Chemistry 360 Dr. Jean M. Standard

Problem Set 6 Solutions 1. Three moles of an ideal gas expand isothermally and reversibly from 90 to 300 L at 300 K.

(a) Calculate

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ΔUm ,

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ΔSm , w per mole, and q per mole.

For an ideal gas,

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ΔU = CvΔT , or in terms of molar quantities,

€

ΔUm = Cv,mΔT . Since the process is isothermal,

€

ΔT = 0 and so

€

ΔUm = 0 . For an isothermal reversible process, the work is

€

w = − nRT ℓn V2 /V1( ) . Therefore, the work per mole is

€

wm = − RT ℓn V2 /V1( ) . Substituting,

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wm = − RT ℓn V2 /V1( )

= − 8.314 J mol−1K−1( ) 300 K( ) ℓn 300 L90 L

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wm = − 3003 J/mol.

From the first law,

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ΔU = q+ w or in terms of molar quantities,

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ΔUm = qm + wm . Since

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ΔUm = 0 for this isothermal process, that means that

€

qm = −wm . Thus, the heat per mole is

€

qm = − wmqm = 3003 J/mol.

Finally, since

€

dS = dqrevT , for an isothermal process, the molar entropy change is

€

ΔSm = qmT

= 3003J/mol300 K

ΔSm = 10.01J mol−1K−1.

(b) If the expansion is carried out irreversibly by allowing the gas to expand rapidly into a vacuum,

determine

€

ΔUm ,

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ΔSm , w per mole, and q per mole. For an ideal gas, even if the process is irreversible,

€

ΔUm = Cv,mΔT . Since

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ΔT = 0 in this process, then

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ΔUm = 0 . This also must be true since U is a state function and the process involves the same initial and final states as part (a). The work is by definition

€

dw = − PextdV , or in terms of molar quantities,

€

dwm = − PextdVm . Since the expansion is into a vacuum,

€

Pext = 0 , and therefore,

€

dwm = 0 or

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wm = 0 . From the First Law,

€

ΔUm = qm + wm . Since

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ΔUm = 0 and

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wm = 0 , we have that

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qm = 0 . Since S is a state function, the change in entropy for the irreversible process must be the same as for the reversible process in part (a) since the initial and final states are the same. Thus,

€

ΔSm = 10.01J mol−1K−1 .

2

2. A normal breath has a volume of about 1 L. The pressure exerted by the lungs to draw air in is about 758 torr. Assuming that the outside air pressure is 760 torr, calculate the change in entropy of a breath of air when it is inhaled into the lungs. Assume that the air remains at a temperature of 25°C and that it behaves ideally.

In this process, the air expands from a pressure of 760 torr to 758 torr. Assuming that the gas behaves ideally and that the process is reversible, we have that the entropy change in an isothermal reversible process is

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ΔS = n R ln V2V1

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Unfortunately, we are given the pressure change rather than the volume change. However, if the process is isothermal then we can use the ideal gas equation to get

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V2V1

= P1P2

.

Substituting this expression into the equation for the entropy change yields

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ΔS = n R ln P1P2

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&

' ( .

Now in order to use the formula, we need to determine the moles in 1 L of air at 25°C and an initial pressure of 760 torr (=1 atm). Using the ideal gas law,

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n = PVRT

= 1atm( ) 1L( )

0.08205Latm/molK( ) 298 K( )n = 0.0409 mol.

Substituting, the entropy change is

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ΔS = n R ln P1P2

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= 0.0409 mol( ) 8.314 J mol−1K−1( ) ln 760 torr758 torr

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ΔS = 0.0009 J/K .

3

3. The molar heat capacity of solid gold is given by the relation

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Cp,m = 25.69 − 7.32×10−4 T + 4.58×10−6T 2 , in units of J mol–1K–1. Calculate the entropy change for heating 2.50 moles of gold from 22.0°C to 1000°C at constant pressure.

The entropy change for heating at constant pressure is

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ΔS = CpTT1

T2∫ dT ,

or since

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Cp = nC p,m ,

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ΔS = nC p,mTT1

T2∫ dT .

Substituting the heat capacity for gold and integrating yields

€

ΔS = n 1T

25.69 − 7.32×10−4 T + 4.58×10−6T 2( )T1

T2∫ dT

= 25.69n 1TT1

T2∫ dT − 7.32×10−4 nT1

T2∫ dT + 4.58×10−6 nT1

T2∫ T dT

ΔS = 25.69n ln T2T1

&

' (

)

* + − 7.32×10−4 n T2 −T1( ) + 4.58×10−6 n T2

2

2−T1

2

2

&

' ( (

)

* + + .

