generalchemistry精品教學（王錦芝）chapter6-answer

1、chapter 6 rreaction heat, direction and extent of chemical reactions習題2、(1) au = q + w = -2.5 kj 500 j = -3.0 kj(2) au 二 q + w 二-650 j + 350 j = -300 j4、(1) rh = 40v - mop1(2) =-2x= 80v - mop1(3) .hm3 = g, + w 二 20v mor5、略6、arh/298.15:29&15k下化學反應的標準摩爾焙變。氣態(tài)水的標準摩爾生成焰。a戶/(比,g)：氫氣的標準摩爾燃燒焰。s/,298.15(

2、h2，g)： 298.15k下氫氣的標準摩爾嫡。亠s/八溫度t下化學反應的標準摩爾爛變。gk溫度t下化學反應的標準摩爾吉布斯自由能變。 /g/(cq,g)：co2氣體的標準摩爾生成吉布斯自由能。k。：化學反應的標準平衡常數(shù)。7、(1)-(2)x6-(3)x3,得：c6h6(/) = 6c(gra) + 3h2(g)=49.2燈 mol'1因此反應 6c(gs) + 3/(g)二 c6/6(z)的、.h< = 49.2v - molx8、2a2/74(/) + n2o,(/) = 3弘(g) + 4/0(/)川/=3亠./7/(他，<?)+ 4亠/(/。/)-2亠碼氣他/，/

3、)- 月/(mo/)=3 x 0 + 4 x (-285.8aj m()r )-2x 50.633 mp/_1-9.16u- mol'=-l 253.6v mr;/-19、(1) x 3 - (2) - (3) x 2,得:6fw(s) + 6co2(g) t 6feo(s) + 6co(g)則有：月二4 = w-亠弘，2 - 2"九,3 二 _ 1 °燈.mol- 6q代:二 3xrgni ,1 - rgm ,2 - 2x ag,”，3 = gkj mol6曲4 =0 0=mm * 一 rhm * = j6 j. q mol t10、(1)、.h j =卜 f h

4、 f (ch 4, g)七片卜 f h f io?, g) 卜 f h j (ch 3oh ,1) =74.6燈-mol'') + 0 (239.2燈 mol) = 164.6v- mol'1j = s j (ch x g) +r (o“ g) _ s r (ch qh ,1)= 186.3八 k_ 加ol + 丄x205.2八 k" 加ol -126.8j-_1 movx =162aj k'x mol 2g, =,./於/ =164.6 燈加o/t 298.15kx 162. l/.k-i 加o/t=116.3v>0在298.15k的標準狀態(tài)下

5、此反應不能自發(fā)進行。(2)亠6/ =亠日/-必0/ =164.6 燈加-tx162.1 八 k" 加。廣*0t>1015k在標準狀態(tài)下此反應的溫度應高于1015k才能自發(fā)進行。11、=fh/(h2yg) + fhmco2,g)-fhmh2o,g)-afhltf(co,g) =0-393.5aj mor) 一(一241.8)r mol'1 一(一1 0.5kj - mol'1)=-41.23 mollarg/=a/gj(/f2,5) + a/g/(co2,5)-azg/(w2ag)-a/g/(cag)=0 - 394.4v mok') - (-228.6)

6、v -mz?/-1 -(-137.2v- mol')-28.6kj - mol4s/=s/(h2，g) + s/(co2，g)-s/(h2°，g)-s/(co,g)= 130.7j-/c_1 -/w?/-1 +213.8j-morx -188.8j k' mo' 一9=j k_' mol= -42丿kt mol''12、 (1)arg/=a/g/(/2o,)-a/g/(h2)-|a/g/(o2)=-228.6aj mol- -0-0 = 228.62 mol < 0 反應正方向自發(fā)。-22&ke=ee 8.314/kt.廠

7、】x29&i5k := l.lxlo40 g & =my152kj mo' >0,反應逆方向自發(fā)。k& =2.1x10® g & =r m-505.2zrj -mor < 0 ,反應正方向自發(fā)。k“=3.2xl0 g & 二 r一344.8燈 m()rx < 0 ,反應正方向自發(fā)。k&=2.6xl0"(5) g 0 =r m-227.8kj - mol <0,反應正方向自發(fā)。k°=8.2x10'9136300r 250gx3.0 燈gj50gxl2rgt'50.0glx

