Mot So Dang Bai Tap Giai Hoa Tren May Tinh Casio

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<p>Cc dng bi tp - Thi gii ton trn my tnh cm tay Casio1. Cu to nguyn t: Quan h s p, s n, s eTh tch v Bn knh nguyn t tnh ra Mng tinh th (s nguyn t v cnh trong n v c s) Ht nhn v phng x (hng s phng x, nin i vt c)2. Cu to phn t: Khong cch ca cc nguyn t trong ng phn hnh hcMomen lng cc3. ng hc:Cn bng ha hcTc phn ng4. Nhit ha hc:Nhit phn ngChiu din bin ca phn ng5. Dung dch in li: Nng dung dchpH ca dung dch6. in ha hc:Pinin phn7. Lp cng thc phn t v xc nh nguyn t8. Xc nh thnh phn % ca hn hp XUT1. Ti 250C, phn ng:CH3COOH + C2H5OH CH3COOC2H5+H2Oc hng s cn bng K = 4Ban u ngi ta trn 1,0 mol C2H5OH vi 0,6 mol CH3COOH. Tnh s mol este thu c khi phn ng t ti trng thi cn bng.CH3COOH + C2H5OH CH3COOC2H5+H2O Phn ng x x[ ] 1 x0,6 x x xK = [ ] [ ][ ] [ ]3 2 5 22 5 3CH COOC H H OC H OHCH COOH 2x(1 x)(0,6 x) = 4 3x2 6,4x + 2,4 = 0 x1 = 0,4855 v x2 = 1,64&gt; 1 Vy, s mol este thu c khi phn ng t ti trng thi cn bng = 0,48551. Ti 4000C, P = 10atm phn ngN2(k)+3H2(k) 2NH3 (k) c Kp = 1,64 104.Tm % th tch NH3 trng thi cn bng, gi thit lc u N2(k) v H2(k) c t l s mol theo ng h s ca phng trnh. N2(k)+3H2(k) 2NH3 (k) Theo PTHH:2 22 2N NH HP n1P n 3 Theo gt: P3NH +P2N+ P2H= 10P3NH +4P2N= 10DBV Ta c: Kp = 32 22NH3N H(P )(P )(P ) =2NH33N N2 2(P )(P )(3P ) = 1,64 104 32NH2NP(P ) 6,65 102. Gii pt cho:6,65 102(P2N )2 + 4P2N10 = 0 P2N = 2,404 v P2N = 62,55 &lt; 0Vy, P2N = 2,404 P3NH =10 4P2N= 0,384 atm chim 3,84%2. Hn hp 3 kim loi Fe, Al, Cu nng 17,4 gam. Nu ho tan hn hp bng axit H2SO4 long d th thot ra 8,96 dm3 H2 ( kc). Cn nu ho tan hn hp bng axit H2SO4 c nng, d th thot ra 12,32 dm3 SO2 ( ktc). Tnh khi lng mi kim loi ban u.Cu khng tan trong H2SO4 long . Fe + H2SO4 FeSO4 + H2 2Al + 3H2SO4 Al2(SO4)3 + 3H2H2SO4 c nng ho tan c 3 kim loi :2Fe + 6H2SO4 Fe2(SO4)3+ 3SO2+ 6H2O2Al + 6H2SO4 Al2(SO4)3+ 3SO2+ 6H2OCu + 2H2SO4 CuSO4+ SO2+ 2H2OS mol H2 = 0,4 ; s mol SO2 = 0,55 H 3 phng trnh : 56x + 27y + 64z = 17,4 x + 1,5y = 0,41,5x + 1,5y + z = 0,55Gii h phng trnh cho : x = 0,1; y = 0,2 ; z = 0,1Lng Fe bng 5,6gam ; Al = 5,4gam ; Cu = 6,4gam2.Hn hp 3 kim loi Al, Fe, Cu. Ho tan a gam hn hp bng axit sunfuric c nng va th thot ra 15,68 dm3 SO2 (kc) v nhn c dung dch X. Chia i X, 1 na em c cn nhn c 45,1 gam mui khan, cn 1 na thm