74
MNOBr = 110 g·mol-1 nNOBr = 1.10/110 = 0.0100 (mol) 2NOBr 2 NO + Br2
平衡时n/mol 0.0100(1-2? ) 0.0200? 0.0100? 平衡时总物质的量= 0.0100(1+? ) mol
pV30.0?1.00 (1) 0℃ n总=1?= 0.0134 (mol)
RT18.31?273 0.0100(1+? ) = 0.0132 ? = 0.320
p/p??4?330.0?4?0.3203 = 0.230 ??2?22(1?2?)(1??)0.360?1.320pNOBrppV35.0?1.00 (2) 25℃ n'总 =2?= 0.0141 (mol)
RT28.31?298K?p1=
pNOpBr22 0.0100(1+? ') = 0.0141 ? '=0.413
30.0?4?0.4133? Kp2== 2.31
0.1742?1.413 lgK?p2K?p1??rHmT?T1?(2) 2.30RT2?T1? ?rHm = 62 kJ·mol-1
75
(1) 此反应为吸热反应。 (2) lgK?p2K?p1??rHmT?T1?(2) 2.30RT2?T1??rHm1.001173?973?() lg3.00?10?22.30?8.31?10?31173?973? ?rHm = 166 kJ·mol-1
76
(C) 77
Kp =pCO2= 9.51 ? 10-3 ? 100 kPa 所以VCO2/V总 = pCO2/p总
所以体积分数为：(9.51 ? 10-1/100) ? 100 % = 0.951 % 78
Fe2++ Co3+ Fe3++ Co2+ ------------① K1= 1.1 ? 1018 Cu+ + Co3+ Cu2++ Co2+ ------------② K2= 1.6 ? 1025 ②-① Cu+ + Fe3+Cu2++ Fe2+
K21.6?10257
K == 1.5 ? 10?K11.1?1018 所以Fe3+可氧化Cu2+，而且反应相当完全。 79
根据反应式 Kp = pBe
又根据lgKp = -?rHm/2.303RT + c 两式对照可得?rHm/2.303R = 1.97 ? 104 K 所以?rH-1m= 377 kJ·mol 80
(B) 81
(C) 82
(1) ← (2) ← (3) ← (4) → 83
nCOC2l= 0.631 / 98.91= 0.00638 (mol) p0 = nRT / V= (0.00638 ? 8.31 ? 900) / 0.472 = 101 (kPa) COCl2(g) CO(g) + Cl2(g) 初始分压力/100 kPa p0 0 0 平衡分压力/100 kPa p0(1-? ) ? p0 ? p0 所以pt = p0 (1-? ) + ? p0 +? p0 = p0 (1+? )
所以? = (pt / p0 ) – 1 = (187.2 / 101) – 1 = 0.853 分解了 85.3 % pCOC2= lp0 (1-? ) = 0.997 ? 100 ? (1 - 0.878) = 12.2 (kPa) 84
(A) 85
(A) 86
(D) 87
(D) 88
(B)
设分解率为?