(1) 在此 K?p= p(C2H5OH, g)/100 kPa
所以25℃乙醇的蒸气压力为0.085 ? 100 kPa = 8.5 kPa
??rHm42.2? (2) 在沸点Tb,?rGm = 0 Tb = = 349(K) ??0.121?rSm
42
(1) H2O(l, 100 kPa) H2O(g, 100 kPa)
? ?rGm = -228.59 - (-237.18) = 8.59 (kJ·mol-1) (2) H2O(l, 饱和蒸气压)H2O(g, 饱和蒸气压) ?rGm= 0 (此时为平衡态)
??H2O(g, 1.33 kPa) (3) H2O(l, 1.33 kPa)?? ?rGm=?rGm + 2.30RT lg(1.33/100) = 8.59 + 2.30 ? 0.00831 ? 298 ? (-1.88) = -2.10 (kJ·mol-1) 过程(1): H2O(g)向H2O(l)转变,水蒸气凝聚为水。 过程(2): 平衡态。
过程(3): H2O(l)向H2O(g)转变,水蒸气变成水蒸气。 43
? ?rGm = 3 (-228.59) - (-741.0) = 55.2 (kJ·mol-1) ???rGm-10
= -9.70, 所以 K?p= 2.0 ? 102.30RT 在298 K,水的饱和蒸气压为3.17 kPa,pH2= 101.3 - 3.17 = 98.1(kPa)
lgK?p=
Qp =
pH2OpH233= 3.37 ? 10-5
Qp >K?p,可判断含饱和水蒸气的H2不能将Fe2O3还原为Fe。 44
? ?rHm = 3 ? (-241.83) - (-822.1) = 96.6 (kJ·mol-1)
? ?rSm = 3 ? 188.72+2 ? 27.2-3 ? 130.59-90.0 = 138.8 (J·mol-1·K-1) ??? ?rGm = ?rHm - T ?rSm= 96.6 - 1000 ? 0.139 = -42.2 (kJ·mol-1) 2? lgK?p= 2.21 Kp= 1.6 ? 10
Qp =
pH2OpH233= 3.37 ? 10-5
    因为Qp      (C)  46      (B)  47      (B)   48      (C)  49      (D)  50      C6H6(l) + NH3(g)C6H5NH2(l) + H2(g)  ?    ?rGm = 153.2 + 0 - (118.5 - 16.63) = 51.3 (kJ·mol-1)  ?    ?rGm = -2.30RT lgK?p  ???rGm?51.3?1000?    = -9.01  2.30RT2.30?8.31?298-10     K?p= 9.8 ? 10  lgK?p=     平衡常数很小,说明在常温下由苯与氨直接作用制备苯胺基本上不可能。  51      (C)  52      (D)  53      (C)  54      (D)  55      (C)  56      (A)  57      起始分压商 Qp = (pNH3/p?)(pN2/p?)1/2(pH2/p?)3/22.0?10?2-2 = 1.0 ? 10 ?1/24.0?1.0    在673 K,Qp     (A)  59      (B)  60          C2H5OH(l)C2H5OH(g)  ?    ?rHm = -235.4 - (-277.6) = 42.2 (kJ·mol-1)  ?    ?rSm = 282 - 161 = 121 (J·mol-1·K-1)      在12000 m ,乙醇的蒸气压为 0.600 ? 100 kPa ?     所以此时的K?p = p(C2H5OH, g) /p= 0.600 ?    所以?rGm = -2.30 RT lg0.600  ???        ?rGm = ?rHm - T ?rSm      所以 4.22 ? 104- T ? 121 = -2.30 ? 8.31 ? T ? lg0.600      所以 Tb = ?rU?m= 337 (K)   61      (C)  62      CaSO4·2H2O(s) CaSO4·12H2O(s) + 32H2O(g)  -1?    ?rGm = -1435.2 +3(-228.6) - (-1795.7) = 17.6 (kJ·mol) 2???rGm-4     = -3.09,  K?p= 8.1 ? 10 2.30RT3/2-1     K?p= (pH2O/100 kPa), pH2O= 8.7 ? 10 kPa  lgK?p= pV8.7?10?1?1000?    n(H2O,g) == 0.35 (mol) RT8.31?298    n(CaSO4·2H2O) = 0.23 mol      0.23 ? (136.2+18 ? 2) = 40 g   CaSO4·2H2O脱水变成CaSO4·1H2O。 2 63      (B)  64  ?    (1) ?rHm= 9.2 - 2 ? 33.2 = -57.2 (kJ·mol-1)  ?       ?rSm= 304 - 2 ? 240 = -176(J·K-1·mol-1)  ???       ?rGm=?rHm-T ?rSm= -57.2 + 298 ? 176/1000 = -4.75 (kJ·mol-1)     所以该反应在298K,标准状态下自发进行。 ???    (2) ?rGm = -RT lnK?p    lnKp=1.92     Kp=6.8                      2NO2N2O4     平衡摩尔分数     ?          1-?       K?p= pN2O4/p?(pNO2/p?)2?(1??)?2.0= 6.8 22??2.0    解得 ? = 0.24 = 24%  (NO2)          1-? = 76%      (N2O4)  65      平衡混合物的平均摩尔质量     M = 0,18?0.083?1273= 95 (g·mol-1)  0.202P2(g)      又设平衡时P2(g)为2x mol      则                     P4(g)     平衡时物质的量/mol      1-x         2x        所以 n总= 1+x       1?x2x(4 ? 