(第二版)趙士鐸主編習題答案】w(h2o2)?3.0%?1000g?l-1-1c(h2o2)???0.88mol?lm(h2o2)34g?mol-1b(ho)?w(h2o2)/m(h2o2)?3.0%/0.034kg?mol?0.91mol?kg-11?w(h2o2)x(h2o2)?n(h2o2)b(h2o2)??1.6%n(h2o2)?n(h2o)b(h2o2)?1/m(h2o)b=1.17mol?kg-1?tb=kbb=0.52k?kg?mol-1?1.17mol?kg-1=0.61ktb=?tf=kfb=1.86k?kg?mol-1?1.17mol?kg-1=2.18ktf=270.87km/mrtvmrt0.40g?8.31kpa?l?mol-1?k-1?300km???2.0?104g?mol-1v?0.10l?0.499kpa化合物中c、h、o原子數比為21：30：2kfmb5.12k?kg?mol-1?0.100gm???310g?mol-1?tbma0.33k?5.00g故該化合物的化學式為c21h30o21.8m{(nh2)2co}/m{(nh2)2co}mb/mb?m(h2o)m(h2o)?mb?342.4g?mo-1l(7)錯誤8）錯誤(1/2)n2(g)+(1/2)o2(g)=no(g)??fhm?(no,g)=(1/4){?rhm?(1)-?rhm?(2)}=(1/4)[-1107kj?mol-1-(-1150kj?mol-1)]=90kj?mol-12.3（1/4）[反應式（3）-反應式(4)+3?反應式(2)-n2(g)+2h2(g)=n2h4(l)?????(5)??fhm?(n2h4,,g)=(1/4){?rhm?(3)-?rhm?(4)+3??rhm?(2)-?rhm?(1)}=(1/4){-143kj?mol-1-(-286kj?mol-1+3?(-317kj?mol-1)-(-1010kj?mol-1))=50.5kj?mol-12?反應式（4）-反應式（5）得：n2h4(l)+)o2(g)=n2(g)+2h2o(l)?rhm?=2??rhm?(4)-?rhm?(5)=2?(-286kj?mol-1)-50.5kj?mol-1=-622.5kj?mol-12.4?rhm?=2?fhm?(co2,g)+3?fhm?(h2o,l)+(-1)??fhm?(ch3och3,l)+(-3)?fhm?(o2,g)??fhm?(ch3och3,l)=2?fhm?(co2,g)+3?fhm?(h2o,l)-?rhm?=-183kj?mol-12.5co(g)+(1/2)o2(g)知，?rhm?(1)0,?rhm?(2)0,?rhm?(3)0??rhm?(1)=?rhm?(2)+?rhm?(3)?rhm?(1)-?rhm?(3)=?rhm?(2)0即：以碳直接作燃料時放熱較多2.6c(s)+h2o(g)=co(g)+h2(g)?rhm?=?fhm?(co,g)+(-1)?fhm?(h2o,g)=-110.5kj?mol-1-(-)241.8kj?mol-1=131.3kj?mol-1co2(g)+h2o(g)?rhm?(2)?rhm?(3)co(g)+h2(g)+o2(g)??rhm?(1)=?rhm?(2)+?rhm?(3)??rhm?(1)-?rhm?(3)=?rhm?(2)02.10不可以。因為物質的sm?,不是指由參考狀態的元素生成該物?h?:-40kj-40kj?g?:-38kj-38kj?s?:-6.7j?k-1-6.7j?k-12.12sn(白錫)=sn(灰錫)?rhm?(298k)=?fhm?(灰錫)+（-1）??fhm?(白錫)=-2.1kj?mol-10?rsm?(298k)=sm?(灰錫)-sm?(白錫)=-7.41j?mol-1k-10??rgm?(t)??rhm?(298k)-t?rsm?(298k)t?[?rhm?(298k)/?rsm?(298k)]?283k2.132fe2o3(s)+3c(s)=4fe(s)+3co2(g)?rhm?(298k)=3?fhm?(co2,g)+(-2)??fhm?(fe2o3,s)=467.87kj?mol-1?rsm?(298k)=4?sm?(fe,s)+3sm?(co2,g)+(-2)?sm?(fe2o3,s)+(-3)?sm?(c,s)=557.98j?mol-1?k-1反應在標準狀態下自發進行：?rgm?(t)??rhm?(298k)-t?rsm?(298k)t[?rhm?(298k)/?rsm?