EVALUACIÓN DEL TEST DE PERSONALIDAD VALORACIÓN

7 A (a) isolated; <b> closed: (c) isolated; id) open; (e) closed: (f) open

7.3 (a) W ork is given by w = - P t A t ' . The applied external pressure is

known, but we must calculate the change in volume given the physical dimensions of the pump and the distance, c l the piston in the pump

moves:

A h = - ; r r ~ d = ,7(1 . 5 cm): (20. cm) 1 ---- — — — | = -0 .1 4 L

i 1 0 0 0 cnv j

A V is negative because the air in the pump is compressed to a smaller volume; work is then;

■1013^ I ’ vt* = -(2 .0 0 atm)( 0.14 L )! ~ - - - --..- = 28 J

l L • a i m

( b ) W ork on the air is positive by convention as work is done on the air, it

is compressed.

7.5 The change in internal energy A t is given simply by summing the two energy terms involved in this process. We must be careful, however, that the signs on the energy changes arc appropriate. In this case, internal energy w ill be added to the gas sample by heating and through compression. Therefore the change in internal energy is;

A U - 524 k j + 340 kJ = S64 k j

7,7 (a) The internal energv increased by more than the amount o f heat added. Therefore, the extra enemy must have come from work done on the system.

7.9 To get the entire internal energy change, we must sum the changes due to heat and work. In this problem, q - +5500 kJ. Work will be given h>

w = -/JxlAF because it is an expansion against a constant opposing pressure:

w = — 750 Torr f 1846 m L - 345 m l.' ^ 760 Torr • atm 1J 1000 m L - L 1 J

= --1.48 T atm

To convert to J we use the equivalency o f the ideal gas constants:

u =-(1.48L-atm ) 8.314 J • K ! • mol" ;- 1 .5 0 x 1 0 ' J 0.08206 L • atm • K"' • mol_: j

AU = q + w = 5500 kJ - 0.150 kJ=5500 kJ

The energy change due to the work term turns out to be negligible in dm problem.

7.11 Using A U = q + w , where AU = -2573 kJ and q ~ '947 j<J . - 2573 kJ = -947 kJ + m

Therefore, w = -1626 kJ.

1626 kJ o f work can be done by the system on its surroundings.

7.13 (a) true if no work is done; (b) always true; (cl always false; /dl true only if w = 0 (in which case AU — q — 0); (e) always true

7.15 (a) During melting heat is absorbed and q is positive.

Since the change is occurring at constant temperature AE 0. Therefore work is done by the system and w is negative.

(b) During condensation heat is released and q is negative. Since the change is occurring at constant temperature A t 0. Therefore w'ork is done on the system and vc is positive.

7.17 (a) The heat change will be made up o f two terms: one term to raise the temperature o f the copper and the other to raise the temperature o f the water:

q = (400.0 g)(4.18 J • (°C) 1■g-1)(100.0°C -2.2.0°C) + (500.0 g)(0.38 J • ( °C )_1 • g_I)(100.0°C - 22.0°C) = 1.30xl05 J +1.48 x 104J = 1.45 x10s J =1.4x10" kJ

(b ) The percentage o f heat attributable to raising the temperature o f water will be

f

v

1.30x10' J '

1.45 x 1 O' J j(l 00) = 90%

7.19 heat lost by metal = heat gained by waiter (20.0 g)(7j-inill - 100.0°C) (0.38 J • (°C) ' - g ’ ) = - (50.7 g )(4 .18 J * (°C)

_■s >('4,.,i-

22.0°C)

(4,

- 100.0°C)(7.6 J *(°C) ’ ) = - (212 .

i'(°crl)(7fi,

lai - 22.0X ) - 100.0°C = - 28(7;. U1| - 22.0°C) ^ l i n a l + 28 T[hm] = 100.0°C + 616°C 29 7;I n a l = 216°C = 25°C

7.21

C' = 22.5 kJ 23.97°C- 22.45°C = 14.8kJ-(°C)_

7.23 (a) The irreversible w ork o f expansion against a constant opposing pressure is given by

w= P.x A V

w = -(1.00 atm ) (6.52 L - 4.29 L)

= -2.23 L • atm

(b) An isothermal expansion will be given by vr = - nRTV2

K

n is calculated from the ideal gas law:

_ PV_ ________ (1.79 atm)(4.29 L)_________

RT ~ (0.082 06 L.■ atm • K"' • mol-1) (305 K)

w = - (0.307 mol)(8.314 J ■ K _l • mol 1 )(305 K ) In = -326 J

Note that the work done is greater when the process is carried out reversibly.

