·3.超额焓HE(excess enthalpy): HE=△mixHre一△mix Hid (6) ·对于理想溶液：△mixHid=-O,故有： ● HE=AmixHre=-T2((GE/T)/OT)p ● =-T2(@(RTXn;lmy;/T)/OT)p HE=-RT2Xn;(olnyi /OT)p (• 3. 超额焓HE(excess enthalpy): • HE=mixHre－mixHid (6) • 对于理想溶液: mixHid=0,故有: • HE=mixHre=－T2 ((GE/T)/T)p • =－T2 ((RT∑ni lni /T)/T)p • HE=－RT2∑ni (lni /T)p (7)