F g H g HF -1reaction mol kJ 4.568ΔH

The average H-F bond enthalpy is then:

-1bondavrg mol kJ 4.568H

And the average H-F bond free energy:

1-


mol kJ 9.565


P4.19) Given the data in Table 4.3 and the data tables, calculate the bond enthalpy and energy of

the following:

a. The CH bond in CH4

b. The CC single bond in C2H6

c. The CC double bond in C2H4

Use your result from part (a) to solve parts (b) and (c).

a) g H 4g C g CH4

1-1-1-1-

4fffreaction

mol kJ 3.1663mol kJ 4.67mol kJ 16.77mol kJ 18.024

g,CHHgC,HgH,H 4H

The average C-H bond enthalpy is then:

1--1


4

mol kJ 30.1663H

And the average C-H bond free energy:

1-


mol kJ 4.1653


1--1

ave mol kJ4.413 4

l1653.4kJmoΔU bond

b) g H 6g C 2 g HC 62

1-1-1-1-

62fffreaction

mol kJ 4.2825mol kJ 4.08mol kJ 16.772mol kJ 18.026

g,HCHgC,H 2gH,H 6H

enthalpy bond C-Centhalpy bond H-C6H reaction

1

-1-1reaction

6.330

)l(415.8kJmo6-)ol(2825.4kJmenthalpy bond H-C6Henthalpy bond C-C

kJmol

1-


mol kJ 0.2808


energy bond C-Cenergy bond H-C6Ureaction

1

-1-1reaction

6.327

)413.4kJmol(6-)ol(2808.0kJmenergy bond H-C6Uenergy bond C-C

kJmol

c) g H 4g C 2 g HC 42

reaction f f f 2 4

-1 -1 -1 -1

H 4 H H,g 2 H C,g H C H ,g

4 218.0 kJ mol 2 716.7 kJ mol 52.4 kJ mol 2253 kJ mol

enthalpy bond CCenthalpy bond H-C4Hreaction

1

-1-1reaction

8.589

)l(415.8kJmo4-)(2253kJmolenthalpy bond H-C4Henthalpy bond CC

kJmol


-1


2240 kJ mol

energy bond CCenergy bond H-C4Ubond

1-1-1 4.586)l(413.4kJmo4-)(2240kJmolenergy bond CC kJmol

P4.21) Benzoic acid, 1.35 g, is reacted with oxygen in a constant volume calorimeter to form

H2O(l) and CO2(g). The mass of the water in the inner bath is 1.240 103 g. The temperature of

the calorimeter and its contents rise 3.45 K as a result of this reaction. Calculate the calorimeter

constant.

The reaction is, with n = ½:

OH 3g CO 7 g O2

15s COOHHC 22256

1-1-1-2

11-

combustioncombustion

mol kJ 76.3225K 298.15mol K J 8.314472mol kJ 3227

T Rn ΔHΔU

1-3

1-1-1-

31-

1-

OHOH

OHR

s

s

rcalorimete

C J 1015.5

C.453

C.453mol K J 5.2917mol g 18.02

g 101.240mol kJ 3225.76

mol g 122.13

g 1.350

ΔT

ΔTCM

mΔU

M

m

C2

2

2

P4.29) A good yield of photosynthesis for agricultural crops in bright sunlight is 2000. kg of

carbohydrate (e.g., sucrose) per km2 per hour. The net reaction for sucrose formation in

photosynthesis is:

12CO2 g 11H2O l light C12 H22O11 s 12O2 g

a. Use standard enthalpies of formation to calculate H reaction for the production of 1 mol of

sucrose at 298 K by the preceding reaction.

b. Calculate the rate at which energy is stored in carbohydrates (e.g., sucrose) per km2 as a

result of photosynthesis. (Note: 1 watt = 1 joule/s.)

c. Bright sunlight corresponds to radiation flux at the surface of the earth of about 1 kW/m2.

What percentage of this energy can be stored in the form of carbohydrates as a result of

photosynthesis?

a) The enthalpy of reaction at 298 K is:

1-

1-1-1-

2f2f112212freaction

mol kJ7.5639

mol kJ 393.5-12mol kJ 8.85211mol kJ 1.2226

g,COΔH 12O,HΔH 11s,OHCΔHK 298ΔH

b) The stored energy is obtained as:

2-62-1-6

113

-13-1-2

stored kmW 10155.9 km s J10155.9 h s 6003mol kg10342.23

mol J107.5639h km kg 0002E

c) The percentage of energy that can be stored is:

% 0.9137

m W 101

m W 155.9

m W 101

kmW 10 155.9%

2-3

-2

2-3

-26

Energystored,

P4.31) The figure below shows a DSC scan of a solution of a T4 lysozyme mutant. From the

DSC data, determine Tm, the excess heat capacity CP and the intrinsic and transition excess heat

capacities at T = 308 K. In your calculations, use the extrapolated curves, shown as dashed lines

in the DSC scan.

