604 ChaPter 19

We will need the following values from Table 19.2:

l,H"[cor(s)] : -393.509 kJ'mol-l arH"[H2o(1)] : -285.83 kJ'mol-l

Af H"[N2H4(1)] : *50,6 kJ.mol-' af H'[cHjoH(l)] : -239.1kJ'mol-tA.H'[Nr(g)J : o

a. Using Hess's law,

L,H" - | ara'[p.oducts] - I or"'lreactants]

: 2(-285.83 kJ) + (-393.5 kJ) - (-239.1 kJ)

/ -126.t kI \ /_1.91_) : _.. , kJ.-s-,: \ ro1 r.rr-'"*1 ) \tz.o+z g 1

b. Again, bY Hess's law,

L,H" - | nrA'lp.oductsl - I or"'lreactants]

: 2(-285.83 kJ) - (+50.6 kJ)

/ -622.3 kJ \ / I mol \:(ffi) (tr*G):-re.tkJg-More energy per gram is produced by combusting methanol.

19-40. Using Table 19.2, calculate the heat required to vaporize 1.00 mol of CCl.,(l) at298 K.

CC14(l) ----+ CClo(S)

We can subrract ArH'[CCl4(1)] from ArH'[CC14(g)] to find the heat required to vaporize CClo:

AuuoH : -102.9kJ + 135.44 kJ : 32.5 kJ

19-4'1. Using the Arfl' data in Table 19.2, calculate the values of A.H" for the following:

a. C2H4(g) + HrO(l) -+ CTHTOH(I)

b. CH4(g) + 4 Clr(s) '---+ CClo(l) + a HCl(g)

In each case, state whether the reaction is endothermic or exothermic'

a. Using Hess's law,

A,Ho : -271.69 kJ - (-285.83 kJ + 52.28kJ) : -44.14kI

This reaction is exothermic.

b. Again, by Hess's law,

A.Ilo : 4(-92.31kJ) - 135.44 kJ - (-74.81 kJ) : -429.81kJ

This reaction is also exothermic.

19-42. Use the fbllowing data to calculate the value of Au"oH' of water at 298 K and compare

your answer to the one you obtain from Table 79.2: L,^rH" at 313 K:40.1 kJ'mol-l;Er{t; :'75.2 J.mol-i.K-t; Cp(c) : 33.6J'mol-r'K-r.

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624 Chapter 20

20-8. Calculate the value of AS if one

from 10.0 dm3 to 20.0 dm3. Explain

gas is expanded reversibly and isothermally

PdV.Then

mole of an ideal

the sign of AS.

For an isothermal reaction of an ideal gas, 6u.'

AS: /J

Using Z from the ideal gas equation gives

fnRAS : /';O'

: nR1n2.00 : s.76J.K-r

The value of AS is positive f"ru*" the gas is expanding.

As in the previous problem, because the reaction is isothermal ,3q : PdV.Fot an ideal gas,

,Yap:_Ya,dv : __0,

P

so we write AS as

f P r v [_!!op-__nRln0.l:te.lJ.Klls: / ;dv- l-;ae:JIJIJPThe value of AS is positive because the gas expands'

- -3q, so dq -u+:

I+r,

20-9. Calculate the value of AS if one mole of an ideal gas is expanded reversibly and isothermaily

from 1.00 bar to 0.100 bar. Explain the sign of AS.

20-10. Calculate the values of q."u and

gas whose equation of state is given

Examole2O-2'

I un*"'o*u

AS along the path D + E in Figure 20.3 for one mole of a

in Example 20-2. Compare your result with that obtained in

Path D * E is the path described by (P, ,Vt,T) - (P' ,V2,\) - (P\.,V2,T)' For the first step,

6Q,",: dU - lu't : Cv(T)dT - Pdv

and for the second step (because the volume remains constant)

5e,",: dU : Cv(T)dT

Then

rT, f\ t:l'cv(Ttdr-lJT, JV,

fv::- | P,dv:-P,(

fv,ASr"u,p+r : - |

JV,

77,

Plv+ I'c,{DarJT.

v2- v)

P,-JdVI

ancl

626 Chapter 20

This is (by no coincidence) the same value as that found for ASo, ASu*., and A,so*u.