NaOH d ri lc kt ta nung trong khng kh n lng khng i cn nng 12 gam. Tm a v khi lng mi kim loi. 2Fe + 6H2SO4 Fe2(SO4)3+ 3SO2+ 6H2O 2Al + 6H2SO4 Al2(SO4)3+ 3SO2+ 6H2O Cu + 2H2SO4 CuSO4+ SO2+ 2H2O Lng 3 mui sunfat = 45,1 x 2 = 90,2 gam v s mol SO2 = 0,7mol Fe2(SO4)3 + 6NaOH 2Fe(OH)3+ 3Na2SO4 Al2(SO4)3 + 6NaOH 2Al(OH)3+ 3Na2SO4CuSO4 + 2NaOH Cu(OH)2+ Na2SO4 Al(OH)3 + NaOH NaAlO2 + 2H2OKt ta lc c ch cn Fe(OH)3 v Cu(OH)22Fe(OH)3 Fe2O3 + 3H2O Cu(OH)2 CuO + H2O 12 x 2 = 24gam l tng lng 2 oxit Fe2O3+CuOH 3 phng trnh: 1,5x + 1,5y + z = 0,7200x + 171y + 160z = 90,2 80x + 80z = 24 Gii h cho x = 0,2 ; y = 0,2 ; z = 0,1 Suy ra lng Fe = 11,2gam ; Al = 5,4gam ; Cu = 6,4gamCTDB(cn cc bi 40, 64, 79, 94 Sch bi dng ha hc THCS )2. Hn hp gm FeCl3, MgCl2, CuCl2 ha tan trong nc c dung dch X. Cho X tc dng vi Na2S d tch ra mt lng kt ta m1. Nu cho mt lng d H2S tc dng vi X tch ra mt lng kt ta m2. Thc nghim cho bit m1 = 2,51m2. Nu gi nguyn lng cc cht MgCl2, CuCl2 trong X v thay FeCl3 bng FeCl2 cng lng ri ha tan trong nc th c dung dch Y.Cho Y tc dng vi Na2S d tch ra mt lng kt ta m3. Nu cho mt lng d H2S tc dng vi Y tch ra mt lng kt ta m4. Thc nghim cho bit m3 = 3,36m4. Xc nh % khi lng mi mui trong hn hp ban u.MgCl2 + Na2S + 2H2O Mg(OH)2 + H2S + 2NaCl 2FeCl3 + 3Na2S2FeS + S + 6NaCl CuCl2+ Na2S CuS +2NaCl MgCl2 + H2Skhng phn ng 2FeCl3 + H2S2FeCl2+ S + 2HCl CuCl2+ H2S CuS +2HCl t s mol cc mui ln lt l x, y, z. Ta c:y58x88y32 96z216y96z+ + ++= 2,5158x + 63,84y = 144,96z(1)S mol FeCl2 = 162,5y127 = 1,28y FeCl2 + Na2S FeS + 2NaCl FeCl2 + H2Skhng phn ng58x88 1,28y96z96z+ += 3,36 58x + 112,64y = 226,56z(2)Gii (1) v (2) cho 48,8y = 81,6zCoi z = 18,8 th y = 48,8 v x = 32,15%MgCl2 = 95 32,15100%95 32,15 162,5 81,6 135 48,8 + + = 13,3%Tnh tng t c: %CuCl2 = 28,76% v%FeCl3 = 57,95%2. Mt hn hp bt kim loi c kh nng gm Mg, Al, Sn. Ha tan ht 0,75 gam hn hp bng dung dch HCl d thy thot ra 784 ml H2 (o ktc). Nu t chy hon ton 0,75 gam hn hp trong oxi d th thu c 1,31 gam oxit. Xc nh % khi lng mi kim loi trong hn hp. Mg+ 2HCl MgCl2 + H2 2Al + 6HCl 2AlCl3 + 3H2 Sn + 2HCl SnCl2 + H2 2Mg + O2 0t 2MgO 4Al + 3O2 0t 2Al2O3 Sn + 2O2 0t SnO2 S mol H2 = 0,035H pt:24x + 27y + 119z = 0,75(x, y, z l s mol tng kim loi)x +32y + z= 0,035 40x + 102y2+ 183z = 1,31Gii h pt cho:x = 0,02;y = 0,01;z = 0Vy, hn hp khng c Sn v % Mg = 0,02 24100%0,75 = 