1、完整版无机化学天津大学第四版答案化学反应中的质量关系和能量关系 习题参考答案1.解:1.00吨氨气可制取 2.47吨硝酸。2.解:氯气质量为 2.9 X 103g。3.解:一瓶氧气可用天数n (p pjy (13.2 103-1.01 103)kPa 32L n2 P2V2 101.325kPa 400L_d4解:t MPVnR mR=318 K 44.9 C5.解:根据道尔顿分压定律nip(N2)= 7.6 104 Pa p(O2)= 2.0 104 Pa p(Ar) =1 103 Pa6.解:(1) n(CO2) 0.114mol; p(C02) 2.87 104 Pa(2) p(N2)
2、p PQ2) p(CO2)3.79 104Pa(3) 迥以空 267 10:Pa 0.286n p 9.33 104Pa7.解:(1) p(H2)=95.43 kPa8.解:(1)=5.0 mol(2)=2.5 mol结论:反应进度()的值与选用反应式中的哪个物质的量的变化来进行计算无关,但与反应式的与法有关。(2) m(H2)=pVMRT=0.194 g9.解: U = Qp p V = 0.771 kJ10.解: (1) V1 = 38.3 10-3 m3= 38.3LpV2(2)T2 = - = 320 KnR(3)W = ( p V) = 502 J(4)U = Q + W = -75
3、8 J(5)H = Qp = -1260 J5 311.解:NH3(g) + 4O2 2标8辔 NO(g) + 空山。) 屮皿=226.2 kJ mol12.解:rHm = Qp = 89.5 kJrU m = r H m nRT=96.9 kJ13.解:(1) C (s) + O2 (g) t CO2 (g)rH m = f H m (CO2, g) = 393.509 kJ mol 1112CO2(g) + 护(s) t CO(g)rHm =:86.229 kJ mol 1CO(g) +1 21 Fe2O3(s) t 2 Fe(s) + CO2(g)r H m =:8.3 kJ mol 1
4、各反应 rHm之和rH m = 315.6 kJ mol 1o(2)总反应方程式为31 3 22 C(s) + O2(g) + 3 Fe2O3(s) t CO2(g) + - Fe(s)rH m = 315.5 kJ mol 1由上看出:与计算结果基本相等。所以可得出如下结论:反应的热效应只与反应 的始、终态有关,而与反应的途径无关。14. 解: rH m (3) = rHm (2)x 3- rHm (1)x 2= 1266.47 kJ mol 115. 解:(1) Qp = r H m = 4 f H m (Al 2O3, s) -3 f H m (Fe3O4, s) = 3347.6 kJ
5、 mol 1(2) Q = 4141 kJ mol 116. 解:(1) rH m =151.1 kJ mol 1 (2) 屮皿=905.47 kJ mol 1 (3) 屮皿=71.7 kJ mol 117.解:rHm=2 fHm(AgCI, s)+ fHm (H2O, l) fHm(Ag20, s) 2 fHm (HCl, g)f Hm (AgCl, s) = 127.3 kJ mol 118.解:CH4(g) + 2O2(g) t CO2(g) + 2H 2O(l)rHm = fHm(CO2, g) + 2 fHm (H2O, l) fHm (CH4, g)=890.36 kJ mo 1Q
6、p = 3.69 104kJ第2章 化学反应的方向、速率和限度 习题参考答案1.解: rHm = 3347.6 kJ mol 1; Sm = 216.64 J mol 1 K 1; Gm = 3283.0kJ mol 1 v 0该反应在298.15K及标准态下可自发向右进行。2.解: rGm = 113.4 kJ mol 1 0该反应在常温(298.15 K)、标准态下不能自发进行。(2) rHm = 146.0 kJ mol 1; rSm = 110.45 J mol 1 K 1; rGm = 68.7 kJ mol 1 0该反应在700 K、标准态下不能自发进行。3.解: rH m = 7
7、0.81 kJ mol 1 ; rSm43.2 J mol 1 K 1;rGm = 43.9 kJ mol 1(2)由以上计算可知:rHm (298.15 K) = 70.81 kJ mol1.rSm (298.15 K)=43.2 J mol 1 K 1rGm亡上彈空=1639 KrSm (298.15 K)4.解:(1) Kcc(CO) c(H2)3c(CH4)c(H2O)Kpp(CO)P(H2)3p(CH4)p(H2O)(2)Kc(3)Kcp (CO) / ppH)/ Pp(CHJ/p p(H2。)/ p1 3c(N2)2 c(H2)2c(NH3)1p(N2)/p 2 P(H2)/pKp
