=-241.818k △n(g)=-0.5mol △U1=△H1-△n(g)RTi =-241.818+0.5×8.3145×298.15×10-3kJ =-240.579kJ 过程(2)为反应产物的恒容升温过程： △U2=2nBCv,m(B,产物)(t-t1) =In(HO,g)Cv.m(H2O,g)+n(O)Cv.m(O)+n(N2)Cv.m(N2)I(t-t1) =(37.66+0.25×25.1+2.8214×25.1)(t℃-25)J =114.752×(t/℃-25)×10-3kJ △U=△U1+△U2=-240.579kJ+114.752×(t/℃-25)×10-3k=0 由上式可得最高温度 240.579 t=（114.752×10+25)℃=2121.5t 2nB(反应物，g)=n1(O2)+n(N2)+n(H2) =(0.75+2.8214+1)mol =4.5714mol nB(产物，g)=n2(O2)+n()+n(HO,g) =4.0714mol 因为p1V=∑nB(反应物，g)RT1,pV=∑nB(产物，g)RT s(产物.g)T 所以设物27×=49器×10 =715.3kPa 解法2：末态温度也可由AH来计算。 前已求出：△H1=-241.818k △，(pV)=∑nB(产物，g)RT-∑nB(反应物，g)RT1 =(4.0714×8.3145T/K-4.5714×8.3145×298.15)J =(33.852T/K-11332.36)×10-3k 因为△U=0 所以△，H=△，U+△.(V)=△，(pV)=△H1+△H2 AH2 =In(H2O,g)Cp.m(H2O)+n2(Oz)Cp.m(O2)+ n(N2)Cp.m(N2)(T-Ti) ={(37.66+8.3145)+(2.8214+0.25)×(25.1+ 8.3145)1(T/K-298.15J =148.604×(TK-298.15)J