(298k)],即t839k2cuo(s)+c(s)=2cu(s)+co2(g)?rhm?(298k)=-79.51kj?mol-10?rsm?(298k)=189j?mol-1?k-10?反應在標準狀態,任意溫度均2.152ash3(g)=2as(s)+3h2(g)?rhm?(298k)=-132.8kj?mol-10?rsm?(298k)=15.71j?mol-1?k-10?標準狀態,任意溫度下ash3的分解反應均自發。加熱的目的是加3.1（1）正確2）錯誤3）錯誤。{c(fe2?)/c?}{p(h2)/p?}k?{c(h?)/c?}2?{k?(3)}4k?{k?(1)}2{k?(2)}2?nrt0.20mol?8.31kpa?l?mol-1?k?1?698kp???116kpav10ln(i2)?n(h2)?c(i2)v?0.0021mol?l-1?10l?0.021mol0.021mol?116kpa?12.2kpa0.20molp(hi)?p?p(i2)?p(h2)?91.6kpap(i2)?p(h2)?x(i2)?p?3.5pv=nrt故對于反應h2(g)+i2(g)=2hi(g){p(hi)/p?}2k??56{p(h2)/p?}{p(i2)/p?}?c(o2)/c?k?p(o2)/p??1.38?10-3mol?l-1/1mol?l-1k(293k)??1.37?10?3101kpa/100kpa?o2(g)=o2(aq)c(o2)/c?k(293k)??1.37?10?3?【篇二：同濟大學普通化學第一章、二章習題答案(詳txt>第一章化學反應的基本規律(習題p50-52）16解（1）h2o(l)==h2o(g)?1?fh?m/kj?mol?285.83?241.82?1?1s?/j?mol?k69.91188.83m?1?rh?=44.01kj?mol?1m(298k)=[?241.82?(?285.83)]kj?mol?1?1?1?1?rs?(298k)=(188.83?69.91)j?mol?k=118.92j?mol?kmw=?p??v=?nrt=?2?8.315j?k?1?mol?1?298k=?4955.7j=?4.956kj(或?4.96kj)∴?u=qp+w=88.02kj?4.96kj=83.06kj17解（1）n2（g）+2o2（g）==2no2(g)?1?fh?0033.2m/kj?mol?1?1s?191.6205.14240.1m/j?mol?k?1∴?rh??2=66.4kj?mol?1m(298k)=33.2kj?mol?1?1?1?1?1?1?rs?m(298k)=(240.1j?mol?k)?2?(205.14j?mol?k)?2?191.6j?mol?k=?121.68j?mol?1?k?1(2)3fe(s)+4h2o(l)==fe3o4(s)+4h2(g)?1?fh?0?285.83?1118.40m/kj?mol?1?1s?m/j?mol?k27.369.91146.4130.68?1?1∴?rh?m(298k)=[?1118.4?(?285.83?4)]kj?mol=24.92kj?mol?1?1?rs?j?mol?k=(669.12?361.54)j?mol?1?k?1=307.58j?mol?1?k?1?1?fh?(298k)/kj?mol?824.2m?1?1s?m(298k)/j?mol?k87.45.7427.3?1?fg?(298k)/kj?mol?742.2m??∵?rg?m=?rhm?t??rsm∴301.32kj?mol?1=467.87kj?mol?1?298k??rs?m?1?1∴?rs?m=558.89j?mol?k?1?1?1?1?1?1?∴?rs?=3s(co(g)298k)+27.3j?mol?k?4?87.4j?mol?k?2?5.74j?mol?k?32mm?1?1?1?1∴s?m(co2(g)298k)=1/3(558.89+192.02?109.2)j?mol?k=213.90j?mol?k??fh?m(298k,c(s,石墨))=0?fgm(298k,c(s,石墨))=0??fh?m（298k,fe(s)）=0?fgm（298k,fe(s)）=0???rh?m=3?fhm(298k,co2(g))?2?fhm(298k,fe2o3(s))?1?467.87kj?mol?1=3?fh?m(298k,co2(g))?2?(?824.2kj?mol)?1?1∴?fh?m(298k,co2(g))=1/3(467.87?1648.4)kj?mol=?393.51kj?mol??