8、1 3P(N2)2 p(H2) 2P(NH3)p(NH3)/ p= c(CO2) Kp=p (CO2 )=p (CO2) / p(4)Kc3c(H2。)33c(H2)Kp3p (H 2O) 3PE)P(H2O)/ p3P(H2)/p5.解:设rSm基本上不随温度变化。rGmrGm(298.15 K) = 233.60 kJ molrGm(298.15 K) = 243.03 kJ mollgK(298.15 K) = 40.92,故 K(298.15 K) = 8.3 1040Ig K (373.15 K) = 34.02,故 K (373.15 K) = 1.0 10346.解:(1) rG
9、m =2 fGm(NH3, g) = 32.90 kJ mol 1 v 0该反应在298.15 K、标准态下能自发进行。(2) lg K (298.15 K) = 5.76, K (298.15 K) = 5.8 1057.解:(1) rGm (l) = 2 fGm (NO, g) = 173.1 kJ mol 1-lg K1= 2m 11丿=30.32,故K1=4.810 312.303 RT(2)rGm (2) = 2f G m(N2O, g) =208.4 kJmol 1lg K2= fGm 36.50,故K2=3.210 372.303 RT(3)rGm (3) = 2fGm(NH3,
10、 g)=32.90 kJ mol1lgK3=5.76,故K3=5.8 105由以上计算看出:选择合成氨固氮反应最好。8. 解:rGm = fGm (CO 2, g) fGm (CO, g) f Gm (NO, g)=343.94 kJ mol 1 0.168 = J,故反应正向进行。12.解:(1)NH 4HS(s)NH3(g) + H2S(g)平衡分压/kPaK = p(NH 3) / p p (H 2S) / p = 0.070贝 U x= 0.26 100 kPa = 26 kPa平衡时该气体混合物的总压为52 kPa(2) T不变,K不变。NH 4HS(s)平衡分压/kPaNH 3(g
11、) + H 2S(g)25.3+ y y(25.3 y) / p y /p = 0.070y= 17 kPa13.解:(1)PCl5(g)PCl3(g) + Cl 2(g)平衡浓度/(mol L-1)0.70 0.500.500.50Kc2.02.02.0c(PCl3)c(Cl2)C(PCl5)=0.62mol L-1,(PCl5) = 71%平衡分压PCl5(g) PCl3(g) + Cl 2(g)RT RT0.20 0.5 V VRTp(PCl3)/p p(Cl2)/pP(PCl5”P=27.2(2)PCl5(g)新平衡浓度/ (mol L 1)0.10 + yPCl3(g) + Cl 2
12、(g)0.100.25 y 0.25 + y2Kc= (0.25 y)(030 y)mol L(0.10 y)=0.62mol L -1 (T 不变,Kc 不变)(3)平衡浓度/(mol L-1)y =0.01 mol L 1,PCl5(g)0.35 z(PCI5)= 68%PCl3(g) + Cl 2(g)0.050 + z(0.050 z)z = 0.62 mol L 10.35 zz= 0.24 mol L 1, (PCl5) = 68%比较(2)、(3)结果,说明最终浓度及转化率只与始、终态有关,与加入过程无关。14.解:平衡浓度/ (mol L - 1)N2(g) + 3H 2(g)
13、1.0 0.502NH3(g)0.50Kc =2c(NH3)c(N2)c(H2)若使N2的平衡浓度增加到新平衡浓度/(mol L-1)Kc = 2.0(mol L 1) 21.2mol L 1,设需从容器中取走 x摩尔的H2。N2(g) + 3H2(g)1.2 0.50+(3 0.2) x2NH3(g)0.50 2 0.20(0.50 2 0.20)21.2 (0.50 3 02 x)31(mol L )2 1 22 = 2.0(mol L 1) 2x=0.9415.解:(1 )a(CO) =61.5%; (2)a(CO) =86.5%; (3)说明增加反应物中某一物质浓度可提高另一物质的转化
14、率;增加反应物浓度,平衡向生成物方向移动。16.解:2NO(g) +O2(g)2NO2(g)平衡分压/ kPa101 79.2 = 21.8286 79.2/ 2 = 24679.2(673K)2P(NO2)/p2P(NO) / p p(O2)/p=5.36rGm = 2.303RTlg KrGm (673 K) = 9.39 kJ mol 117.解:rGm (298.15 K) = 95278.54 J mol 1rGm (298.15 K) = rHm (298.15 K) 298.15 K “(298.15 K) rSm (298.15 K) = 9.97 J mol 1 K 1, G