7.25 N O ,. The heat capacity increases with molecular complexity— as more atoms are present in the molecule, there are more possible bond vibrations that can absorb added energy.

7.27 (a) The molar heat capacity o f a monatomic idea! gas at constant pressure is C P m = 4 R- The heat released will be given by

---- - (25.0°C - 97.6°C)(20.8 J - m o l1 •

(°C y’ )

= -90.6 J ' y 83.80 g • mol- )

(b) Similarly, the molar heat capacity o f a monatomic ideal gas at constant volume is C, m = \R. The heat released will be given by

f ---5-° - ? , 1 (2 5 .0 °C - 9 7 .6 °C )(1 2 .5 J ■ moP1- ( “C ) ' 1) = -5 4 .4 J 83.80 g • mol J

7.29 (a) IICN is a linear molecule. The contribution from molecular motions will be 5/2 R.

(b) C ,H fi is a polyatomic, nonlinear molecule. The contribution from molecular motions will be 3R.

(c) Ar is a monatomic ideal gas. The contribution from molecular motions to the heat capacity will be 3/2 R.

(d) HBr is a diatomic, linear molecule. The contribution from molecular motions will be 5/2 R.

7.31 The strategy here is to determine the amount of energy per photon and the amount o f energy needed to heat the water. Dividing the latter by the former will give the number of photons needed. Energy per photon is given by:

A

( 6.626 x ] O' 34 J -s)(2.9979x 10" m -s~ 4.50 x 10 3 m

= 4.41 x 10” ° J ■ photon 1

The energy needed to heat the water is:

350 g (4.184 J • g ! • ° C ! ) (100.0 °C - 25.0 "C) = U 0 x 105 J The number o f photons needed is therefore:

1.10 x 105 J

--- --- - = 2.49 x|0- photons 4.41 x 10 ^ J • photon

7.33 (a) Using the estimation that 3R = C: 3/? = C = (0.392 J - K1 -g'1 )(>/)

3 R

M = — - 63.6 g mol

0.392 J • K. -g '

This molar mass indicates that the atomic solid is Cu(s)

(b) From Example 5.3 we find that the density of a substance which forms

4

M

a face-centered cubic unit ceil is given bv:c/ = -rrr--- r . Therefore, wc

8’ " A7

, r

expect the densit\ o f copper to be: 4 (63.55 g • moT1) c/ =

7.35 (a) A //Nap

22.45 g 46.07 g • mol j

7.37 This process is composed o f two steps: melting the ice at 0°C and then raising the temperature o f the liquid water from 0°C to 25°C :

Step 2: AH = (80.0 g ) (4.18 J • (°C)"1 • g“ ' ) (20.0°C - 0.0°C) = 6.69 kJ Total heat required = 26.7 kJ + 6.69 kJ = 33.4 kJ

7.39 The heat gained by the water in the ice cube will be equal to the heat lost by the initial sample o f hot water. The enthalpy change for the water in the ice cube will be composed o f two terms: the heat to melt the ice at 0°C and the heat required to raise the ice from 0°C to the final temperature.

heat (water) = (400 g)(4.184 J - (°C)"1 • g"1 >(7; -4 5 °) = (1.67xl03 J • (°C)_1 )(7^ -4 5 °)

Setting these equal:

- (1.67 X 1 o3 J • (°C) 1)/ ;. + 7.5 X 1 o4 J - 1.67 X1 o4 J + (209 J • (°C)~! )7;. Solving for 7j :

T = --- x '°J J

= 3 r c

' 1.88 x 10’ J ' ( ° C r ‘ Step 1: AH =( 80.0 g * *v 18.02 g • mol" (6.01 kJ- m o l1) = 26.7 kJ + (50.0

g)<4.