64%;%Al = 36% 3.Mt mu than ly t hang ng ca ngi Plinxian c ti Ha Oai c tc l 13,6 phn hy 14C trong 1 giy tnh vi 1,0 gam cacbon. Bit trong 1,0 gam cacbon ang tn ti c 15,3 phn hy 14C trong 1 giy v chu k bn hy ca 14C l 5730 nm . Hy cho bit nin i ca mu than ?Hng s phng x: k = 12ln2t = 0,6935730 Nin i ca mu than t = 0tN 1 5730 15,3ln lnk N 0,693 13,6 = 973,88 (nm)3. Mt mu than ly t hang ng vng ni vi tnh Ha Bnh c 9,4 phn hy 14C. hy cho bit ngi Vit c i to ra mu than cch y bao nhiu nm? Bit chu k bn hy ca 14C l 5730 nm, trong kh quyn c 15,3 phn hy 14C. Cc s phn hy ni trn u tnh vi 1,0 gam cacbon, xy ra trong 1,0 giy.Hng s phng x: k = 12ln2t = 0,6935730 Nin i ca mu than t = 0tN 1 5730 15,3ln lnk N 0,693 9,4 = 3989,32 (nm)4000 (nm)Ngi Vit c i to ra mu than cch y khong 4000 nm 4. Cu hnh electron ngoi cng ca nguyn t ca nguyn t X l 5p5. T s ntron v in tch ht nhn bng 1,3962. S ntron ca X bng 3,7 ln s ntron ca nguyn t thuc nguyn t Y. Khi cho 4,29 gam Y tc dng vi lng d X thu c 18,26 gam sn phm c cng thc XY. Xc nh in tch ht nhn ca X, Y v vit cu hnh electron ca Y. Cu hnh y ca X l [36Kr] 5s24d105p5.s ZX = 53 = s protonMt khc: xxnp 1,3692 nX = 74 AX = pX+ nX = 53 + 74 = 127</p> <p>xynn= 3,7nY = 20X + Y XY4,2918,26DBCTDBY X Y4,29 18,26+Y 127Y4,29 18,26+ Y = 39 AY = pY+ nY 39 =pY+ 20 pY= 19hay ZY = 19Cu hnh electron ca Y l [18Ar] 4s14. Mi phn t XY3 c tng cc ht proton, ntron, electron bng 196; trong , s ht mang in nhiu hn s ht khng mang in l 60, s ht mang in ca X t hn s ht mang in ca Y l 76.a) Hy xc nh k hiu ho hc ca X,Y v XY3 .b) Vit cu hnh electron ca nguyn t X,Y. a)K hiu s n v in tch ht nhn ca X l Zx , Y l Zy ; s ntron (ht khng mang in) ca X l Nx , Y l Ny . Vi XY3 , ta c cc phng trnh: Tng s ba loi ht: 2 Zx + 6 Zy + Nx + 3 Ny = 196 (1) 2 Zx + 6 ZyNx 3 Ny =60(2) 6 Zy2 Zx=76(3)Cng (1) vi (2) v nhn (3) vi 2, ta c: 4 Zx +12 Zy = 256(a)12 Zy4Zx =152(b)Zy=17;Zx=13Vy X l nhm, Y l clo.XY3 l AlCl3 .b) Cu hnh electron:Al : 1s22s22p63s23p1;Cl : 1s22s22p63s23p5 5. Mt loi khong c cha 13,77%Na; 7,18%Mg; 57,48%O; 2,39%H v cn li l nguyn t X v khi lng. Hy xc nh cng thc phn t ca khong . Hm lng %X = 100 13,77 7,18 57,48 2,39 = 19,18%Cn bng oxi ha kh trong hp cht: 13,77 7,18 57,48 2,39 19,181 2 2 1 y23 24 16 1 X + + + = 0 X = 5,33yLp bng xt:Y 1 2 3 4 5 6 7 8X 5,33 10,66... ... ... 