《最新普通高中课程标准实验教科书--化学选修4[江苏版]优秀名师资料.doc》由会员分享，可在线阅读，更多相关《最新普通高中课程标准实验教科书--化学选修4[江苏版]优秀名师资料.doc（50页珍藏版）》请在三一文库上搜索。
1、普通高中课程标准实验教科书-化学选修4江苏版-4 11-1-1 12 31 2 3 - 1 - 451,2HkJ/mol 3H0 4- 2 - 能能反应物 量 生成物 量 ?H0 反应物 生成物 反应过程 反应过程 121 2 5 - 3 - 1 2 3- 4 - 12 1101kPa25 23H+ 4H 51mol 6HKJ/molH- 5 - 31 23- 6 - 1 ( ) AB C DE 2 1molHHQkJ 11molO = OQkJ 21molHOQkJ 3A2 Q + Q 4 Q123 B2 Q + Q 4 Q123 CQ + Q 2Q 1A.H(g)+Cl(g)=2HCl(g
2、); H=- 22Q1 1/2 H (g)+1/2 Cl (g)=HCl(g);H=- Q222 BC(s)+1/2O(g)=CO (g); H=- Q 21C(s)+ O (g)=CO (g); H=- Q222 C2 H (g)+ O (g)=2HO(l); H=- Q2221 2 H (g)+ O (g)=2 HO (g); H=- Q2222 DS(g)+ O (g)=SO (g); H=- Q221 S(s)+ O (g)= SO (g); H=- Q 222- 8 - 41mol C1mol 1mol CO 1mol H 131.5 kJ 25CH0.5 4mol CHCOHO 4
3、22445 kJA2CH( g )4O ( g ) 2CO( g ) 4 22 4HO ( l ) H = +890 kJ/mol 2BCH 2 O CO HO 422 2H = 890 kJ/mol CCH ( g )2 O( g ) CO ( g ) 2 42 2HO ( l ) H = 890 kJ/mol 2D1/2CH( g )O ( g ) 1/2CO ( g ) 4 22HO ( l ) H = -890 kJ/mol 2- 9 - 6 1H ( g )1/2 O ( g )HO ( g )222 H = a kJ/mol 122 H ( g )O ( g )2 HO ( g )
4、222 H = b kJ/mol 23H ( g )1/2 O ( g )HO ( l )222 H = c kJ/mol 34 2 H ( g )O ( g )2 HO ( l )222 H4= d kJ/mol 1abcdAacd0 C2a=b0 - 10 - -4 21-1-2 12. 123 4 - 11 - 1 1mol 1mol 57.3kJ/mol + - H+ OH= HO2+-H(aq) + OH(aq) HO(l)H =-57.3kJ 21molHO57.3KJ/mol 22 1 0.50 mol/L0.50 mol/L0.50 mol/L0.50 mol/L 23 ,34
5、.1810mt,，,HKJmol/ 0.025- 12 - 4 1 2 50 mL 0.50 mol/L50 mL 0.55 mol/LNaOHNaOH3 NaOH,4 5 57.4 kJ/mol- 13 - 6 1. 2 3 4.5.6.12NaOH3450mL0.55mol/LNaOH- 14 - 4NaOH60 mL 0.50 mol/L50 mL 0.55 mol/L NaOH 150 mL 0.50 mol/L50 mL 0.50 mol/L100 mL 0.50 mol/L100 mL 0.50mol/LNaOH- 15 - 21L1mol/LNaOHHH H 123H H H 1
6、32 ? C(s)+1/2 O (g)=CO(g) H=? 21CO(g)+1/2 O (g)= CO(g) H222 =-283.0kJ/mol C(s)+ O (g)= CO(g) 22 H=-393.5kJ/mol 3+ = H + H =H123 H =H- H 132=-393.5kJ/mol+ 283.0kJ/mol =-110.5kJ/mol - 16 - HHHH 12H=H+H12 1NHNO242NHO 22N (g)+2O(g)=2 NO 222(g)H=+67.2kJ/mol 1NH (g)+ O (g)= N (g)+2 HO (l) 24222- 17 - H=-5