184 J • ( “C T1

-g-‘

)(7j. - 0°) = 1.67x10“ J = (209 J - ( ° C r ') ( r f - 0°)

7,41 Based on the A//(u, = i U.O kJ mol' ! and A//,;;r • 20.0 kJ m o l"1 and a constant heating rate, melting should occur twice as fast as vaporization

w h i c h is observed in ( h i (c) and <d). Heating curve (d) can be eliminated

because the slope lo r the solid, liquid and gas are all the same. Curve (c) can also be eliminated because the heating slope for the solid and liquid are the same (yet they have different heal capacities). T h is leaves (b) as the best match.

7.43 (a) L \ H = (1.25 mol)(+358.8 kJ • mol""') - 448 kJ

(b) 197 a C 12.01 a • mol 1 C 358,8 kJ : 4 m o l

C

J !()’ kJ (e) A H = 4 1 5 kJ = (/?, s ) 358.8 kJ • mol } 4ntoTcS^ ;

n ( v - 4.63 mol CS, or (4.63 m ol)(76.13 g • mol ! ) -- 352 g CS,

7.4?

(a) (12 ft x 12 ft x 8 ft)' — IS™. ; =

3,26 x

SO cm’

I ft 1

The heat capacity o f air is 1.01 J - i cO ' ■ mol 1 and the average molar mass of air is 28.97 g • mol-1 (see Table 4.1). The density of air can be

calculated from the ideal gas lav :

_ P _ 1.00 atm.

M R T (28.97 g • mol )(0.082 06 L ■ atm - K " 1 • mol-11(277.6 K } d = 0.001 52 e cm

4 0 °F = 4.44°C. 78~F - 2 T 5 ^ C

A T - 25.55°C - 4.44: (; -- 2 1.1 :C

The heat required is

( 3 . 2 6 x 1 0 cnT )(0.O1.' i 52 g • cm ") {1.01 J •( ° C i 1 • mol 1) (21.1 )

The mass o f octane required to produce this much heat will be given by ' -1.056 xlQ :’ kJ ,-5471 kJ-mol'' (1 14.22 g-m ol’ 1 ) = 22.0465 g 2.2 x I0; g (b) AH = (1.0 gal)(3.785 x IQ3 mL-gal~')(0.70 g ■ mL'! ) 114.22 g- mol' 1 "J r -10942 kJ v2 mol octane J = -1.3x10" kJ

7.49 From AH = AU + PA V at constant pressure, or AU - AH - i'AJ . Because w - - P A V - +22 kJ, we get -15 kJ + 22 kJ = AU - - 7 kJ.

7.51 Net production o f 1 mole of gas:

An — +1 mol P V = nRT PAV = AnRT - P A V = -A n R T = -(lmol)(8.314 J.K l.m o f1)(298K) = -2.48 kJ A u = q + w = AH - PA V = -318 kJ - 2.48 kJ = -320 k J

7.53 To determine the enthalpy o f the reaction we must start with a balanced chemical reaction and determine the limiting reagent:

2HCl(aq) + Zn(s) —> I L (g ) + ZnCl2(a q ). 0.800 L - 0.500 M HC1 - 0.400 mol HC1

8 5 g

--- --- r = 0.130 mol Zn 65.37g • mol'1

Examining the reaction stoichiometry and the initial quantities o f HC1 and Zn, we note that Zn is the limiting reagent (0.260 mol o f HCI is needed to completely react with 0.130 moles o f Zn). The enthalpy o f reaction may be obtained using tabulated enthalpies o f formation:

AH, --153.89 kJ mol kJ kJ - + 2 -167.16— 1-2! -167.16 mol,' V mol - 0 mol

This is the enthalpy per mole o f Zinc consumed. Therefore, the energy released by the reaction o f 8.5 g o f Zinc is:

( r j 3

I -153.89---- (0.130 mo!) = -20.0 kJ

l mol J '

