復(fù)旦 物理化學(xué) 第一章 習(xí)題答案.doc

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第一章習(xí)題解答 1. 體系為隔離體系, DU=0 W=Q=0 2. (1) W=pDV=p(Vg-Vl)pVg=nRT=18.314373.15=3102 J (2) W=pDV=p(Vs–Vl) 3. (1)恒溫可逆膨脹 (2)真空膨脹 W = 0 (3)恒外壓膨脹 W = p外(V2–V3) = = 2327 J (4)二次膨脹 W=W1 + W2 4. DH=nDHm,汽化=40670 J DU=DH–D(pV)=DH–p(Vg-Vl)=40670–101325(30200–1880)10–6 =40670–3058=37611 J 5. Cp,m=29.07–0.836103T+2.0110–6T2 (1) Qp=DH =20349–380+625=20.62 kJ (2) QV=DU=DH–D(pV)=DH–(p2V1–p1V1) V2=V1 \ QV=DH–nR(T2–T1)=20.62–R(1000-300)10–3=14.80 kJ (3) 6．（1）等溫可逆膨脹 DU =DH = 0 Q =W （2）等溫恒外壓膨脹 DU =DH = 0 Q = W = p2 (V2–V1) = p2V2–p2V1= p1V1–p2V1= (p1–p2)V1 =(506.6-101.3)103210–3 = 810 J 7. (1) p1T1=p2T2 (2) DU=nCV,m(T2–T1)= DH=nCp,m(T2–T1)= (3) 以T為積分變量求算: pT=C(常數(shù)) 也可以用p或V為積分變量進(jìn)行求算。 8． DU=nCV,m(T2–T1)=20.92(370–300)=1464 J DH=nCp,m(T2–T1)=(20.92+R)(370–300)=2046 J 始態(tài)體積 體積變化： 壓力 W=W1+W2=p2(V2–V1)+0=821554(0.003026–0.0246)=–17724 J Q=DU+W=1464–17724=–16260 J 9. 雙原子分子 W=–DU=–nCV,m(T2–T1) 10. (1) (2) DU=nCV.m(T2–T1)=n (28.8–R)(224.9–298)= –263 J DH=nCp.m(T2–T1)=n 28.8(224.9–298)= –369 J 11. 證明 U=H–pV 12. 證明 （1） H=f(T,p) V不變，對(duì)T求導(dǎo) 代入(1) 13． nQV+CDT=0 QV=–4807200 J C7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g) Dn =–4 DcHm = QV + DnRT =–4807200–4R298 = –4817100 Jmol–1 =–4817.1 kJmol–1 14．(1) 2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方) Dn =–3 QV =–223.8 kJ DrHm = QV + DnRT = –223.8 + (–3)RT10–3 = –231.2 kJ (2) 2C(石墨) + O2(g) = 2CO2(g) Dn = 1 QV =–231.3 kJ DrHm = QV + DnRT = –228.8 +RT10–3 = –228.8 kJ (3) 2H2(g)+Cl2(g) = HCl (g) Dn =0 QV =–184 kJ DrHm = QV =–184 kJ 15. (1) x=4 mol (2) x=2 mol (3) x=8 mol 16． 2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g) DfHm(kJmol–1) –411 –811.3 –1383 –92.3 DrHm=S(nDfHm)產(chǎn)物–S(nDfHm)反應(yīng)物= (–1383–292.3)–(–811.3–2411) = 65.7 kJ DrUm =DrHm–DnRT=65.7–2RT10–3=60.7 kJ 17. DrHm=S(nDcHm) 反應(yīng)物–S(nDcHm) 產(chǎn)物= (–2283–4285.8)–(0–1370)=339.2 kJ 18．生成反應(yīng) 7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l) DcHm(kJmol–1) –394 –286 –3230 DrHm=S(nDcHm) 反應(yīng)物–S(nDcHm) 產(chǎn)物= [7(–394) + 3(–286)] – (–3230)= –386 kJ 19. 反應(yīng) C(石墨) C(金剛石) DcHm(kJmol–1) –393.5 –395.4 DrHm=DcHm，石墨–DcHm，金剛石 =–393.5–(–395.4)=1.9 kJ 20. 反應(yīng) CH4(g)+2O2(g) = CO2(g) + 2H2O(l) DfHm(kJmol–1) –74.8 –393.5 –285.8 DrHm=S(nDfHm)產(chǎn)物–S(nDfHm)反應(yīng)物=[–393.5+2(–285.5)]–(–74.8)=–890.3 kJ 21. 反應(yīng) (COOH)2(s)+2CH3OH(l) = (COOCH3)2(l) + 2H2O(l) DcHm(kJmol–1) –251.5 –726.6 –1677.8 0 DrHm=S(nDcHm) 反應(yīng)物–S(nDcHm) 產(chǎn)物=[–251.5+2(–726.6)]–(–1677.8)=–26.9 kJ 22. 反應(yīng) KCl(s) K+(aq, ) + Cl–(aq, ) DfHm(kJmol–1) –435.87 ? –167.44 DrHm=17.18 kJ DrHm=S(nDfHm)產(chǎn)物–S(nDfHm)反應(yīng)物 17.18=[DfHm (K+,aq, )–167.44]–(–435.87) DfHm (K+,aq, )=–251.25 kJmol–1 23. 生成反應(yīng) H2(g) + 0.5O2(g) = H2O(g) DrH298=–285.8 kJmol–1 Cp,m(JK–1mol–1) 28.83 29.16 75.31 DCp=75.31–(28.83+0.529.16)=31.9JK–1 =–285.8+31.9(373–298)10–3=–283.4 kJmol–1 24. 反應(yīng)N2(g) + 3H2(g) = 2NH3(g) DrH298=–92.888 kJmol–1 a b103 c107 N2(g) 26.98 5.912 –3.376 H2(g) 29.07 –0.837 20.12 NH3(g) 25.89 33.00 –30.46 D –62.41 62.599 –117.904 =–92880+[–6241+2178–144]=97086 J H2(g) + I2(s) H2(g) + I2(g) DrH291=49.455 kJ 2HI(g) 2HI(g) Cp,m=55.64 JK–1mol–1 T=386.7K D熔Hm=16736 J Cp,m=62.76 JK–1mol–1 Cp,m=7R/2 JK–1mol–1 T=457.5K D蒸Hm=42677 J sl l g DrH473=？ DHHI Cp.m=7R/2 DHI2 DHH2 Cp.m=7R/2 25. 按圖示過(guò)程計(jì)算： DHH2=nCp,mDT=3.5R(473–291)=5296 J DHHI=nCp,mDT=23.5R(473–291)=10592 J DHI2=DH1(s,291386.7K) + DH2(sl) + DH3(l,1386.7457.5K) + DH4(lg) + DH5(g,457.5473K) =55.64(386.7–291)+16736+62.76(457.5–386.7)+42677+3.5R(473–457.5) =69632 J DHH2+DHI2+DrH473=DrH291+DHHI 5296+69632+DrH473=49455+10592 DrH473=–14881 J