7、34kJ/mol 22NH(g)+2NO(g)=3N(g)+4HO(l) 24222H=-1135.2kJ/mol 2H(g)+1/2 O (g) HO (g) ; H = 2212241.8kJ/mol H(g)+ 1/2 O (g) HO (l) ; H = 2222285.8kJ/mol HO l) HO (g) ; H 223HHH 123H=H+H123 3 C(s)+ O (g) CO (g) H = 22395.4 kJ/mol C(s)+ O (g) CO(g) H = 22393.5 kJ/mol H= +1.9 kJ/mol - 18 - -4 31-1-3 12. 12
8、3- 19 - 1101kPal mol. 101kPal g. 2 101kPa 1molH HS2SOS 2CCOCOH22HO (l)HO (g) 22H(g)+ 1/2 O(g) = HO(l); H =285.8 222kJ/mol 31 mol - 20 - - 21 - 4 A B1mol2mol C1mol2mol DH(g)+1/2O(g)=HO(g) H = 222-241.8 kJ/mol 2 H (g)+ O (g)=2 HO (g) H = -483.6 222kJ/mol H (g)+1/2 O (g)= HO (l) H = 222-285.8 kJ/mol 2
9、H (g)+ O (g)=2 HO (l) H = 222-571.6 kJ/mol 285.8 kJ/mol 5 10gSO2500gHO18262.4- 22 - 1 4.18J -1297 kJ?mol S(s)O(g)SO (g) H22 -1 297kJ?mol4 6 H - 23 - 12 3451 CCH 81823 SO2- 24 - 45 - 25 - 1COCH4 CO(g)+1/2 O (g)= CO(g) 22 H=-283.0kJ/mol CHg+2Og COg+2HO4222lH =890kJ/mol 1molCO3molCH4 A2912kJ B2953kJ C3
10、236kJ D3867kJ 2 H(g)=2H(g) 2H+Q1 1/2O(g)=O(g) 2H+Q2 2H(g)+O(g)=HO(g) 2H-Q3 HO (g)= HO (l) 22H-Q4 HO(g)+1/2 O (g)= 22HO (l) H-Q25 QQQQQ 5123432 H(g)+ O (g)=2 HO (l)H = -571.6 222- 26 - kJ/mol HO2233.6L _kJ 4 C(s)+1/2 O (g)=CO(g) H =2110.5kJ/mol C(s)+ O (g)= CO(g) H=-393.5kJ/mol 22 CO3/4 5NHNO242 NO (
11、g) H22 N (g)+2O(g)=2 NO (g)222H=+67.2kJ/mol 1NH (g)+ O (g)= N (g)+2 HO (l) 24222H=-534kJ/mol 21molNO2kJ - 27 - -4 41-2-1 1 1 2 2 3 4 511 ,. 2+2+: Zn+ Cu Zn + Cu 22 ,. -2+ : Zn-2e=Zn2+-Cu+2e=Cu - 28 - 2+2+Zn+ Cu Zn + Cu 3 4 1 2 3 5UKCl-+ClKZnCuZnCuPrimary Cell 62+-Cu+2e=Cu -2+ Zn-2e=Zn- 29 - 2+2+Zn+
12、Cu Zn + Cu,2+2+ (-)Zn|Zn(C) Cu(C)|Cu(+)12|- +3+-1,(-)Pt|H(100KPa)|H(1.0molL)Cr(1.0mo2+2-1-2-1-1l,L),H(1.010mol,L),CrO(10mol,L27)|Pt(+) -+ : H -2e = 2H2+-3+2- : CrO+14 H+6e=2 Cr+7 27HO2+3+2- : CrO+3 H=2 Cr+7 + 8 H272 HO 2- 30 - 1. - 31 - 2Zn + 2AgNO=Zn(NO)+2Ag 3323+2+ 2Fe+Fe=3Fe32:Cu+2AgNO=Cu(NO)+2Ag
