人卫版物理化学(第六版)课后习题答案详解 下载本文

?G2?0?G3?nRTln?G4?0?G5??Vdp?0?G??G1??G2??G3??G4??G5??326.7J18.计算下列恒温反应的熵变化:

298Kp22.28?1?8.314?268ln??326.7Jp12.64

??C2H6(g) 2C(石墨)+3H2(g)??已知25 oC时的标准熵如下:C(石墨)5.74J?K?1?mol?1;H2130.6J?K?1?mol?1;

C2H6229.5J?K?1?mol?1。

??C2H6(g) 解:2C(石墨)+3H2(g)???????rSm??SC?2S?3SHCH262298K?229.5?2?5.74?3?130.6??173.78J?K19.计算下列恒温反应(298K)的??Gm:

??1

KC6H6(g)?C2H2(g)?298???C6H5C2H3(g)

已知25 oC时C6H5C2H3的?fHm?147.36kJ?K解:由附表查得:C6H6?g?:

??1??mol?1,Sm?345.1J?K?1?mol?1

?fH298?82.93kJ?mol?1S298?269.2J?K

??1?mol?1

C2H2?g?:

?fH298?226.7kJ?mol?1S298?200.8J?K?mol??1?1?????rHm??fHC??H??H?147.36??82.93?226.7???162.27kJHCHfCHfC6523662H2?????1???rSm?SC?S?S?345.10?269.2?200.8??124.9J?KC6H6C2H26H5C2H3????rGm??rHm?T?rSm??162.27?298???124.9???125kJ????

20.25 oC、101.325kPa时,金刚石与石墨的规定熵分别为2.38J?K其标准燃烧热分别为?395.4kJ?mol和?393.5kJ?mol?值,并说明此时哪种晶体较为稳定。 ??Gm?1?1?1?mol?1和5.74J?K?1?mol?1;

。计算在此条件下,石墨?金刚石的

解:

????rHm??CH石墨??CH金刚石??393.5?395.4?1.9kJ????rSm?S金刚石?S石墨?2.38?5.74??3.36J?K?1????rGm??rHm?T?rSm?2898J

21.试由20题的结果,求算需增大到多大压力才能使石墨变成金刚石?已知在25 oC时石墨和金刚石

的密度分别为2.260?10kg?m和3.513?10kg?m。

3?33?3C石墨?25C,p????C金刚石?25OC,p?O?G'解:??G1

?GC石墨?25OC,101.325kPa????C金刚石?25OC,101.325kPa?设25 oC,压力为p时,石墨和金刚石正好能平衡共存,则

?G'?0?G1??V1dp?V1P??P

P?G2????Vdp?V?P?P?PP??P?22?G'??G1??G2??Gp?p????G??V2?V1?G?11??M??????1??2?1527.2?106Pa

p?1527.1?106Pa

22.101325Pa压力下,斜方硫和单斜硫的转换温度为368K,今已知在273 K时,S(斜方)? S(单斜)的?H?322.17J?mol,在

?1273~373K之间硫的摩尔等压热容分别为

Cp.m?斜方??17.24?0.0197TJ?K?1?mol?1;Cp.m?单斜??15.15?0.0301TJ?K?1?mol?1;试求

(a)转换温度368K时的?Hm;(b)273K时转换反应的?Gm

?Cp?15.15?0.0301T??17.24?0.0197T???2.09?0.0104T368K:?rHm??Hm,273???CpdT?322.17??273368368273??2.09?0.0104T?dT?446.93J?mol?1解:273K:

?H368446.93?rS368???1.214J?K?1?mol?1T368273?Cp273??2.09?0.0104T??rS273??S368??dT??S368??dT368T368T273?1.214???2.09?ln?0.0104??273?368??0.85J?K?1?mol?1368?rGm??rHm,273?T?rS273?322.17?273?0.85?90.12J?mol?1

23.1mol水在100oC、101.3KPa恒温恒压汽化为水蒸气,并继续升温降压为200oC、50.66KPa,求整

-3-1-10

个过程的△G。(设水蒸气为理想气)。已知CP,H2O=30.54+10.29×10TJ﹒Kmol,SH2O(=188.72 (g)(g)298K)

-1-1

J﹒Kmol

解:1mol 1mol 1mol

100 oC 100 oC 200oC 101.3KPa △G1 101.3KPa △G2 50.66KPa H2O(l) H2O(g) H2O(g) △S △ G1=0 △