32thy ch c y = 6 l tha mn X = 32 S (lu hunh) Na : Mg : O : H : S =13,77 7,18 57,48 2,39 19,18: : : :23 24 16 1 32 = 2 : 1 : 12 : 8 : 2Cng thc khong:Na2MgO12H8S2Na2SO4.MgSO4.4H2O5.Mt khong cht c cha 20,93% Nhm; 21,7% Silic v cn li l oxi v Hidro (v khi lng). Hy xc nh cng thc ca khong cht ny. t % lng Oxi = a th % lng Hidro = 57,37 aTa c: t l s nguyn tAl : Si : O : H = 20, 93 21, 7 a: : : (57, 37 a)27 28 16Mt khc: phn t khong cht trung ha in nn DBCTCT</p> <p>20, 93 21, 7 a3 4 2 (57, 37 a) 027 28 16 + + Gii phng trnh cho a = 55,82Suy ra, Al : Si : O : H =20, 93 21, 7 55,82: : :1, 5527 28 16 = 2 : 2 : 9 : 4Vy cng thc khong cht Al2Si2O9H4hayAl2O3.2SiO2.2H2O (Cao lanh)6. Tinh th ng kim loi c cu trc lp phng tm din. a) Hy v cu trc mng t bo c s v cho bit s nguyn t Cu cha trong t bo s ng nyb) Tnh cnh lp phng a() ca mng tinh th, bit nguyn t Cu c bn knh bng 1,28 c) Xc nh khong cch gn nht gia hai nguyn t Cu trong mngd) Tnh khi lng ring ca Cu theo g/cm3 a) Mng t bo c s ca Cu (hnh bn)Theo hnh v, s nguyn t Cu l tm nh lp phng = 8 18 = 1 6 mt lp phng = 6 12 = 3Vy tng s nguyn t Cu cha trong t bo s ng = 1 + 3 = 4 (nguyn t)b) Xt mt lp phng ABCD ta c: AC = a 2 = 4 rCu a = 0Cu4 r 4 1,28A2 2 3,63 c) Khong cch ngn nht gia 2 nguyn t l on AE:AE = AC a 22 2= 2,55 d) Khi lng ring: + 1 mol Cu = 64 gam + Th tch ca 1 t bo c s = a3cha 4 nguyn t Cu+1 mol Cu c NA = 6,02 1023 nguyn t Khi lng ring d = mV= 4 23 8 3646,02 10 (3,63 10 ) = 8,88 g/cm36. St dng (Fe) kt tinh trong mng lp phng tm khi, nguyn t c bn knh r = 1,24 . Hy tnh:a) Cnh a ca t bo s ng b) T khi ca Fe theo g/cm3.c) Khong cch ngn nht gia hai nguyn t FeCho Fe = 56 a) Mng t bo c s ca Fe (hnh v)Theo hnh v, s nguyn t Fe l tm nh lp phng = 8 18 = 1ABCDaEABCDaD CABEEaD CABECTDB tm lp phng = 1Vy tng s nguyn t Cu cha trong t bo s ng = 1 + 1 = 2 (nguyn t)b) T hnh v, ta c: AD2 = a2 + a2= 2a2xt mt ABCD: AC2 =a2 + AD2 = 3a2mt khc, ta thy AC = 4r = a3 nn a = 4r3 = 4 1,243 = 2,85 c) Khong cch ngn nht gia 2 nguyn t l on AE:AE = AC a 32 2= 2,85 32 = 2,468 d) Khi lng ring: + 1 mol Fe = 56 gam + Th tch ca 1 t bo c s = a3cha 2 nguyn t Fe+1 mol Fe c NA = 6,02 1023 nguyn t Khi lng ring d = mV= 2 23 8 3566,02 10 (2,85 10 ) = 7,95 g/cm37. Cho rng ht nhn nguyn t v chnh nguyn t H c dng hnh cu. Ht nhn nguyn t hiro c bn knh gn ng bng 1015 m, bn knh nguyn t hiro bng 0,53 1010 m.Hy xc nh khi lng ring ca ht nhn v nguyn t hiro.