無機(jī)化學(xué)課后習(xí)題答案(天津大學(xué)第四版)

1、第2章化學(xué)反應(yīng)的方向、速率和限度習(xí)題參考答案1.解：=?3347.6kJmol?1=?216.64Jmol?1K?1=?3283.0kJmol?1?mrH??mrS??mrG?＜0該反應(yīng)在298.15K及標(biāo)準(zhǔn)態(tài)下可自發(fā)向右進(jìn)行。2.解：=113.4kJmol?1＞0?mrG?該反應(yīng)在常溫(298.15K)、標(biāo)準(zhǔn)態(tài)下不能自發(fā)進(jìn)行。（2）=146.0kJmol?1=110.45Jmol?1K?1=68.7kJmol?1＞0?mrH??mrS

2、??mrG?該反應(yīng)在700K、標(biāo)準(zhǔn)態(tài)下不能自發(fā)進(jìn)行。3.解：=?70.81kJmol?1=?43.2Jmol?1K?1=?43.9kJmol?1?mrH??mrS??mrG?（2）由以上計算可知：(298.15K)=?70.81kJmol?1(298.15K)=?43.2Jmol?1K?1?mrH??mrS?=?T≤0?mrG??mrH??mrS?T≥=1639KK)(298.15K)(298.15mrmr??SH??4.解：（1）==

3、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（3）==cK)(CO2cpK)(CO2p=?K?pp)(CO2（4

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?==0.070?K????S)(H)(NH23??pppp則=0.26?100kPa=26kPax平衡時該

5、氣體混合物的總壓為52kPa（2）T不變，不變。?KNH4HS(s)?NH3(g)H2S(g)平衡分壓kPa25.3yy==0.070?K????)25.3??pypy?（=17kPay13.解：（1）PCl5(g)?PCl3(g)Cl2(g)平衡濃度／(molL?1)0.250.070.0?0.250.00.250.0==0.62molL?1，(PCl5)=71%cK)PCl()Cl()PCl(523ccc?PCl5(g)?PCl3(

6、g)Cl2(g)平衡分壓0.200.50.5VRTVRTVRT=?K?????????pppppp)(PCl)(Cl)(PCl523=27.2（2）PCl5(g)?PCl3(g)Cl2(g)新平衡濃度／(molL?1)0.100.25?0.25yyy?210.0=molL?1=0.62molL?1(T不變，不變)cK)10.0()30.0)(25.0(yyy???cK=0.01molL?1，(PCl5)=68%y?（3）PCl5(g)?

7、PCl3(g)Cl2(g)平衡濃度／(molL?1)0.050z?35.0zz==0.62molL?1cKzzz??35.0)050.0(=0.24molL?1，(PCl5)=68%z?比較(2)、(3)結(jié)果，說明最終濃度及轉(zhuǎn)化率只與始、終態(tài)有關(guān)，與加入過程無關(guān)。14.解：N2(g)3H2(g)?2NH3(g)平衡濃度／(molL?1)1.00.500.50=cK????32223)H()N()NH(ccc=21)Lmol(0.2??若