Tm = 304 K

Cp = 1.25 J K-1 g-1

Cptrs = 0.83 J K-1 g-1

Cpint = 0.42 J K-1 g-1

Tm = 304 K

Cp = 1.25 J K-1 g-1

Cptrs = 0.83 J K-1 g-1

Cpint = 0.42 J K-1 g-1

P4.32) Using the protein DSC data from Problem P 4.31, calculate the enthalpy change between

the T=288 K and T=318 K. Give your answer in units of kilojoules per mole. Assume the

molecular weight of the protein is 14000. grams. Hint: You can perform the integration of the

heat capacity by estimating the area under the DSC curve and above the dotted baseline in

Problem P 4.31. This can be done by dividing the area up into small rectangles and summing the

areas of the rectangles. Comment on the accuracy of this method.

Dividing the area up into small rectangles and summing the areas of the rectangles as

-1-1p g K JΔC KΔT we obtain:

-1-1-1 mol kJ 2.214 mol g 40001g J 15.3K 288 -K 318ΔH

The method is not very accurate since it depends on the size of the rectangles chosen by eye,

which is not small enough to give an accurate answer.

Answers (Q1-Q3 first, then textbook problems): Q1. Enthalpy is a state function and therefore an infinitesimal change in H can be written: δH = (δH/δT)p,n δT + (δH/δP)T,n δP + (δH/δn)T,p δn (δH/δP)T,n = -CpμJ-T and (δH/δT)p,n = Cp

The term (δH/δP)T,n is the pressure dependence of Enthalpy at constant temperature and constant number of particles. It can be found with the heat capacity at constant pressure, Cp, and the Joule-Thompson coefficient. Likewise, (δH/δT)p,n is the definition of heat capacity at constant pressure.

Q2. We calculate the reaction enthalpy at temperature T and P = 1 bar for the following reaction:

2B + C X+3Y The path for the reaction is given in the following 4 steps:

(1) 2B + C (at T) (2) 2B+C (at 298.15K) (3) X+3Y ( at 298.15K) (4) X+3Y (at T)

For the “reaction” (or the change of the state) from (1) to (2), 2 moles of B and 1 mole of C is subject to the temperature change from T to 298.15K at 1 bar. Thus, the enthalpy change in this process is given by

2)ΔH(1 dTCCdTCC mCPmBPmCPmBP )2()2( ,,,,

T

298.15K

,,,,

298.15K

T

For the change of the state from (2) to (3), the reaction takes place at the standard condition (T =298.15K and 1 bar) as 2B + C X + 3Y. The reaction enthalpy at the standard condition is given by

(C)(B)2(Y)3(X)

n(C)n(B)n(Y)n(X)

ΔHΔHΔHΔH

ΔH

3) ΔH(2

0000

0B

0Y

0X

0

CBYX

298.15K reaction,

ffff

Cffff

HHHH

HHHH

,

Now, our system is made of 1 mole of X and 3 mole of Y (X + 3Y) at T = 298.15K. Then, for a change of the state from (3) to (4),

4)ΔH(3 dTCC mYPmXP )3( ,,,,

T

298.15K

')(T'ΔCΔH

23

(C)(B)2(Y)3(X)

4)ΔH(33)ΔH(22)ΔH(1ΔH

T

298.15

P298.15K reaction,

15.298 15.298,,

15.298,

15.298,

0000

T reaction,

dT

dTCdTCdTCdTC

HHHH

T T

mPCmPB

T

mPY

T

mPX

ffff

where )'()(T'ΔC,P TCnmPjj

k

Q3. The path for the reaction is given in the following 4 steps:

(1) 2B + C (at P) (2) 2B+C (at 1 bar) (3) X+3Y ( at 1 bar) (4) X+3Y (at P)

For the the change of the state from (1) to (2), 2 moles of B and 1 mole of C is subject to pressure change from P to 1 bar at 298.15 K. Thus, the enthalpy change in this process is given by

2)ΔH(1

PCCdPCCdPCC TJmCPmBPTJmCPmBPTJmCPmBP )2(')2(-')2(- ,,,,,,,,

P

bar 1

,,,,

bar 1

P

For the change of the state from (2) to (3), the reaction takes place at the standard condition (T =298.15K and 1 bar) as 2B + C X + 3Y. The reaction enthalpy at the standard condition is given by

(C)(B)2(Y)3(X)

ΔH

3) ΔH(2

0000

298.15K reaction,0

ffff HHHH

,

Now, our system is made of 1 mole of X and 3 mole of Y (X + 3Y) at T = 298.15K. Then, for a change of the state from (3) to (4),

4)ΔH(3 PCCdPCC TJmYPmXPTJmYPmXP )3(')3(- ,,,,,,,,

P

1bar

P

PCCCC

HHHH

TJ

TJmPCmPBmPYmPX

ffff

(298.15K)ΔCΔH

23

(C)(B)2(Y)3(X)

4)ΔH(33)ΔH(22)ΔH(1ΔH

P298.15K reaction,

,,,,

0000

T reaction,

where )'()(T'ΔC,P TCnmPjj

k

Homework5 Answer (Due Friday, 02/23/2011) - Wetalldid – · PDF file ·...

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