同理?rg?m=3?fgm(298k,co2(g))?2?fgm(298k,fe2o3(s))?1?301.32kj?mol?1=3?fg?m(298k,co2(g))?2?(?742.2kj?mol)m(298k,co2(g))=1/3(301.32?1484.4)kj?mol=?394.36kj?mol19.解6co2(g)+6h2o（l）==c6h12o6(s)+6o2（g）?1?fg?m(298k)?kj?mol?394.36?237.18902.90?1?1∴?rg?m(298k)=[902.9?(?237.18?6)?(?394.36?6)]kj?mol=4692.14kj?mol0所以這個反應不能自發進行。20．解(1)4nh3(g)+5o2(g)==4no(g)+6h2o(l)?1?fg?(298k)/kj?mol?16.4086.57?237.18m?1?1∴?rg?m(298k)=[(?237.18)?6+86.57?4?(?16.4)?4]kj?mol=?1011.2kj?mol0∴此反應能自發進行。(2)2so3(g)==2so2(g)+o2(g)?1?fg?m(298k)/kj?mol?371.1?300.190?1?1∴?rg?(298k)=[(?300.19)?2?(?371.1)?2]kj?mol=141.82kj?mol0m21．解（1）mgco3(s)==mgo(s)+co2(g)?1?fh?(298k)/kj?mol?1111.88?601.6?393.51m?1?1s?m(298k)/j?mol?k65.627.0213.8?1?fg?(298k)/kj?mol?1028.28?569.3?394.36mm(298k)=[?601.6+(?393.51)?(?1111.88)]=116.77kj?mol?1?1?rs?m(298k)=[213.8+27.0?65.6]=175.2j?mol?k?1?rg?m(298k)=[(?394.36)+(?569.3)?(?1028.28)]=64.62kj?mol?1?1?1??(2)?rg?m(1123k)=?rhm(298k)?t??rsm(298k)=116.77kj?mol?1123k?175.2j?mol?k=116.77kj?mol?1?196.75kj?mol?1=?79.98kj?mol?1又∵rtlnk?(1123k)=??rg?m(1123k)∴8.315j?mol?1?k?1?1123k?lnk?(1123k)=?(?79.98)kj?mol?1m(t)=?rhm(298k)?t??rsm(298k)=0116.77kj?mol?1t???666.5k??1?1?rsm(298k)175.2j?mol?k?rhm(298k)?22.解法一:k?(298k)=5.0?1016??1?116?1∴?rg?(298k)=?rtlnk(298k)=?8.315j?mol?k?298k?ln(5.0?10)=?95.26kj?molm??∵?rg?m(298k)=?rhm(298k)?298k??rsm(298k)∴?95.26kj?mol?1=?92.31kj?mol?1?298k??rs?m(298k)?1?1∴?rs?m(298k)=9.90j?mol?k??∴?rg?(500k)=?h(298k)?500k??srrmmm(298k)=?92.31kj?mol?1?500k?9.90j?mol?1?k?1=?97.26kj?mol?1??1?1?而?rg?m(500k)=?rtlnk(500k)=?8.315j?mol?k?500k?lnk(500k)??rgm(500k)97.26?103j?mol?1∴lnk(500k)=?==23.40?1?1rt8.315j?mol?k?(500k)????rhm(298k)11k?(500k)?)?(∵ln?=500298rk(298k)?202?(?92.31?103j?mol?1))k=?15.05?(=500?2988.315j?mol?1?k?1k?(500k)∴?=2.9?10-7k(298k)?1?fh?(298k)/kj?mol00?45.9m?1?1s?m(298k)/j?mol?k191.6130.68192.8?1?1∴?rh?m(298k)=2?(?45.9)kj?mol=?91.8kj?mol?1?1?1?1s?m(298k)=(2?192.8?191.6?3?130.68)j?mol?k=?198.04j?mol?k???rg?m(t)=?rhm(298k)?t??rsm(298k)=0=?91.8kj?mol?1?t?