20-13. Show that

AS:C^lnl'7,for a constant-pressure process if C" is independent of temperature. Calculate the change in entrop..of 2.00 moles of H,O(l) (e o : 7 5.2 J.K-1. mol-l ) if it is heated from l0"C to 90"C.

Because AS is a state function, we can calculate it using a reversible process. For a constant-pressu:.reversible process (Equation 19.31), 6e,., : d H : C pdT , and so

o, : ? : 1,,' ?0, - ,e ,t" (?)For 2.00 mol of HrO,

AS : (2.00 mol) (7 5.2J. K-r . mol-r i fn I : 3J .4 I.K-l283

r)"f :iti,>i;

ii'l:-5 "

,l ,

,1,

ffr:-<t

20-'14. Show that

AS:e,. nL+ntn-%.'Ttvl

if one mole of an ideal gas is taken from 7,, 4 to T, V, assuming that C utemperature. calculate rhe value of AS if one mole of Nris) is expanded fromto 300 dm3 at 400 K. Take C o : 29.4 J.K-r .mol-r.

is independent20.0 dm3 at 2-:

For the path ( f, . V, ) --*write AS as

(7, Vr),6w :

AS:

:

Rcnrrrcef -C -

P for an ideal gas, we can write this as

For Nl

LS : (29.4 J.mol-r.K I

: 30.6 J'K-r.mol I

A5: (?^ - R) In L * *rnL'Ttvl

- 8.314J'K-'.mot- r; lnfl+ 8.314J.K-l

: CvdT + Pdv. We cai

. t(_,mol .ln

--20

, -PdV andSq: dU - lta

:U ?,,. 1 +,,)re ., r R

J zo'+ l idvT^VC,,lnL * Rlnl'TrVl

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60

634 Chapter 20

Because this is an isothermal reversiblc expansion, 5q : -5* : PdV. We then use the ideal gasequation to write

AS...: t+: ffov: [*or:,nh!.I.JfJtJv"Vl

For an isothermal cxpansion of an ideal gas, P, V, : Pr.%. We can then write the change of entropyof the gas as

At," : (l moi)(8.3145 J.mol-r.K r; 1n5.00 : +13.4 J.K-l

For a reversible expansion, AS,o, : 0, so AS,uo : -Atr. : -13.4 J.K-l.

20-26. Redo Problem 20-15 fbr an expansion into a vacuum, with an initial pressure of 10.0 bar anda final nressure of 2.00 bar.

As in Problem20*25, AS,r, :a vacuum, AS,u* : 0, so AS,o,

13.4 J.K '. However, because this is an irrevcrsible expansion into:73.4 J.K 1.

rt000 F

^t:nl* 7o'4 l0O{)

: (1 mol)R;f nn

: lg2"tgJ.K'

20-27, The molar heat capacity of 1-butene can be expressed as

e rfrl1 n: 0.05641 + (0.04635 K')r - e.39zx t0-s K r)Tt+ (4.80 x 10-e K,)2,over the temperature range 300 K < f < 1500 K. Calculate the change in entropy when one moleof 1-butene is heated from 300 K to 1000 K at constant Dressure.

At constant pressure, 5q : d H : nT ,dT. Then Equati on 20.22 becomes (assuming a reversibleprocess)

[o.oso+rr-i + (o.o+035 K-r) - (.2.392 x l0 5 K-')T

+(4.80x 10 eK-\T2ldT

20-28. Plot A..*S against ], for the mixing of two ideal gases. At what value of ), is A.."S a

rnaximum? Can you give a physical interpretation of this result?