15、m (500 K)97292 J mollg K (500 K) = 0.16 , 故 K (500K )=1.4 1010或者lnK2rH m (298.15K) T2 T1K1(500 K) = 1.4 101018.解:因 rGm (298.15 K) = rGm (1) + rGm (2) = 213.0 kJ mol 1 0,说明该耦合反应在上述条件可自发进行。第3章酸碱反应和沉淀反应 习题参考答案解:(1) pH=-lg c(H+)=12.00(2) 0.050mol L-1HOAc 溶液中,HOAc - H+ + OAc-c 平/(mol L-1) 0.050-x x xKa 住
16、)C(OAC ) x J 1.8 10 5a c(HOAc ) 0.050 xc(H+) = 9.5 10mol L-1 pH = -lg c(H+) = 3.022.解:(1) pH = 1.00 c(H+) = 0.10mol LpH = 2.00 c(H+) = 0.010mol L-1 -等体积混合后:c(H+) = (O.IOmol L-1+0.010mol L-1) /2 = 0.055 mol L-1 pH = -Ig c(H +) = 1.26(2) pH = 2.00 c(H+) = 0.010mol L-1 pH =等体积混合后:13.00 pOH = 14.00-13.0
17、0 = 1.00, c(OH-) = 0.10mol L-1 -c(H ) O.SOm? L 0.0050mol L-11-、 0.10mol L -1c(OH ) 0.050mol L2酸碱中和后:H + + OH- t H2Oc(OH-) = 0.045mol LpH =12.653.解:正常状态时pH = 7.35 c(H+) = 4.5 10mol L-1pH = 7.45 c(H+) = 3.5 10mol L-1患病时pH = 5.90 c(H+) = 1.2 10mol L-11.2 10 -6 mol L-1274.5 10 -8 mol L-11.210-6 mol L-18
18、3.5 10 8molL-1344. 解:一兀弱酸 HA , pH = 2.77 c(H+)=1.7 W-3mol L-1患此种疾病的人血液中c(H+)为正常状态的2734倍。HA+ A-H +c 平/(mol L-1) 0.10-1.71 区3 1.7X0-31.7 X0-3Kac(H )c(Ac(HA)3 2(1.7 10 )50.10 1.7 10 3 2.9 10(X =1.7 10 30.10100%1.7%5. 解:溶液的 pH = 9.00, c(H+) = 1.0 1Q9mol L-1c(OH-) = 1.0 10 Ksp(Mg(OH) 2)时开始有 Mg(OH) 2沉淀出。c
19、(OH;Ksp(Mg(OH) 2)c(Mg 2+)5.61 5.01010121.0 10 5 mol L-1(2)c(Al 3+) c(OH-)3 = 4.0 W22 Ksp(AI(OH) 3),所以还有 Al 3+可被沉淀出。c(Fe3+) -c(OH)3 = 2.0 W22 Ksp 0(Fe(OH) 3),所以还有 Fe3+可被沉淀出。11.解: Cd2+ + Ca(OH) 2Ca2+ + Cd(OH) 2 JCd(OH)2(s) Cd2+ + 2OH- Ksp0=7.2 X0154若使 c(Cd2+)v 0.10mg L1 = 1.Q 10 g 1 gL-1 =8.9 M07mol L
20、-1112.41g gmol c(OH)Ksp(Cd(OH) 2) c(Cd2+ )/c7.2一10,8.9 101579.0 10 5mol L-1pH (14.00-pOH) = 10.0设OH-浓度为12.解:(1)混合后:c(Mn2+) = 0.0010mol L-1 - c(NH3 H2O) = 0.050 mol L-1 - x mol L -1NH3 H2ONH4+ OH-c 平/(mol L-1)0.050-x0.050 x1.8 10 5x2 = 9.0 W7,即 c(OH-)2 = 9.0 W7c(Mn2+) c(OH-)2 =9.0 杓-10 Ksp0(Mn(OH) 2)
21、 =1.9 10-13所以能生成Mn(OH) 2沉淀。(2)已知(NH4)2SO4的相对分子质量为 132.15c(NH 4)2SO4)= .495 1000 mol L-1 = 0.25mol L -1132.15 15c(NH4-) = 0.50 mol L-1 -设OH-浓度为x mol L -1NH3 H2O NH4+ + OH-c 平/(mol L-1) 0.050-x 0.50+x xc(NH)gc(OH -)c(NH 3W2O)Kb(NH 3CH2O)0.50x0.0501.8 10(0.50 x)x0.050 x1.810x=1.8 W6c(OH)=1.8 W6mol L-1c(Mn2+) c(OH-)2 = 3.2 W-
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