S3=

?nCP,MTdT

30.54?10.29?10?3dT=6.856+0.772 =?298T373=7.628J﹒K

00-1S373=7.628+ S298=7.254+1.029+5.76=14.04 J﹒K

-1

S473=196.35+14.04=210.39 J﹒K

△G2=△H2-△(TS)=-23.676KJ △G=△G1+△G2=-23.676J

24.计算下述化学反应在101.325KPa下,温度分别为298.15K及398.15K时的熵变各是多少?设在该温度区间内各CP,M值是与T值无关的常数。

000

C2H2(g,P)+2H2(g,P)=C2H6(g,P)

0-1-1

已知:Sm(J﹒Kmol) 200.82 130.59 229.49

-1-1

CP,M(J﹒Kmol) 43.93 28.84 52.65

0-1

解:△rS298.15=229.49-2×130.59-200.82=-232.51J﹒K △S

0

r398.15

-1

=△rS298.15+

0

?P-1

?298.15TdT=-232.51-20.12×0.2892=-238.3J﹒K

398.1525.反应CO(g)+H2O(g)=C2O(g)+H2(g),自热力学数据表查出反应中各物质△fHm,Sm,及CP,M,

000

求该反应在298.15K和1000K时的△fHm,△fSm和△fGm。 解:各物质热力学数据如下表: 数据 -1-1-100

CO(g) H2O(g) -241.818 188.825 31.80 4.47 5.10 CO2(g) -393.509 213.74 22.59 56.15 -24.85 H2(g) 0 130.684 28.45 1.2 0.42 △H(J﹒Kmol) -110.525 S(J﹒Kmol) 197.674

a×(J﹒Kmol) 3-1-1 28.70 -1 0.14 b×10×(J﹒Kmol) -1 4.64 c×10×(J﹒Kmol) 6-1-1

-1

△rH298=-393.509-(-110.525-241.818)=-41.14KJmol

-1-1

△rS298=213.74+130.684-197.674-188.825=-42.05J﹒Kmol

3-1

△rG298=-41.17×10+298.15×42.08=-28.62J﹒Kmol

-1-1

△a=-9.46J﹒Kmol

-3-2-1

△b=52.94×10J﹒Kmol

△c=-34.17×10J﹒Kmol

-1

△rH1000=-34.87J﹒Kmol

-1-1

△rS1000=-32.08J﹒Kmol

-1

△rG1000=-2.79J﹒Kmol

26.指出下列式子中哪个是偏摩尔量,哪个是化学势?

-6

-3

-1

??

??A??; ???ni?T,P,nj??G??; ??n???i?T,V,nj??H???n?i??; ??T,P,nj??U???n?i??; ??S,V,nj??H???n?i???V??;???S,P,nj??ni???A???;; ???n???T,P,nj?i?T,V,nj???V??;???T,P,nj??ni??; ??T,P,nj解:偏摩尔量:

????A???H??; ????ni?T,P,nj??ni

化学势:

??U???n?i

???H??;???S,V,nj??ni???A???;; ???n???S,P,nj?i?T,V,nj27.对遵从范德华气体方程(P+证明:?a)(v-b)=nRT的实际气体。 v2a??U???(2)

V??V?T证明:Du=TdS-PdV

??U???S????T???P 由dA= -SdT-pdV ??V?T??V?T=T???P???S???P???P (1) 得?? ?=???T?V??V?Tj??T?V(P+

a)(v-b)=nRT两边对T微分 v2(v-b)???p??=nR ??T?VjnRTaa??U??P?P?2?P?2 ??VV??V?TV?b将上式代入(1)则?

??U???H???????V?S??P?S28.对理想气体,试证明:??nR

?U??????S?V?pv??H???U???U?证明:由麦克斯韦关系得:? ??V,??????P,??P?ST?V??S??s??V理想气体:PV=nRT

??U???H??????V?P?pv??S??S 所以 ??nR ??nR,则

T??U????S??V

???A?????T????????U

29.试导出亥姆霍兹能A的吉布斯-亥姆霍兹公式,即:?T2??T?????V证明:??A??U?A?U???A???() (1) ????S??TTTT??V1???A??A?U???2?2 T??T?VTT 由(1)得:

???A?????T????????U

所以?T2??T?????V

30.有一个水和乙醇形成的溶液,水的物质的量分数为0.4,乙醇的偏摩尔体积为57.5cm3mol-1,溶液的密度为0.8494KgL-1,求此溶液中水的偏摩尔体积。 解:设水的物质的量 n=0.4,则乙醇物质的量为n=0.6

0.4?18?10?3?0.6?46?103?103 0.6?57.5+0.4?0.8494VH2O=16.175cm3mol-1

31.25oC,n摩尔NaCl溶于1000g水中,形成溶液体积V和n之间关系可表示如下: V(cm)=1001.38+16.625n+1.7738n+0.1194n2 试计算1mNaCl溶液中H2O及NaCl的偏摩尔体积。

3

32