We can use Equation 20.30 for two gases:

A.,*3 : - R (1, ln )r .1, ln -yr)

Because ], * 1,,, : 1, we can write this as

A.i"S/R: --Vr tn,lr 1t -.r',) ln(l - y,)

640

From Problem20-35' we can

Chapter 2o

20-38. consider one mole of an ideal gas confined to a volume v' calculate the probability that all the

No molecules otttris ideal gas wilibe found to occupy one half of this voiume' leaving the other

half emPtY.

write the ProbabilitY as

(+)^^:(i)"^=o

20-39. Show

that f p,[hat S.r,,.. given by Equation 20'40 is a maximum when all the p

' ate equal' Remember

: 1, so that

f p,ln p -- p,lnP, I P.lnPri "'/r'l 'l

+(1-p,-p,

See also Problem J-i0

Begin with Equation 20'40'

* p,;\n pn ,

p,,-,) ln(1 - Pt - Pz- "' - P,-,)

tl

!ilcll

]I

Itl

5

Pr-,)]

1

vlI

^/^ I

s.u.,"' : -k, t Piln Pi

Substituting the expression given for !, P,ln p,' we hnd

s,y,t". : -'t<ulrrln pr * P2ln pt+ "' + Pn-tln Pn-t

+ (l - Pr - Pz P,-') ln(1 - Pt - Pz- '

-t= -1npi+l-ln(1- Pt-"'- Pi-,- P,.r - P,-,\-dp,

0 - ln p; - ln(l - Pr - "'- Pi-, - Pi+t - "'- P'-')

F,:|- Pr-"'- P1-t- P,.l - Pn-t

Because P, canbe any of p, to P,-r' andthe above equality holds for all p'' all the p- must be

equal.

2o_4o.UseEquation20.45tocalculatethemolarentropyofkryptonat2gS.2Kandonebar,andcompare your result with the experimental value of 164'1 J'K-r'mol-t'

This problem is like Example 20-6' We use Equation 20'45'

{ r - ' -..r2s : ;R-L R,"

L(11#4 )

Assuming ideal behaviot'at298'2 K and one bar

No NoP

-:-VRT(6.022 x l0rr mol I)(l bar)

tO.OS3 la,ltr'bar'mol I'K -r )t298.2 K)-l2.429 x 1022 dm-3 :2.429 x 1025 m

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,,.5 6 Chapter 21

21-18. Use the following data to calculate the standard molar entropy of cyclopropane at 298.1 K.

C}[C.Hu(s)JlR:_ 1.921+(0'1508K_')r-e.670x10_aK-)T'(2.694 x 10-6 6-:;7:15K<T<145.5K

c;lc.H6(l)l lR : 5.624 + (4.493 x 10-2 K-')r - (1.340 x 10-a 11-2)72

145.5K <T <240.3K

C![C.H6(S)] I R : _ 1.'/g3 + (3.21'7 x 10 2 K-')r - 0.326x 10 5 11-2172

240.3K<T < 1000K

Tr,"- 145.5 K, 4,.p :240.3 K,At".H :5.44 kJ mol-r., A",nF:20'05 kJ'moi-l, and Oo

130 K' The correctitn for nonideality : 0'54 J'K I 'mol-r '

c/T)'a

,1LZl( R

5T

/T\\%/

n l5I_I

JO

,, - ["" e"lC.Hutstl )r -L Ltu,ho,* J,, -- T-o,- u;r5K

* I,':,','

--+!f dr + #** L:,r,r

cPLCiHo(s)ldr

* correction

: 0.995 J.K-r .mol-r + 66.1J'K-r'mol-r + 31 .4 J'K r'mol-r

+3g.5 J.K-r.mol-r + g3.4 J.K r.mol-r + 10.g J.K-r.mol-l

+0.54 J'K-r'mol-r

:237.8 J'K-r'mol-r

This compares very well with the literature value of 231 '5 J'K-1 'mol r'

2.,-1g. Use the data in Problem 21-18 to plot the standard molar entropy of cyclopropanc from 0 K

to 1000 K.

Do this in the same manner as Probiem 21_15, using the appropriate values from Problem 21-18

and changing the limits of integration as required'

300

200

100

0600

TIK

I

lc4

200 400 800 I 000

o I-All-J6o= loob-Ti,r. l0'

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