(?198.04j?mol?1?k?1)=0?91.8?103j?mol?1∴t==463.5k?198.04j?mol?1?k?124.解1）?（2）得：co2(g)+h2(g)?co(g)+h2o(g)平衡常數為:???根據?rg?(t)=?rtlnk(t)=?h(t)?t??srrmmm(t)???rhm?rsm?)lnk(t)??(rtr????rhmk1(t)11假設?rh(t),?rs(t)不變，則ln???(?)rt1t2k2(t)?m?m)習題答案w(h2o2)?3.0%?1000g?l-1-1c(h2o2)???0.88mol?lm(h2o2)34g?mol-1b(ho)?w(h2o2)/m(h2o2)?3.0%/0.034kg?mol?0.91mol?kg-11?w(h2o2)x(h2o2)?n(h2o2)b(h2o2)??1.6%n(h2o2)?n(h2o)b(h2o2)?1/m(h2o)b=1.17mol?kg-1?tb=kbb=0.52k?kg?mol-1?1.17mol?kg-1=0.61ktb=?tf=kfb=1.86k?kg?mol-1?1.17mol?kg-1=2.18ktf=270.87km/mrtvmrt0.40g?8.31kpa?l?mol-1?k-1?300km???2.0?104g?mol-1v?0.10l?0.499kpa化合物中c、h、o原子數比為21：30：2kfmb5.12k?kg?mol-1?0.100gm???310g?mol-1?tbma0.33k?5.00g故該化合物的化學式為c21h30o21.8m{(nh2)2co}/m{(nh2)2co}mb/mb?m(h2o)m(h2o)?mb?342.4g?mo-1l(7)錯誤8）錯誤2.2（1/4）[反應式（1）-反應式(2)]得：(1/2)n2(g)+(1/2)o2(g)=no(g)??fhm?(no,g)=(1/4){?rhm?(1)-?rhm?(2)}=(1/4)[-1107kj?mol-1-(-1150kj?mol-1)]=90kj?mol-12.3（1/4）[反應式（3）-反應式(4)+3?反應式(2)-n2(g)+2h2(g)=n2h4(l)?????(5)??fhm?(n2h4,,g)=(1/4){?rhm?(3)-?rhm?(4)+3??rhm?(2)-?rhm?(1)}=(1/4){-143kj?mol-1-(-286kj?mol-1+3?(-317kj?mol-1)-(-1010kj?mol-1))=50.5kj?mol-12?反應式（4）-反應式（5）得：n2h4(l)+)o2(g)=n2(g)+2h2o(l)?rhm?=2??rhm?(4)-?rhm?(5)=2?(-286kj?mol-1)-50.5kj?mol-1=-622.5kj?mol-12.4?rhm?=2?fhm?(co2,g)+3?fhm?(h2o,l)+(-1)??fhm?(ch3och3,l)+(-3)?fhm?(o2,g)??fhm?(ch3och3,l)=2?fhm?(co2,g)+3?fhm?(h2o,l)-?rhm?=-183kj?mol-12.5co(g)+(1/2)o2(g)知，?rhm?(1)0,?rhm?(2)0,?rhm?(3)0??rhm?(1)=?rhm?(2)+?rhm?(3)?rhm?(1)-?rhm?(3)=?rhm?(2)0即：以碳直接作燃料時放熱較多2.6c(s)+h2o(g)=co(g)+h2(g)?rhm?=?fhm?(co,g)+(-1)?fhm?(h2o,g)=-110.5kj?mol-1-(-)241.8kj?mol-1=131.3kj?mol-1co2(g)+h2o(g)?rhm?(2)?rhm?(3)co(g)+h2(g)+o2(g)??rhm?(1)=?rhm?(2)+?rhm?(3)??rhm?(1)-?rhm?(3)=?rhm?(2)02.10不可以。因為物質的sm?,不是指由參考狀態的元素生成該物?h?:-40kj-40kj?g?:-38kj-38kj?s?:-6.7j?k-1-6.7j?k-12.12sn(白錫)=sn(灰錫)?rhm?(298k)=?fhm?(灰錫)+（-1）??fhm?(白錫)=-2.1kj?mol-10?rsm?(298k)=sm?(灰錫)-sm?(白錫)=-7.41j?mol-1k-10??rgm?(t)??rhm?(298k)-t?rsm?(298k)t?[?rhm?(298