For 2.5 moles of gold being heated from 295 K to 1273 K, the entropy change is

ΔS = 25.69n ln T2

T1

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&' − 7.32×10−4 n T2 −T1( ) + 4.58×10−6 n T2

2

2−T1

2

2

"

#$

%

&'

= 25.69 2.5mol( ) ln 1273K295K

"

#$

%

&' − 7.32×10−4 2.5mol( ) 1273K − 295K( )

                                                 + 4.58×10−6 2.5mol( )1273K( )2

2−

295K( )2

2

"

#$$

%

&''

= 93.91 J/K − 1.79 J/K + 8.78J/KΔS = 100.9 J/K.

4

4. A quantity of ice at 0°C is allowed to melt in a large body of water also at 0°C. Calculate the standard molar entropy change for this process. Since the process described involves a phase change, we can use the equation

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ΔS fus! =

ΔH fus!

T fus.

The molar enthalpy of melting is also known as the enthalpy of fusion,

€

ΔH fus! , which can be found in your

textbook or in the CRC.

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ΔS fus! =

ΔH fus!

T fus

= 6008 J/mol273.15K

ΔS fus! = 22.0 J mol−1K−1 .

5. Calculate the change in entropy when one mole of aluminum is heated from 600°C to 700°C. The melting

point of aluminum is 660°C, the enthalpy of fusion is 393 J/g, the heat capacity of solid aluminum is 31.8 J mol–1K–1, and heat capacity of liquid aluminum is 34.4 J mol–1K–1. For heating a solid with a phase change from a solid to a liquid, the entropy change is

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ΔS = Cp s( )TT1

Tfus∫ dT + n ΔH fus

!

T fus +

Cp ℓ( )TTfus

T2∫ dT .

Expressing the entropy change in terms of standard molar heat capacities (since that is what is given in the problem), we have

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ΔS = nC p,m! s( )TT1

Tfus∫ dT + n ΔH fus

!

T fus + n

C p,m! ℓ( )TTfus

T2∫ dT .

Assuming that the heat capacities are independent of temperature (they are listed in the problem as constants), the expression becomes

€

ΔS = nC p,m! s( )TT1

Tfus∫ dT + n ΔH fus

!

T fus + n

C p,m! ℓ( )TTfus

T2∫ dT

= n C p,m! s( ) 1

TT1

Tfus∫ dT + n ΔH fus

!

T fus + n C p,m

! ℓ( )Tfus

T2∫ 1TdT

ΔS = n C p,m! s( ) ln

T fusT1

$

% &

'

( ) +

n ΔH fus!

T fus + n C p,m

! ℓ( ) ln T2T fus

$

% & &

'

( ) ) .

5

5. Continued Substituting,

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ΔS = n C p,m! s( ) ln

T fusT1

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' ( +

n ΔH fus!

T fus + n C p,m

! ℓ( ) ln T2T fus

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$ % %

&

' ( (

= 1mol( ) 31.8 J mol−1K−1( ) ln 933.15K873.15K

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' ( +

1mol( ) 26.98 g/mol( ) 393J/g( )933.15K

+ 1mol( ) 34.4 J mol−1K−1( ) ln 973.15K933.15K

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ΔS = 14.94 J/K .

6. A mixture of 40% N2O and 60% O2 may be used in a dentist office as an anesthetic. Assuming that the

gases behave ideally, determine the entropy of mixing to produce 1 mole of the mixture. The entropy of mixing,

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ΔSmix , is given as

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ΔSmix = − R ni ℓn xii∑ ,

where

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ni is the number of moles of each species and

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xi is the mole fraction. One mole of the mixture would correspond to 0.40 moles N2O and 0.60 moles of O2. The mole fraction of N2O is 0.40 and the mole fraction of O2 is 0.60. Substituting,

€

ΔSmix = − R ni ℓn xii∑

= − R n1 ℓn x1 + n2 ℓn x2( )= − 8.314 J/molK( ) 0.40 mol( ) ℓn 0.40 + 0.60 mol( ) ℓn 0.60[ ]= − 8.314 J/molK( ) −0.6730 mol( )

ΔSmix = 5.60 J/K .

