解：△，H(298K)=∑aYgA,H(B,298.15K) =(-84.67-52.28+0)kJmol-1 =-136.95 kJ-mol-1 △，SR(298.15K)=∑BVgS8B,298.15K) =(229.6-219.6-130.7)Jmol-1K-1 =-120.7Jmol-1K-1 △，GR(298.15K)=A,HR(298.15K)-T△，S(298.15K) =-136.95 kJ-mol-1-298.15K×(-120.7×10-3 kJmol-1.K-1) =-100.96kJmo1 K2815K)-er- G(298.15K) RT -exp- -100.96×103 =4.88×1017 .314×298.15解： (298 ) (B,298.15K) O B B f m O DrHm K = å n D H ＝（－84.67－52.28＋0）kJ·mol－1 ＝－136.95kJ·mol－1 (298.15K) (B,298.15K) O B B m O DrSm = å n S ＝（229.6－219.6－130.7） J·mol－1 ·K－1 ＝－120.7 J·mol－1 ·K－1 (298.15K) (298.15K) (298.15K) O r m O r m O DrGm =D H -TD S ＝-136.95kJ·mol-1－298.15K×(-120.7 ×10-3 kJ·mol-1·K-1) ＝-100.96 kJ·mol-1 ( ) ( ) 17 3 O O r m 4.88 10 8.314 298.15 100.96 10 exp 298.15K 298.15K exp = ´ ÷ ÷ ø ö ç ç è æ ´ - ´ = - ÷ ÷ ø ö ç ç è æ D - RT G K =