天津大學(xué)無機(jī)化學(xué)第五版習(xí)題答案

1、第1章化學(xué)反應(yīng)中的質(zhì)量關(guān)系和能量關(guān)系化學(xué)反應(yīng)中的質(zhì)量關(guān)系和能量關(guān)系習(xí)題參考答案習(xí)題參考答案1解：1.00噸氨氣可制取2.47噸硝酸。2解：氯氣質(zhì)量為2.9103g。3解：一瓶氧氣可用天數(shù)331111222()(13.2101.0110)kPa32L9.6d101.325kPa400LdnppVnpV?????????４解：pVMpVTnRmR??=318K℃44.9?５解：根據(jù)道爾頓分壓定律iinppn?p(N2)=7.6?104Pap

2、(O2)=2.0?104Pap(Ar)=1?103Pa６解：（1）0.114mol2(CO)n?2(CO)p?42.8710Pa?（2）222(N)(O)(CO)pppp???43.7910Pa??（3）4224(O)(CO)2.6710Pa0.2869.3310Panpnp?????7.解：（1）p(H2)=95.43kPa（2）m(H2)==0.194gpVMRT8.解：（1）?=5.0mol（2）?=2.5mol結(jié)論:反應(yīng)進(jìn)度(?

3、)的值與選用反應(yīng)式中的哪個物質(zhì)的量的變化來進(jìn)行計(jì)算無關(guān)，但與反應(yīng)式的寫法有關(guān)。9.解：U=Qp?pV=0.771kJ??10.解：（1）V1=38.3103m3=38.3L?（2）T2==320KnRpV2（3）?W=?(?p?V)=?502J（4）?U=QW=758J（5）?H=Qp=1260J11.解：NH3(g)O2(g)NO(g)H2O(g)=?226.2kJmol?145298.15K?????標(biāo)準(zhǔn)態(tài)23?mrH?12.解：=

4、Qp=?89.5kJmrH?=??nRTmrU?mrH?=?96.9kJ13.解：（1）C(s)O2(g)→CO2(g)3.解：=?70.81kJmol?1=?43.2Jmol?1K?1=?43.9kJmol?1?mrH??mrS??mrG?（2）由以上計(jì)算可知：(298.15K)=?70.81kJmol?1(298.15K)=?43.2Jmol?1K?1?mrH??mrS?=?T≤0?mrG??mrH??mrS?T≥=1639KK)(

5、298.15K)(298.15mrmr??SH??4.解：（1）==cK??O)H()(CH)(H(CO)2432ccccpK??O)H()(CH)(H(CO)2432pppp=?K????????????ppppppppO)H()(CH)(H(CO)2432（2）==cK????)(NH)(H)(N3232212cccpK????)(NH)(H)(N3232212ppp=?K???????pppppp)(NH)(H)(N3232212

6、（3）==cK)(CO2cpK)(CO2p=?K?pp)(CO2（4）==cK????3232)(HO)(HccpK????3232)(HO)(Hpp=?K????3232)(HO)(H??pppp5.解：設(shè)、基本上不隨溫度變化。?mrH??mrS?=?T?mrG??mrH??mrS?(298.15K)=?233.60kJmol?1?mrG?(298.15K)=?243.03kJmol?1?mrG?(298.15K)=40.92故(29

7、8.15K)=8.3?1040?Klg?K(373.15K)=34.02故(373.15K)=1.0?1034?Klg?K6.解：（1）=2(NH3g)=?32.90kJmol?1＜0?mrG??mfG?該反應(yīng)在298.15K、標(biāo)準(zhǔn)態(tài)下能自發(fā)進(jìn)行。（2）(298.15K)=5.76(298.15K)=5.8?105?Klg?K7.解：（1）(l)=2(NOg)=173.1kJmol?1?mrG??mfG?==?30.32故=4.8?10