13、 332,4,2Zn-HSO-Cu34. . 1(); 2; 3 .- 32 - 1(:),; ,; 2,- 33 - -4 51-2-2 1 :( ) :( ) : 2 :( ) :( ) : 3(:) ():( ) (HSO):( ) 24():( ) - 34 - (KOH):( ) 4 :( ) :( ) : () 1 -Zn+2OH-2e = ZnO+HO2-AgO+HO+2e = 2Ag+2OH 22AgO+Zn 2Ag+ZnO 22 -Zn+2OH-2e = Zn(OH)2-2MnO+2HO+2e = 2MnOO H +2OH 22Zn+2MnO+2HO 22Zn(OH)+2Mn
14、OO H 23(:) -+():( H ) 2H-4e=4H 22(HSO):( O ) 242+-O+4H+4e=2HO 22():( H ) 2-H+2OH-2e=2HO 22(KOH):( O ) 2-1/2O+HO+2e=2OH 22- 35 - 4 -2-:( Pb ) Pb+SO-2e=PbSO 44:( PbO ) 2-2-PbO+4H+SO+2e=PbSO+2HO 2442: () PbO+Pb+2HSO 224放电,2PbSO+2HO ,42充电5 Ni-CdCd + 2NiO(OH) + 2HO = 22Ni(OH) + Cd(OH)22 Ni-FeFe + 2NiO(OH
15、) + 2 HO = 22Ni(OH) + Fe(OH)22 6 LiAlCl 4SOCl 28Li + 3SOCl = 6LiCl + LiSO+ 2S 2230-56.7C- 0+71.1C 7 - 36 - 1991-3+4Al 12e = 4Al -3O + 6H O + 12e = 12OH 224Al + 3O + 6H O = 224Al(OH) 320-508 CO-CO-LiCO + NaCO-CO232322CO + O = 2CO22 0650C -2-2CO + 2CO -4e= 4CO32 - 37 - -2-O + 2CO + 4e = 2CO 22312H 2+O
16、 =2HO22_ _ _ _ 2a- 38 - 3,.LiCONaCO2323,CO,CO20,650C,: : , 2- :O+2CO+4e=2CO223: . 1 2+-HOHHO23- 39 - -4 61-2-3 12341_ 2_- 40 - _ 3_ 4- 41 - 1_ _ 2_ _ _ 3_ 5 6 1_- 42 - _ 2 Pt C_CuPtC_,_2S I Br Cl OH F23Ag Hg Fe 22222Cu H Pb Sn Fe Zn 322Al Mg Na Ca K 7_ 8- 43 - 123+HOH4pH HSO 24 HO NaOH 2NaNO 3 HBrHB
17、r - 44 - CuClCuCl2 2HClNaCl CuOCuSO 4CuCO3 1-10.01mol?L-AOH BC D- 45 - -2+2e-4 = 4e-2HO+O+4e=4OH 22-OH B 2ABCD+H 3+2+2+2Fe+Cu=Cu+2Fe - 46 - BC 3 1996KIO 31_ 2电解,KI+3HOKIO+3H?232_ -1IIO3I 2-2I-IOHOH322- 47 - -2I - 2e=I 2-2HO+2e=2OH+H?I222-3I+6OH=5I + IO+3HO 23215KI+KIO+6HCl=3I+6KCl+3HO 322-2I+6OH - 6
18、e = IO + 3HO 32-6HO-6e = 3H +6OH? 22- 48 - 1 2 3 4 1- 49 - 235ZnC CCu2 Zn-2e = Zn 2Cl2e = Cl? 22 H+2e = H? Cu 22 +2e = Cu 2 2Zn +2 H= Zn + H? Cu 2+2Cl= Cu + Cl? 261 2 7 8- 50 - -4 71-3-1 123 12.34 5 6- 51 - 1-失e2 ,氧化反应-n+M - ne = M 45 - 52 - OH- -Fe - 2e = 2+Fe -2HO + O + 4e= 4OH 226 - 53 - 7 -2+-2+
