(1) ÔÚ´Ë K?p= p(C2H5OH, g)/100 kPa
ËùÒÔ25¡æÒÒ´¼µÄÕôÆøÑ¹Á¦Îª0.085 ? 100 kPa = 8.5 kPa
??rHm42.2? (2) ÔڷеãTb£¬?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, ±¥ºÍÕôÆøÑ¹)H2O(g, ±¥ºÍÕôÆøÑ¹) ?rGm= 0 (´ËʱΪƽºâ̬)
??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) ¹ý³Ì(1)£º H2O(g)ÏòH2O(l)ת±ä£¬Ë®ÕôÆøÄý¾ÛΪˮ¡£ ¹ý³Ì(2)£º ƽºâ̬¡£
¹ý³Ì(3)£º H2O(l)ÏòH2O(g)ת±ä£¬Ë®ÕôÆø±ä³ÉË®ÕôÆø¡£ 43
? ?rGm = 3 (-228.59) - (-741.0) = 55.2 (kJ¡¤mol-1) ???rGm-10
= -9.70£¬ ËùÒÔ K?p= 2.0 ? 102.30RT ÔÚ298 K£¬Ë®µÄ±¥ºÍÕôÆøÑ¹Îª3.17 kPa£¬pH2= 101.3 - 3.17 = 98.1(kPa)
lgK?p=
Qp =
pH2OpH233= 3.37 ? 10-5
Qp >K?p£¬¿ÉÅжϺ¬±¥ºÍË®ÕôÆøµÄH2²»Äܽ«Fe2O3»¹ÔΪFe¡£ 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
ÒòΪQp (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= ƽºâ³£ÊýºÜС£¬ËµÃ÷ÔÚ³£ÎÂÏÂÓɱ½Óë°±Ö±½Ó×÷ÓÃÖÆ±¸±½°·»ù±¾Éϲ»¿ÉÄÜ¡£ 51 (C) 52 (D) 53 (C) 54 (D) 55 (C) 56 (A) 57 Æðʼ·ÖѹÉÌ Qp = (pNH3/p?)(pN2/p?)1/2(pH2/p?)3/22.0?10?2-2 = 1.0 ? 10 ?1/24.0?1.0 ÔÚ673 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) ÔÚ12000 m £¬ÒÒ´¼µÄÕôÆøÑ¹Îª 0.600 ? 100 kPa ? ËùÒÔ´ËʱµÄK?p = p(C2H5OH, g) /p= 0.600 ? ËùÒÔ?rGm = -2.30 RT lg0.600 ??? ?rGm = ?rHm - T ?rSm ËùÒÔ 4.22 ? 104- T ? 121 = -2.30 ? 8.31 ? T ? lg0.600 ËùÒÔ 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¡¤2H2OÍÑË®±ä³ÉCaSO4¡¤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) ËùÒԸ÷´Ó¦ÔÚ298K£¬±ê׼״̬ÏÂ×Ô·¢½øÐС£ ??? (2) ?rGm = -RT lnK?p lnKp=1.92 Kp=6.8 2NO2N2O4 ƽºâĦ¶û·ÖÊý ? 1-? K?p= pN2O4/p?(pNO2/p?)2?(1??)?2.0= 6.8 22??2.0 ½âµÃ ? = 0.24 = 24£¥ (NO2) 1-? = 76£¥ (N2O4) 65 ƽºâ»ìºÏÎïµÄƽ¾ùĦ¶ûÖÊÁ¿ M = 0,18?0.083?1273= 95 (g¡¤mol-1) 0.202P2(g) ÓÖÉèÆ½ºâʱP2(g)Ϊ2x mol Ôò P4(g) ƽºâʱÎïÖʵÄÁ¿/mol 1-x 2x ËùÒÔ n×Ü= 1+x 1?x2x(4 ? 31) +(2 ? 31) = 95 1?x1?x x = 0.31 ¼´ P4(g)·Ö½â°Ù·ÖÂÊ=31£¥ 66 (1) ÎüÈÈ·´Ó¦£¬ÒòTÔö¼Ó£¬K?pÖµÔö¼Ó¡£ -1?(2) ÔÚ940 Kʱ£¬= -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 ƽºâʱn/mol 1-? ? ? n×Ü= 1+? ƽºâ×ÜѹÁ¦Îªpʱ£º? = 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 ζȲ»±ä£¬ÔòKp²»±ä£¬×ÜѹΪp/2ʱ£¬ ËùÒÔ ?¡¯= 0.236 ѹÁ¦Îªp/2ʱ£¬Æ½ºâ×ÜŨ¶ÈΪ0.704 mol¡¤dm-3 0.704 = c(1+?¡¯)£¬c = 0.570 mol¡¤dm-3 ËùÒÔ [Cl2]¡¯= [PCl3]¡¯= 0.570 ? 0.236 = 0.135 (mol¡¤dm-3) [PCl5]¡¯= 0.435 (mol¡¤dm-3) 68 (B) 69 ¼õС ¼õС ÏòÓÒ(NH3ºÏ³É) Ôö´ó 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 ÒòΪ pVtot = ntotRT ntot 2NO2(g)N2O4(g) ³õ̬ÎïÖʵÄÁ¿/mol 1.00 0 ƽºâ̬ÎïÖʵÄÁ¿/mol 1.00-2x x ntot = 1.00 ¨C 2x + x = 1.00 ¨C x = 0.60 (mol) Âx = 0.40 mol (N2O4) ËùÒÔnNO2= 1.00 ¨C 2x = 0.20 (mol)£¬ËµÃ÷80£¥µÄNO2µÞºÏΪN2O4¡£ 72 2? 0.2 mol¡¤dm-3 Fe (phen)3ÓëµÈÌå»ý0.2 mol¡¤dm-3Ce4+ Ïà»ìºÏ£¬Ôò 2? [Fe (phen)3] = 0.1mol¡¤dm-3 = [Ce4+] 2? ÉèÆ½ºâºó£¬[Fe (phen)3] = x£¬ÓУº 2? Ce4+ + Fe (phen)3? Ce3+ + Fe (phen)33 pVtot93.3?15.7= 0.602 (mol) ?RT8.31?293 ƽºâŨ¶È/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? ËùÒÔ 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) ³õʼÎïÖʵÄÁ¿/mol 1.00 0 ƽºâÎïÖʵÄÁ¿/mol 1.00-? 2? ƽºâʱ×ܵÄÎïÖʵÄÁ¿Îª(1.00+? )mol£¬ÉèÆ½ºâʱ×ÜѹΪ 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)´úÈë(1) Kp2 = 1.00??50.0 ?21.00??= 0.021 ? = 0.13 p = 2.4 ? 102 kPa