【用戶】高昇賢
【年級】小三上
【評論內容】dlnKp° = (△H° / RT2) dT∫Kp2Kp1 dlnKp° = ∫T2T1 (△H° / RT2) dTlnKp2° - lnKp1° = (△H° / R) 〔(1/T1) -(1/T2)〕△G = △G° + R·T·lnQ = 0平衡狀態下△G=0平衡常數定義 K = QK(T) = e(-△G°/RT)Kp,298°= e(-4730/(8.314*298))= 0.1482ln(Kp,600°/0.1482) = (57200 / 8.314) 〔(1/298) -(1/600)〕= 11.62Kp,600° = 1.65×10...