19、 Fe-2e=Fe=2Fe2Fe-4e- Fe(- ) 2HO + 2e = O+2HO+4e=4OH222-H?+2OH 2Fe+2HO+O= 2 222C(+) Fe + 2 HO Fe(OH) 2 2= Fe(OH)4Fe(OH) : + H? +2HO+O=4Fe(OH)22222 3FeOO ? nH232 - 54 - 10%1 2 - 55 - - 56 - - - 57 - 1. ,cv,2. ,t 1.0 mol/L 3.0 mol/L20.8 mol/L 2.4 mol/L0.4 mol/L 2 3H+ N = 2NH2 23 mol/L 3.0 1.0 0 2S- 58
20、- mol/L 2.4 0.8 0.4 31 32 123 4 14NH(g)+5O(g)= 4NO(g)+6HO(g)3 2 210L 0.45 molv(x) Av(NH)= 0.010 mol/(Ls) 3Bv(O)= 0.0010 mol/(Ls) 2Cv(NO )= 0.0010 - 59 - mol/(Ls) Dv(HO )= 0.045 2mol/(Ls) 2AH0.1 mol/(Lmin) 2 BNH0.3 mol/(Lmin) 3CN0.2mol/(Lmin) 2DH0.3 mol/(Lmin) 2 32XYZ2425100mLHCl 14.6g5.6g2H1.12L2FeC
21、l2244- 60 - V 24HCl511.A.B2L,:3A()+B()xC()+2D() 5min,D=0.5mol/L,A:B=3:5,C0.1mol/L?min.A= mol/L,B mol/L?min,x . 62L,: 3A+B2C (A.B.C) A.B4mol,A0.12mol/L?s,10sB A.2.8mol B.1.6mol C.3.2mol D.3.6mol - 61 - 1|2- 62 - 33-34 - 63 - 35 36 - 64 - , 1 Cl+H=2HCl 22 HHClClHCl P 34 1- 65 - 234 1 22HIHI22HI 122 -
22、66 - N3H222NH 3 12, Zn+2HClHZnCl22 3 35 1210K- 67 - 24 3, MnO24 36 , - 68 - 2106 1013 34 pH pH pH 5 - 69 - 1 2 3G=H- TS1 2 3 4 . - 70 - 1 : : 2NHCl+ : 4: . 1 2 3 - 71 - S0 S0 1 (rHm0) 298.15K- 72 - 2 1(H)H 0(H0)NH4ClAg2O298.15K NH4Cl(s) ? NH4+(aq) + Cl-(aq) H -1= 14.7 kJ?mol Ag2O(s) ? 2Ag(s) + 1/2O2
23、(g) H = -131.05 kJ?mol CaCO3(H 0) CaCO3(s) ? CaO(s) + CO2(g) -1H = 178.32 kJ?mol 298.15K1123KCaCO3178.32kJ?mol-1,()- 73 - ?NH4C1Ag2ONH4ClNH4+Cl-NH4C1(aq)NH4Cl+-NH4(aq)Cl(aq)NH4C1 ()(S) 2 ()- 74 - S(g)(1)(s)(S)(H)S 0, SS0S3 H S 1 - + 2 + - - 75 - 3 + + 4 - - 42 H-TS0=0- 76 - 122-2-2 1 21 1H+O=HO 2222
24、- 77 - 31 23V V V, V VV?0 2 - 78 - V=V VV?0 CPTV?V: 1 = 2- 79 - m A(g) + n B (g) p C (g) +q D (g) mmolA mmolA nmolB pmolC nmolBqmolD - 80 - (A) ?(B) ?(C) ?(D) = m ? n ? p ? q m + n?p + q m + n =p + qm + n? p + q M m + n = p + q MM Cs+COg2COg2 Ng+3Hg2NHg223 Hg+Ig2HIg22- 81 - 1(T,V)2(T,V)3(T,P) - 82 -