The change in the temperature o f the water is then:

-20000 J ( -4.184 ‘’C’ gJ 800 g ) A T AT - 5.98 °C and T, - 250C + 5.98°C = 31"C

7.55 The enthalpy o f reaction for the reaction

4 C7H5N }0<As) - 21 0 :(g) -> 28 CO:(g) + 10 I K ) ( g ) t 6 N2(g) may be found using enthalpies o f formation:

28 • 393.51 kJ mol,-10 -241.82 — Y mol J -4 k 1 '• _67—— | - -13168 mol J kJ mol This is the energy released per mole o f reaction as w ritten. One fourth o f this amount o f enerex or 3292—— will be released per mole o f TNT

consumed. The energy density in k.1 per L may be found by dividing this amount o f energy with the mass o f one mole o f TN T and then by multiplying with the density o f TNT:

3292 ^ --- ^^+1 1.65 5^ 227.14 - - cnr'y mol 10 cm' 1 L kf -+23.9x10 — L 7.57 (a) CO(g) + H20 (g ) - C 02(g) + H2(g)

(b) Bomb calorimeter means volume is fixed. Therefore, w = 0. A U ~ q + w Need to find q. Use q - CAT

First work out how much heat was released on burning 1.40 g o f carbon monoxide.

^reaction + ^calorimeter — 0 <•/react ion " ^calorimeter

= - CAT

= - (3.00 kJ (°Cr')(22.799 °C - 22.113 °C) = - 2.058 kJ

MW (CO) = 12.0107 g.mol’ 1 + 15.9994 g.mol' 1 = 28.0101 g.mol' 1

moles (CO) - 1.40g

28.0101 g.mol'1= 4.998 x 1 O'2

For 1.00 mol CO(g), _{9 =} - 2.058 LI 4.998 x W 1 mol

= - 41.176 kJ. mol' 1 = - 41.2 kJ, mol

-1

For the combustion o f 1.00 mol CO(g), A U - q = - 41.2 kJ. mof

7.59 The combustion reaction o f diamond is reversed and added to the combustion reaction o f graphite to give the desired reaction:

C(gr) + 02(g )--- >C 02(g) A H ° = -393.51 k.l CO: (g )--- >C(dia) + 0 ,(g ) A//0 = +395.41 kJ C (gr)---> C(dia) AH° - 4-1.90 kJ

7.61 7.63 7.65 7.67 -** 3BaO(s) —> 2Ba(s) + - 0 , ( g ) 2A!(s) + -^0 ;(g) —> A l,0 ,(s)

3Ba0(s) + 2AI(s) --> A l,0 ,(s ) - 3Ba(s)

A/7° = 166 kJ

A//c = -1676 kJ A/7° = - l5 kJ

First, write the balanced equations for the reaction given:

C ,H ,(g ) + 4 0 . (g )--->2 C O ,(g) + 11,0(1) A H 0 = -1300 k.l

C,H,.(g) + 4 0 , (g )--- >2 CO,(g| r 3 H .0(1) A H '° = -1560 k.l

II.(g ) + 4 0 , ( g ) --- >11,0(1) AH° = -286 kJ The second equation is reversed and added to the first, plus 2x the third:

C ,H ,(g ) + 4 0 , ( g ) --- >2 C O ,(g) + 11,0(1) AH ° - - 1300k.l 2CO . (g) + 3 H ,0(1)--->C.H „(g) + 4 0 . (g) AH ° = + 1560 kJ 2[H_,(g) + 4 0 ,(g)--- >H ,0 (l)] 2[AH° = - 286 k.l] C \ H ,(g) + 2 H , ( g ) --- >C; ll„ (g )

A//° = -1300 kJ- mol 1 t 1560 kJ-mol' 1 + 2(-286 kJ • mol 1)

= -312 kJ ■ mol '