6

7. A quantity of 35.0 g of water at 25.0°C is mixed with 160.0 g of water at 86.0°C. Calculate the final temperature and the entropy change of the system. The molar constant pressure heat capacity of water is 75.291 J/molK.

First, we need to get the final temperature of the system. We know that if two systems are placed in contact, their final temperatures will be equal and the heat gained by one system will equal the heat lost by the other system. If we call the water initially at 25.0°C system A, and the water initially at 86.0°C system B, then

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qA = − qB . Assuming that the mixing occurs at constant pressure, we know that

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qp = ΔH = CpΔT = nC p,mΔT .

The result given above assumes that the heat capacity is independent of temperature. Substituting,

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qA = − qBor nA C p,m H2O( ) ΔTA = − nB C p,m H2O( ) ΔTB .

The heat capacity of water cancels out on both sides of the equation. Also, the moles on either side can be written in terms of masses,

€

n = wM

,

where w is the mass and M is the molecular weight. Substituting,

€

nA C p,m H2O( ) ΔTA = − nB C p,m H2O( ) ΔTBnA ΔTA = − nB ΔTB

wAMH 2O

ΔTA = − wBMH 2O

ΔTB .

The molecular weights cancel out, leaving the equation

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wA ΔTA = − wB ΔTB ,

or wA T final −Tinit,A( ) = − wB T final −Tinit,B( ) . Solving for the final temperature,

€

wA + wB( )T final = wATinit,A + wBTinit,B , or

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T final = wATinit,A + wBTinit,B

wA + wB

= 35.0 g( ) 25.0!C( ) + 160.0 g( ) 86.0!C( )

35.0 g + 160.0 g

T final = 75.05!C .

7

7. Continued The entropy change can be calculated as the sum of the entropy changes of systems A and B,

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ΔS = ΔSA + ΔSB . At constant pressure, the entropy change for heating or cooling a system is

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ΔS = CpTTinit

Tfinal∫ dT = nC p,mTTinit

Tfinal∫ dT .

Substituting this into the equation above for systems A and B,

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ΔS = ΔSA + ΔSB

= nAC p,mTTinit ,A

Tfinal∫ dT + nBC p,mTTinit ,B

Tfinal∫ dT

= nAC p,m lnT finalTinit,A

$

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'

( ) ) + nBC p,m ln

T finalTinit,B

$

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'

( ) )

= 35.0 g( )

18.015g/mol( )75.291J mol−1K−1( ) ln 348.20 K

298.15K

$

% &

'

( )

+ 160.0 g( )

18.015g/mol( )75.291J mol−1K−1( ) ln 348.20 K

359.15K

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'

( )

= 22.70 J/K - 20.70 J/KΔS = 2.00 J/K .

8. Which system has the higher absolute entropy?

(a) 1 g solid Au at 1064 K or 1 g liquid Au at 1064 K 1 g liquid Au – According to the Third Law, for a given substance, the liquid will always have higher absolute entropy than the solid at the same temperature.

(b) 1 mole CO at 25°C and 1 atm or 1 mole CO2 at 25°C and 1 atm

1 mole CO2 – At the same temperature and pressure the triatomic molecular has more energy states available to it; therefore, gaseous CO2 has higher entropy than gaseous CO.

(c) 1 mole Ar at 25°C and 1 atm or 1 mole Ar at 25°C and 0.01 atm.

1 mole Ar at 25°C and 0.01 atm – Both samples are at the same temperature, so the gas with the lower pressure will have the larger volume and hence the larger entropy.

8

9. The thermite reaction involves solid aluminum powder reacting with iron(III) oxide (hematite) to produce aluminum oxide and iron. The reaction is so exothermic that the iron produced is molten (liquid phase). Write the balanced chemical reaction. Determine

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ΔSr! for the process using the tables in the

appendix of your textbook and the standard molar entropy of liquid Fe, 34.76 J/molK. Assume standard pressure and 25°C.

The balanced reaction is

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2 Al s( ) + Fe2O3 s( ) → Al2O3 s( ) + 2 Fe ℓ( ) .

For the chemical reaction shown above, the molar entropy change is

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ΔSr! = Sm

! Al2O3( ) + 2Sm! Fe( ) − Sm

! Fe2O3( ) − 2Sm! Al( ) .