25、 12mAg+ nBgpCg+ qDg pqmnK=c(C)?c(D)/c(A)?c(B) 452-7 3 1 K K 2mA(g)+ nB(g)pC(g)+ - 83 - qD(g)pqmnQ=C(C)?C(D)/C(A)?C(B)cQK cQK cQK c3 () () 452-82-9 1 2()C(s)+HO(g)CO(g)+H(g)22K=c(CO)?c(H)/c(HO) 22Fe(s)+CO(g)Fe(s)+CO(g)K=c(CO-2)/c(CO) 2- 84 - 3N(g)+3H(g)2NH(g)223K1/2N(g)+3/2H(g)NH(g)1223KNH(g)1/2N(g)+
26、3/2H(g)2322K 32KK K=K 1212KK K?K=1 23232KK K?K=1 1313该反应物的起始浓度,该反应物的平衡浓度=100% 该反应物的起始浓度该反应物反应中消耗(反应)浓度=100% 该反应物的起始浓度该反应物反应中消耗(反应)的物质的量=100% 该反应物的起始物质的量461247 11L- 85 - 2mol/L3mol/L(:K=0.5625 3.5mol/L6.5mol/L) 2CO(g)+HO(g)CO(g)+H(g)CO222HO2mol/L2K=2.60 1CO502CO3CO2mol/LHO26mol/LCO(:1,261.7% 386.5%)
27、3CO(g)+H(g)CO(g)+HO(g)222Kt t 700 800 830 1000 1200 - 86 - K 0.6 0.9 1.0 1.7 2.6 K= c(CO)?c(HO)/c(CO)?c(H) 222 B A Bc(CO) C Dc(CO) = c(CO) 2EK FCOHO 2c(CO)c(H)=c(CO)c(HO)222830 - 87 - 132-3-1 1 2 12 3vv - 88 - VV VV VV 1 2-2-2CrO()CrO()+HO42721 4mL0.1mol/LKCrO241mol/LHNO_ 3_ - 89 - 2 1mol/LNaOH_ _ 1
28、- 90 - 2CuCl2CuCl 22-2-Cu(HO)()+4Cl CuCl(244)+4HO 2_- 91 - _ _ 2 1P492-212- 92 - N(g)+3H(g) 222NH(g) 3NO(g) 242NO(g) 2FeO(s)+CO(g) Fe(s)+CO(g) 2:aA(g) +bB(g)cC(g)+dD(g) 1a+bc+d_ 2a+b=c+d_ 3a+bc+d_- 93 - _ 3 2-CoClCoCl242+- 2-Co + 4ClCoCl H04CoCl2_,_ _,_ - 94 - _ _ N(g)+3H(g)2NH(g) H=-92.4kJ/mol223-
29、95 - 1CO352OCHO261266CO+12HOCHO+6HO+6O22612622函数的增减性：_互余关系sinA=cos(90A)、cosA=sin(90A)_ 当a越大，抛物线开口越小；当a越小，抛物线的开口越大。2CO集合性定义：圆是平面内到定点距离等于定长的点的集合。其中定点叫做圆心，定长叫做圆的半径，圆心定圆的位置，半径定圆的大小，圆心和半径确定的圆叫做定圆。(4)面积公式:(hc为C边上的高);HC(s)+HO(g)CO(g)+H(g)222- 96 - 61273K(1.5X10Pa)定理：在同圆或等圆中,相等的圆心角所对的弧相等、所对的弦相等、所对的弦心距相等。HOg 2应用题1_ tanA没有单位，它表示一个比值，即直角三角形中A的对边与邻边的比；2?_2.点与圆的位置关系及其数量特征：将2mol SO和3molO在在某一温度下达到化学平衡状态。试回答： 22?若测得SO与O的平衡浓度相等，求NH在平衡混合气中的物质的量分数； 223115.75.13加与减（二）2 P61-63 数学好玩2 P64-67?若测得平衡混合气体对H的相对密度为20时，求SO的转化率。 22- 97 -