The reaction enthalpy for this reaction is given by: A// ° = 12 ( A//° , ( IP O J )) - [4 (AH ° f (l I \ 0 , J )) + 5 (A f f ° ( ( N2H t J ))] = 1 2 ( - 2 8 5 . 8 3 k J - m o P ' ) — [4(-174.10 kJ • m oP1) + 5(+50.63 kJ • mol-1 )j = -2986.71 kJ • mol-1 2 NH4C1(s) — 2 N H a g ! - 2 HCl(g) N2( g ) - 4 H 2( g ) - C H g ! — 2 NH4Cl(s) 2 Nl p (g) —» N -( Q ) -- ? H-(g)_________ H2(g) + CP(g) — : HChg) Ali° -- -184.6 kJ AH° - 2( 176.0) kJ AH° = -628.86 k.J AH° = 92.22 kJ

A//° = 2(+188.32 kJ) A H° +92.22 kJ 7.69 2NIUBr(s) —> 2NH3(g) - 2HBr(g) 2 NH3(g) -> N2(g) + 3 H2(g) N2(g) + 4 H2(g) -r Br2(l) -> 2 NH4Br(s) AI f - -541.66 kJ H2(g) + Br2(]) 2 HBr(g) A I f = -72.80 kJ

7.71 From Appendix 2A, AH°, (N O ) = +90.25 kJ The reaction we want is

N ,(g ) + 4 0 , ( g ) --- > N ,05(g)

Adding the first reaction to half o f the second gives

2 NO(g) + 02( g ) --- > 2 N O ,(g ) AH° = -114.1 kJ 2 N O : (g) + f 0 : ( g ) --- > N205(g) AH ° = -55.1 kJ 2 NO(g) + 4 02( g ) --- > N205(g) -169.2 kJ

The enthalpy o f this reaction equals the enthalpy o f formation o f N205 (g) minus twice the enthalpy o f formation o f NO, so we can write

-169.2 kJ = A H °{ (N ,O ,)-2 (+ 9 0 .2 5 k J ) A//0, ( N , 0 5) = +11.3 kJ

7.73 The enthalpy o f the reaction

PC1,(1) + Cl2( g ) --- > PCl5(s) A//° = - 124 kJ is A//°r = S AH ° f (products) - I AH ° f (reactants)

- 124 kJ = AH ° f (PC15, s) - AH ° f (PCI.,, 1)

Remember that the standard enthalpy of formation o f C L (g) will be 0 by definition because this is an element in its reference state. From the Appendix we find that

M I ° , (PC13,1) = -319.7 k J -m ol1

- 124 kJ = A/7°i (PCl5,s) - (-319.7 kJ) A/7°, (PCl5,s) = -444 kJ • mol 1

H : ( g )^ 4 Q : (g)---w [].()(.])

The enthalpy change can be estimated from bond enthalpies. We w ill need to put in (1 mol)(436 U • mol 1) to break the i I — I 1 bonds in

1 moFH ,(g). (

4

-mol) (496 kJ • moi 1) to break the O— O bonds in

4 mol 0 : (g ); we w ill get back ( 2 m ol) (.463kJ • mol 1) for the formation of 2 mol O— H bonds. T h is w ill give A H ~ -2 4 2 kJ • mol ‘ . T h is value, however, w ill be to produce water in the gas phase. In order to get the value for the liquid, we w ill need to take into account the amount of heat given o ff when the gaseous water condenses to the liquid phase. T h is is

44.0 kJ • m o F1 at 298 K:

7.75 (a) For i l :0 ( i ) , we want to Unci the enthaip> of the reaction

- A H \ ... = -2 4 2 kJ •mol-1 - 44.0 kJ • mol = -2 8 6 kJ • mol

(b) The calculation for methanol is done sim ilarly: C(gr) i- 2 I M g ) + y C M g j---►CH .O H tl)

A H for individual bond contributions:

atomize 1 mol C(gr) (1 m ol)(717 kJ - m ol"1) break 2 mol H - H bonds (2 m ol)(436 kJ - mol 1) break 4- mol O, bonds (4-rnoi )( 4 % k j • mol 1) form 3 mol C— H bonds - ( 3 m olM 412 k j • mol''1)

form 1 mol C— O bonds -(1 mol){360 k j • mol 1) form 1 m olO— H bonds (1 m ol)(463 k j • mol ' )