At 25°C, the standard molar entropies are available in the appendix of the text. Substituting,

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ΔSr! = Sm

! Al2O3( ) + 2Sm! Fe( ) − Sm

! Fe2O3( ) − 2Sm! Al( )

= 50.92 + 2 34.76( ) − 87.40 − 2 28.33( ) in J mol−1K−1( )ΔSr! = − 23.62 J mol−1K−1.

10. Using the tables in the appendix of your textbook, determine the standard molar entropies of formation

of the following compounds at 25°C. (a) H2O (l)

The formation reaction is

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H2 g( ) + 12 O2 g( ) → H2O ℓ( ) .

For the chemical reaction shown above, the molar entropy of formation is

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ΔS f! = Sm

! H2O,ℓ( ) − Sm! H2( ) − 1

2 Sm! O2( ) .

Substituting the standard molar entropies at 25°C from the appendix,

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ΔS f! = Sm

! H2O,ℓ( ) − Sm! H2( ) − 1

2 Sm! O2( )

= 69.91 − 130.68 − 12 205.14( ) in J mol−1K−1( )

ΔS f! = −163.34 J mol−1K−1.

9

10. Continued (b) H2O (g)

The formation reaction is

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H2 g( ) + 12 O2 g( ) → H2O g( ) .

The molar entropy of formation is

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ΔS f! = Sm

! H2O,g( ) − Sm! H2( ) − 1

2 Sm! O2( ) .

Substituting the standard molar entropies at 25°C from the appendix,

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ΔS f! = Sm

! H2O,g( ) − Sm! H2( ) − 1

2 Sm! O2( )

= 188.83 − 130.68 − 12 205.14( ) in J mol−1K−1( )

ΔS f! = − 44.42 J mol−1K−1.

(c) CuSO4 (s)

The formation reaction is

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Cu s( ) + S s( ) + 2 O2 g( ) → CuSO4 s( ) .

The molar entropy of formation is

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ΔS f! = Sm

! CuSO4( ) − Sm! Cu( ) − Sm

! S( ) − 2 Sm! O2( ) .

Substituting the standard molar entropies at 25°C from the appendix,

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ΔS f! = Sm

! CuSO4( ) − Sm! Cu( ) − Sm

! S( ) − 2Sm! O2( )

= 109 − 33.15 − 31.08 − 2 205.14( ) in J mol−1K−1( )ΔS f! = − 366 J mol−1K−1.

10

10. Continued (d) H3PO4 (s)

The formation reaction is

€

32 H2 g( ) + P s( ) + 2 O2 g( ) → H3PO4 s( ) .

The molar entropy of formation is

€

ΔS f! = Sm

! H3PO4( ) − 32 Sm

! H2( ) − Sm! P( ) − 2 Sm

! O2( ) . Substituting the standard molar entropies at 25°C from the appendix,

€

ΔS f! = Sm

! H3PO4( ) − 32 Sm! H2( ) − Sm

! P( ) − 2 Sm! O2( )

= 110.50 − 32 130.68( ) − 41.09 − 2 205.14( ) in J mol−1K−1( )

ΔS f! = − 536.89 J mol−1K−1.

(e) C (s, diamond)

The formation reaction is

€

C s, graphite( ) → C s, diamond( ) .

The molar entropy of formation is

€

ΔS f! = Sm

! C, diamond( ) − Sm! C, graphite( ) .

Substituting the standard molar entropies at 25°C from the appendix,

€

ΔS f! = Sm

! C, diamond( ) − Sm! C, graphite( )

= 2.38 − 5.74 in J mol−1K−1( )ΔS f! = − 3.36 J mol−1K−1.

11

11. Using the tables in the appendix of the textbook, determine the standard molar entropy change

€

ΔSr! at

25°C for the combustion reaction

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CH4 g( ) + 2O2 g( ) → CO2 g( ) + 2 H2O g( ) . For the chemical reaction shown above, the molar entropy change is

€

ΔSr! = Sm

! CO2( ) + 2Sm! H2O,g( ) − Sm

! CH4( ) − 2Sm! O2( ) ..

At 25°C, the standard molar entropies are available in the appendix of your text. Substituting,

€

ΔSr! = Sm

! CO2( ) + 2Sm! H2O,g( ) − Sm

! CH4( ) − 2 Sm! O2( )

= 213.74 + 2 188.83( ) − 186.26 − 2 205.14( ) in J mol−1K−1( )ΔSr! = − 5.14 J mol−1K−1.