近代化学基础第四章习题答案

1、+O =H=?r m22 H+=HH=?222+2O=+r mH=+2H(g)=CH(g)的?rm24H 。?rm. H=HH H?rmrmr= +mr m NH NO N(g)和42422HNHH。?r m224H+NO(g)=+H为242422?rmH(NH,1)=H(NO)=H(HO=?fm2 4?fm2 4?fm2H=4H(HO)-2H(NH)-H(NO)?rmf=- =m2fm24fm2 4S +=+H2232 略 HG?2232f m， + - =GG 0?rmr m = C+O2G = G =0?fmr mC H O +O= +H 222H H O ）?fm 和HO(l)的H?2

2、2f mH=11H(HO)+12H(H(C H O )?rmfm2fm=+12- =2fm 有W W =mgh= n=m = nM +H=H2K =?2a +O= K =?b622 +H=H+O的K?2222K = K/ K = / 6=?a?b =+253G和K?rmK。? 的 G =H=?m5ff m的 G =H =?f m?3fmG K G )=/ = H =?rm?rm?rm H 1 18800011KK2ln( ) () 22.27603rmR T T8.314 298.15 800112K =?23 += K= ，?22H = K?m?rKK H 1 121ln( )R T Trm

3、12K112()4.97 298.15 K=?A+BC ccABccBA =kc c当c =kc c)ABAAB当c =kc) c)BAB/ 得/=2 =kcc2AB CHClCH+252 4Ek s k a12kE 1 1aR T Tln ( )2k112k 11()25.9 s 5 1k k s 212N N+O222 Ek/akAeEa,2/RTee(121300244800)/8.314967 15.362 4.69106(E E )/RT2221kAeEa,1/RT11 在k k=t t2/121 1( )R T TtkkEln ln21at21121 14 11tE Rln ( )

4、 8.314ln () J.mol 1111148 301 278tT T1a22K：? =+322 =+32 O+=+342123 (g) N (g) H (g)32222(p / p ) (p / p )3/2 1/2K HN22(p / p )3 BaCO (s)BaO(s) CO (g)32K = p?2 Fe O (s) 4H (g)3Fe(s) 4H O(g)3422K (p / p )4H OH22OVO2 522O23K? +O = 3220n=?p p? ?(p / p )(0.77/22KSO 29.53(p / p ) (p / p ) (0.23/ (0.615/22SOO22(s)的H S G ?f m3fmm=+32在H S G ；在G K在?rmrmr m?rm2查时H= S= ?.fmmG =的H= S =?m?m?ffmGH- -=?r m?mfS+ - = -=?rm GHS?m?r