(c) 6C(gr) + 3 H 2(g )--->C6H6(1)

Without resonance, we do the calculation considering benzene to have three double and three single C— C bonds:

atomize: 6 mol C(gr) (6 mol) (717 kJ • mo) !)

break: 3 mol H II bonds (3 mol) (436 U • mol ’ } form: 3 mol CRC bonds - (3 mol)(612 kj • mol : i form: 3 mol C— C bonds - (3 mol) (348 kj ■ mo! 1) form: 6 mol C— H bonds - ( 6 mol) (412 ki • mo! )

Total +•258 kJ

t*„Mnc(i) = A//0f.ben/e,K.(g) =+258 kJ • mol" -30.8 k;-mc.

= +227 kJ • m ol'1

(d) 6 C(gr) + 3 H ,(g )--->C6H(>(I)

With resonance, wc repeal the calculation considering benzene to have ?; resonance-stabilized C— C bonds:

atomize: 6 mol C(gr) (6 moi)(717kJ • mol )

break: 3 mol H— H bonds (3 mo!) (436 kJ • mol' ' } form: 6 mol C— C bonds, resonance - (6 mol) (518 kJ • mol form: 6 mol C— TI bonds - (6 mol) (4 i 2 kj mo! 1)

Total -30 VJ

A //°f.benzene< I) = A //°r.ben/ene(g) “ A //°vap = +30 kJ ‘

' “ 30 8

= -1 kJ • mol-1

SM-196

L

A//L = 2 A H °( (Na. g) + AH ° i (O, g) + 21, (Na)

- £ eal( 0 ) - £ ca2( 0 ) - A ^ f (Na20(s) A//, = 2 (107.32 kJ • mol"1) + 249 kj • mol-1 + 2 (494 kJ • mol"1)

- U l k J m o F 1 + 844 kJ • mol"1 + 409 kJ-mol"1 AHl =2564 kJ-moF'

7.79 (a) AH x = A H 0,. (Na, g) + AI f 0, (Cl, g) + /, (Na) - Eea o f C l-A / / r(NaCl(s))

787 kJ-mol"1 =108 kJ-mol 1 +122 kJ-mol 1 + 4 9 4 kJ-mol'1 - 349 kJ • mol"1 - A//r(NaCl(s))

A//r(NaCl(s)) = -412 kJ-mol-1

AH, = A H (K, g) + A l l 0, (Br, g) + /, (K )

- £ ca(B r)-A / / r(KBr(s))

AHl = 89 kJ • moF' + 97 kJ • mol-1 + 418 kJ • inol 1 - 325 kJ • moF1 + 394 kJ • moF!

= 673 kJ • mol 1

(c) AH, = A H °t. (Rb, g) + AH ° t (F, g) + /, (Rb) - Eca (F) - AH f (RbF(s))

774 kJ • mol 1 = AH ° f (Rb, g) + 79 kJ • moF1 + 402 kj • moF1 - 328 kJ • moF1 + 558 kJ • moF1 A//°r(Rb, g) = 63 kJ-mol"1

7.77 For the reaction N a .O (s)---

>

2 N a "(g ) + 0 2“ (g)

7.81 (a) break: form:

3 mol C = C bonds 3(837) kJ • m o P 6 mol C=C bonds -6(518) kJ • mol"1

Total - 597 kJ • m oF1

(b) break: 4 mol C — H bonds 4(412) kJ • mol"1

4 mol Cl Cl bonds 4(242) kJ • moF1 fonn: 4 mol C—Cl bonds -4(338) kJ • moF1

7.83

4 mol H — Cl bonds -4(431) kJ ■ mol'1

Total -460 kJ • mol 1

(c) The number and types o f bonds on both sides o f the equations are equal, so we expect the enthalpy o f the reaction to be essentially 0.