8、l5.63gt'= 6.35 厶15、(1) co?(g) + h2(g) = co(g) + h2o(g)a：/ =0.14(2) coo(s) + w2() = co(5)+ h2o(g) k/ = 67-(1),得：(3) coo(s) + co(g) = co(s) + co? (g)備 4.8x2反應(2)的 arg,/,823 = -/?tln k/ = -28.8v -mol'1反應(3 )的g/.823 = -rtln k； = -42.2u molxco(g)對coo(s)的還原能力比 出)更強。= -rtxk° =-/?tlnadp弘加訂atpq

9、ox 1()一3 y 1 nyin-3atp31.05h mol'1 = -8.314八 kx310.15k in:解得，atp = 1.8xlo_,1(/-ri)gm=rgm0 + rtlnql.oxlo-2-31.05燈加t +8.314丿k 加廠 x31o.15kxln'°xl° xl.oxlo-51.97p 加引 t17、(1) k°h+4cr2/ =a/.g/(m/i2) + a/g/(c/2) + 2a/.gj(h2o,/),gbfgf (mno2，s) 4bfg(,aq) 2'fg (cl 衛(wèi)q)=-228au mol'

10、;1 + 0 - 2x 237.1r wol + 465.1 - 0 + 2x 131.2r mol=25.2kj mo廣' > 0此反應不能自發(fā)進行。由 arcz/ =-rtlnk。得,k” =3.85x10rgm=rgltrtnq1 x 1(3) =25.2kj-mor +8.314八 k_ 加x298kxln 712.04x12.02=-11.8aj - molx < 0反應能自發(fā)進行。川/=2x4h/(so3，g) 2x4h/(so2，g)-4h/(o2，g)18、=2x (-395.7r morx )-2x (-296.81u- molx) 一 0 =一1978防

11、 mor由ina h &r mrf k&900k 一 972kj mo' /900k 800k、intf = 8.314/k"。川 900kx800k)k&9ook = 33-49、二兩)+ 6g/(ct®) sg/(agc/,s)=77.1v inol 3l2kj mol + 109.8v mo' = 55.7 kj- molx55jkj-mor = -8.341 j-加o x29&15kxln kk/agc/)二 k& = 1.75x10"°ga4&(o2，g) + »&

12、;(ch4,g)-6&(chqh,/)21 > = qkj -mor1 -50.5kj mo' + 166.6v -tnolx=116.1v -mor亠 h 加亠比(q,g) + d 比(chg)-亠比(chqhj)=ou mol'1 - 74.6v - mop1 + 239.2v mol'= 164.6kjmol-s：氣小 s：(ch加-s：(chqh,1)= -x205.2j- kj -mol'1 +186.3八 k' molx -126.8八 k' mol 2= 62aj kmorl"斗 1646s 沁 “015k,

13、£ 162aj kl morlkjexercises3.cocm + h2o二 2hc7(g) + co,g),g： = »g/(co2, g) + 2 x azg/(hc/, g) - sfgmc0cl2i g) -/)=-394.4v mol'1 +2x(-95.3v mol) - (-21 okj - mol'1) - (-237.1)v mov4.=-137.9u -mor14. 4n0(g)t 2m0(g)+。2(刃(1) aq= 39.56kj mol2no(g) + q(g) t 2no3(2) ,g/(2) = 69.702wldo the

14、following calculation (2)x2-(1), you get,2o2(g) t ano?(g)2n2o(g)-。2(g), or 2n2o(g) + 3o2(g) t 4no?(g)hence,arg/=2xarg/(2)-arg/(l)=2x (-69.70 kj- mol'' )-(-139.56 kj mo' ) = 0.16u-mo 5. q=、-hf=3.4kjmo'a血=31ww羽94.2j-/c'1 -morx6. according to the isothermal equation of chemical reac

15、tions,.gj =-rtlnk°, substitute the values into the equation,-5079r moll =-834ukt加lx298kxlnk&solve the equation, you get ke =7.92x10*the ke value is large, which means that the forward reaction for the production of methane is favorable, and studying this reaction as a means of methane production is worth pursuing.7. c51hms) + 80o?(g) = 57co,g) + 52比0arw/=57xa/w/(co2) + 52xa/w/(w2o,/)-a/w/(c57wi04o6,5)-80xazw/(o2) 335x1()4qmq=57x(393.52 mo') +