款器题金，板应为袋反应，“ 1=365×2×24h=17520h,xA=30% 所，-d”752m-20s8 1 1 1 6.解△，H(298.15K)=∑"aA,H9(B,298.15KN =[110.06-(-0.125)JkJ-mol-=110.19kJmo △，S9298.15K=∑"。△，SgB,298.15K) =(130.6+278.5-305.3]kJmo'=103.8Jmol1-K A,G9(298.15K)△，H9(298.15K)-T△，S9(298.15K) =(110.19-298.15×103.8×103kJ-mo'-74.29kJmo 由△，G2(298.15K)=-RInK(298.15K) InKe(298.15K)= 79.24×103 298.15×8.314 解得Ke(298.15K)-1.310x1014 由nK9(830.15K)=A,H282(29815K83015K)1 K9(298.15K) R 解得Ke(830.15K)=3.08×102 六、吸附速率为u=k,1-)p 脱附速率=k,82 当吸附达平衡时：U。=心4 k(1-0)2p=k01 令，b=无1k,则1-，由此得0中 根据题意，该反应为一级反应，t﹦ 1 A k ln 1 1 A − x t﹦365×2×24h﹦17520h , x A ﹦30% 所以 k A ﹦ 1 t ln 1 1 A − x ﹦ 1 17520h ln 1 1 0.3 − ﹦2.0358×10-5 (h-1 ) 将 kA 代入 A 1 8938 ln( ) 20.40 h / k T K − = − + , 得 T﹦286.45 K 6.解  r m H (298.15 K)﹦ B v  r m H (B,298.15 K) ﹦[110.06－(－0.125)]kJ·mol -1﹦110.19 kJ·mol-1 r m  S (298.15K) ﹦ B v r m  S (B,298.15 K) ﹦(130.6＋278.5－305.3)]kJ·mol-1﹦103.8 J·mol-1·K-1  r m G (298.15K)= (298.15K) (298.15K)  −  r m r m H T S ﹦(110.19－298.15×103.8×10-3 )kJ·mol-1﹦74.29 kJ·mol-1 由  r m G (298.15K)﹦－RTln K (298.15 K) ln K (298.15 K)﹦－ 3 79.24 10 298.15 8.314   解得 K (298.15 K)﹦1.310×10 -14 由 (830.15K) 1 1 (298.15K) ln ( ) (298.15K) 298.15K 830.15K K r m H K R  = − l 解得 K (830.15K)﹦3.08×10-2 六、吸附速率为 2 a a   = − k p (1 ) 脱附速率 2 d d   = k 当吸附达平衡时：  =  a d 2 2 a d k p k (1 ) − =   令， a d b k k = / , 则 bp(1－ )﹦ 2 ， 由此得  ﹦ 1 bp + bp