(a) break: 1 mol N — N triple bonds (1 mol) (944 kJ - mol ) 3 mol F F bonds (3 m ol)(158 kJ-mol ') form: 6 mol N— F bonds (6 m ol)( -195 kJ - mol-’ )

Total + 248 kJ • m oP1

(b) break: 1 mol C=C bonds (1 mol)(612 kJ-mol"1) 1 mol O H bonds (1 mol) (463 kJ -moP1) form: 1 mol C — C bonds -(1 mol) (348 kJ • moP1)

1 mol C— 0 bonds -(1 mol) (360 kJ-moP1) 1 mol C— H bonds -(1 mol) (412 kJ - moP1)

Total -4 5 kJ • moP1

(c) break: 1 mol C— H bonds (1 mol) (412 kJ - mol"1) 1 mol Cl -Cl bonds (1 mol) (242 kJ • moP1) form: 1 mol C Cl bonds - (1 mol)(338 kJ • m oP1)

1 mol H— Cl bonds -(1 mol) (431 kJ • moP1)

7.85 The value that we want is given simply by the difference between three isolated C=C bonds and three isolated C — C single bonds, versus six resonance-stabilized bonds:

3 O C bonds + 3 C— C bonds = 3(348 kJ) + 3(612 kJ) - 2880 kJ 6 resonance-stabilized bonds = 6(518 kJ) = 3108 kJ

As can be seen, the six resonance-stabilized bonds are more stable by ca. 228 kJ.

7.87 (a) The enthalpy o f vaporization is the enthalpy change associated with the conversion C6H6(1)--- » C(1H6(g) at constant pressure. The value at 298.2 K will be given by

A //°vapon/ational^ K = A//°r (C6H6, g) - AH°( (C 6Hft. 1) = 82.93 kJ - mol 1 - (49.0 kJ • mol-1) = 33.93 kJ-mol-1

(b) In order to take into account the difference in temperature, we need to use the heat capacities o f the reactants and products in order to raise the temperature o f the system to 353.2 K.. We can rewrite the reactions as follows, to emphasize temperature, and then combine them according to Hess’ s law:

Q H „(l> ai:,SK---- > C „H „(g)al2WK AH° = 33.93 kJ

Q H 6<l)a, --- ►C J U I U uk AW° = ( 1 mol)(353.2 K - 298.2 K) (136.1 J-niol-1- K "1) = 7.48 kJ

A//0 = (1 mol)(353.2 K - 298.2 K) (81.67 J-mol-1- K ' 1) = 4.49 kl

To add these together to get the overall equation at 353.2 K, we must reverse the second equation:

AH° = - 7.48 kJ

^“ 6 ^ 6 ^ )at 353.2 K ^ ^ 6 ^ 6 ( O at 298 K

C6H6( g W --- ► C6H6(g)at,5.,2K A tf° = 4.49 kJ

^ (>^ 6 ( 0 ai 353.2 K ^ ^ 6 ^ 6 ( 8 ) a l 353.2 K

AH ° = 33.93 kJ • moP! - 7.48 kJ • moP! + 4.49 kJ • moP' - 30.94 kJ • moP1

(c) The value in the table is 30.8 kJ • mol"1 for the enthalpy o f

vaporization o f benzene. The value is close to that calculated as corrected by heat capacities. At least part o f the error can be attributed to the fact that heat capacities are not strictly constant with temperature.

7.89 For the reaction: A ( 2B —> 3C -I D the molar enthalpy o f reaction at temperature 2 is given by:

A//r°2 = //T, (products) - I I ^ 2 (reactants)

= 3/C, (C) -f //;2 ( D) - h: 2 (A ) - 2YC, (B) = 3[/9" ,(C) + C„.„,(C)(7; - 7J )j + [ / C ,(D ) + CpJ D ) ( T 2-7J)] -[/ / :, (A ) + C „ (A)(7; - 7;)] - 2[h: , (B ) + C„.,„(B)(r2 - Tt)] = 3/C, (C) + (D ) - //;, (A ) - 2 (B) I3C.. ..IO + C,.. (D )- C ,,„ (A > -2C(,,„(B )](r2 -7 ") = A//r°, +[3C (v,1(C) + C/,.„,(D )-C ?.,„(A )-2 C (,.,„(B)](r; -71) Finally, A H r 2° - A I I r ] ° + ACp(P2 - 7j) , which is Kirchhoff s law.