31) +(2 ? 31) = 95 1?x1?x    x = 0.31   即 P4(g)分解百分率=31%  66      (1) 吸热反应,因T增加,K?p值增加。  -1?(2) 在940 K时,= -2.30RT lgK?= -2.30 ? 0.00831 ? 940 ? lg0.500 = 5.41(kJ·mol) ?rGmp     (3) lgK?p2K?p1??rHmT?T1?(2) 2.30RT2?T1??rHm4.61100?()        lg0.5002.30?0.008311040?940?       ?rHm = 180 kJ·mol-1  ??                ?rHm - ?rGm   180-5.41      (4)  ?=?rSm???rHm??rGm180?5.41?= 0.186 (kJ·mol-1·K-1)  T940  67                     PCl5  PCl3 + Cl2  平衡时n/mol    1-?         ?     ?      n总= 1+?       平衡总压力为p时:? = 0.204= 0.169  1.00?0.204?20.1692    Kp=p =p = 0.0294 p 221??1?0.169p?'2?    = 0.0294 p  1??'22           1.00-0.1692      温度不变,则Kp不变,总压为p/2时,      所以 ?’= 0.236      压力为p/2时,平衡总浓度为0.704 mol·dm-3     0.704 = c(1+?’),c = 0.570 mol·dm-3      所以 [Cl2]’= [PCl3]’= 0.570 ? 0.236 = 0.135 (mol·dm-3)          [PCl5]’= 0.435 (mol·dm-3)  68      (B)  69      减小              减小     向右(NH3合成)     增大  70      693K  pHg = 2/3 ? 5.16 ? 104 = 3.44 ? 104 (Pa)           pO2= 1/3 ? 5.16 ? 104 = 1.72 ? 104 (Pa)     723K  pHg = 2/3 ? 1.08 ? 105 = 7.20 ? 104 (Pa)           pO2= 1/3 ? 1.08 ? 105 = 3.60 ? 104 (Pa)      Kp(693K) = pHg2·pO2= (3.44 ? 104)2 (1.72 ? 104)     Kp(723K) = pHg2·pO2= (7.20 ? 104)2 (3.60 ? 104)  ??(723?69)3?1000(7.20?104)2(3.60?104)?rHm   lg  ?4242.303?8.31?693?723(3.44?10)(1.72?10)?    ?rHm = 307 kJ·mol-1   71      p = 93.3 kPa   T = 293 K    Vtot = 15.7 dm3     因为 pVtot = ntotRT     ntot                              2NO2(g)N2O4(g)     初态物质的量/mol          1.00           0     平衡态物质的量/mol        1.00-2x         x       ntot = 1.00 – 2x + x = 1.00 – x = 0.60 (mol)    x = 0.40 mol (N2O4)     所以nNO2= 1.00 – 2x = 0.20 (mol),说明80%的NO2缔合为N2O4。  72  2?    0.2 mol·dm-3 Fe (phen)3与等体积0.2 mol·dm-3Ce4+ 相混合,则  2?    [Fe (phen)3] = 0.1mol·dm-3 = [Ce4+] 2?    设平衡后,[Fe (phen)3] = x,有: 2?                          Ce4+ + Fe (phen)3? Ce3+ + Fe (phen)33  pVtot93.3?15.7= 0.602 (mol) ?RT8.31?293    平衡浓度/mol·dm-3      x      x           0.1-x  0.1-x   ?[Ce3?][Fe(phen)3(0.1?x)253]    K == 1 ? 10 ?4?2?2[Ce][Fe(phen)3]x2?    所以 x = [Fe (phen)3] = 3 ? 10-4 mol·dm-3   73     (1) lgK?p2K?p1?74.8350?rHmT?T1(} ?(2)= 2.30?8.31?10?3773?11232.30RT2?T1        K?p2= 16      (2)                  CH4(g)C(s) + 2H2(g)      初始物质的量/mol    1.00                0     平衡物质的量/mol    1.00-?              2?       平衡时总的物质的量为(1.00+? )mol,设平衡时总压为 p  2?22?2p()(pH2/p)4?21.00????p-------------(1)     Kp2=?21.00??pCH4/p1.00??p1.00??     pV = (1.00 + ? )RT ---------------------------------------------------(2)  4?28.31?1123?    (2)代入(1)    Kp2 =   1.00??50.0      ?21.00??= 0.021     ? = 0.13    p = 2.4 ? 102 kPa