7.91 This process involves Five separate steps: (1) raising the temperature o f the ice from -5.042 °C to 0.00 °C. (2) melting the ice at 0.00°C, (3) raising the temperature o f the liquid water from 0.00°C to 100.00°C, (4) vaporizing the water at 100.00°C, and (5) raising the temperature o f the water vapor from 100.00°C to 150.35°C.

Step l :

Step 2: AH = 42.30 g 18.02 2 - mol'- K6.01 kJ ■ mol-1 ) = 14.1 kJ Step 3: AH = (42.30 g )(4 .18 J • (°C )_I-g "1)(100.00 °C - 0.00°C) = 17.7 kJ r 4230a > , Step 4: A l l = |--- :— s— (40.7 kJ • m ol'1) = 95.5 kJ 118.02 g- m o l"'; Step 5: A H = (42.30 g)( 2.01 J •(°C )~'-g_l )(150.35 °C - 100.00°C) = 4.3 kJ The total heat required

= 0.4 kJ + 14.1 kJ +17.7 kJ +95.5 kJ + 4.3 kJ = 132.0 kJ 7.93 (a) (b) 2H ,S(g) + 3 0 2(g) - » 2 S 0 ,(g) + 2H20(1) 4H2S(g) + 2S02(g) -+ 6S(s) + 4H2Q(1) 6H2S(g) + 3 0 2 (g) -+ 6S(s)+ 6H20(1) or 2H2S(g) + 0 2(g) -> 2S(s) + 2H20(1) 60.0x 103g

32.065 g.tnor' = 1871.2mol = 1.87xl03 mol

A H ° (reaction) = LA//, ° (products) - LA//, ° (reactants)

= 2(—2.85.83 kJ. mol-1) - 2(-20.63kJ. mol-1) = -571.66 kJ.m of1 + 41.26 kJ.m of1 = -530.40 kJ.m of1

Since -530.40 kJ.mof reaction produces 2 moles o f S(s), then the enthalpy change associated with the production o f 60.0 kg o f S(s) would be -4.96xl0-kJ

7.95 Appendix 2A provides us with the heat o f formation o f h (g) at 298K (+62.44 kJ • mol-1) and the heat capacities o f I, (g) (36.90 J - K '1 • m or1) and I2(s )(54.44 J 'K " 1 • mol-1). We can calculate theA//sub° at 298K:

1, ( s )---- > 1, (g) A//sub0 = +62.44 kJ • mol"1 We can calculate the enthalpy o f fusion from the relationship

= AH

but these values need to be at the same temperature. To correct the value for the fact that vve want all the numbers for 298K, we need to alter the heat o f vaporization, using the heat capacities for liquid and gaseous iodine.

1,(1) at )84.3°C---■+ 1, (g ) at 184.3°C A tfv>p0 = +41,96 k.l • mol 1 From Section 6.22, we find the following relationship

&H,,U + & C PJ ( T 2 - T t) = A + 3M7, 5,° + (C PJ (I2,g) - CP,„°(12>I)) (T2 - T t)

=+41.96 kJ-mor'

+ (36.90 J • K~' • mol-1 -80.7 J - K '1 •m or')(2 9 8 K -4 7 5 .5 K ) = +49.73 kJ-mol'1 So, at 298K: +62.44 kJ ■ mol'1 - AH (J + 49.73 kJ • mol-1 AH tJ =+12.71 kJ-mo!''

7.97 First, we’ ll w7ork out how much heat was required to raise the temperature o f the water sample: 9 = gCAT = (150g) (4 .1 8 J .°C V )(5 .0 0 *C ) = 31353 Since both the water and ice samples were at 0.00° C and the water sample took 0.5 h to get to 5.00° C, we can estimate that the ice took 10.0 h to melt. If 3135 J o f heat were transferred in 0.5 h, then the amount o f heat transferred in 10.0 h is,

In document Área Jurídica, Social y Administrativa CARRERA DE ADMINISTRACIÓN DE EMPRESAS (página 142-154)