(cho khi lng proton = khi lng ntron 1,672 1027 kgkhi lng electron = 9,109 1031 kg) Khi lng ht nhn nguyn t hiro chnh l khi lng ca proton = 1,672 1027 kg+ Th tch ht nhn nguyn t hiro bngV =34 4r3 3 3,14 (1015)3 = 4,19 1045 (m3)Khi lng ring ca ht nhn nguyn t hiro bng:D =27451,672 104,19 10 = 3,99 108 (tn/m3)+ Th tch gn ng ca nguyn t hiro l:10 34 3,14 (0,53 10 )3 = 0,63 1030 (m3)+ Khi lng ca nguyn t hiro (tnh c khi lng ca electron) = 1,673 1027 kgKhi lng ring ca nguyn t hiro bng27301,673 100,63 10 = 2,66 103 (kg/m3) = 2,66 103 (g/cm3)7. Tnh bn knh nguyn t gn ng ca Ca 200C, bit ti nhit khi lng ring ca Ca bng 1,55 g/cm3. Gi thit trong tinh th cc nguyn t Ca c hnh cu, c c kht l 74%. Cho nguyn t khi ca Ca = 40,08 Th tch ca 1 mol Ca = 40,081,55 = 25,858 cm3, mt mol Ca cha NA = 6,02 1023 nguyn t CaTheo c kht, th tch ca 1 nguyn t Fe = 2325,858 0,746,02 10 = 3,18 1023 cm3CTT V = 34r3 Bn knh nguyn t Ca = r =3 3V4 =233 3 3,18 104 3,14 = 1,965 108 cm7. Tnh bn knh nguyn t gn ng ca Fe 200C, bit ti nhit khi lng ring ca Fe bng 7,87 g/cm3. Gi thit trong tinh th cc nguyn t Fe c hnh cu, c c kht l 68%. Cho nguyn t khi ca 55,85 = 40 Th tch ca 1 mol Fe = 55,857,87 = 7,097 cm3. mt mol Fe cha NA = 6,02 1023 nguyn t FeTheo c kht, th tch ca 1 nguyn t Fe = 237,097 0,686,02 10 = 0,8 1023 cm3T V = 34r3 Bn knh nguyn t Fe = r =3 3V4 =233 3 0,8 104 3,14 = 1,24 108 cm8. Bit rng mono clobenzen c momen lng cc 1 = 1,53 D.a) Hy tnh momen lng cc o ; m ; p ca ortho, meta, para diclobenzen.b) o momen lng cc ca mt trong ba ng phn c = 1,53 D. Hi l dng no ca diclobenzen? clo c m in ln, 1 hng t nhn ra ngoi ortho meta para= 3 = = 0Cng vect s dng h thc lng trong tam gic a2 = b2 + c2 2bc cos ADn xut ortho: o = 2 2 01 12 2cos60 + = 1 3Dn xut meta: m = 2 2 01 12 2cos120 + = 1Dn xut para: p = 1 1= 0b) Theo u bi=1,53D = 1 l dn xut meta -diclobenzen8. Clobenzen c momen lng cc 1 = 1,53 D ( 1 hng t nhn ra ngoi); anilin c momen lng cc 2 = 1,60D ( 2 hng t ngoi vo nhn benzen). Hy tnh ca ortho cloanilin; meta cloanilin v para cloanilin. clo c m in ln, 1 hng t nhn ra ngoi nhm NH2 c cp e t do lin hp vi h e ca vng benzen hai momen lng cc cng chiu DB ortho meta paraCng vect s dng h thc lng trong tam gic a2 = b2 + c2 2bc cos ADn xut ortho: 2O = 21 + 22 2 1 2 cos 600 =21 + 22 1 2 = 2,45 o =2,45 = 1,65DDn xut meta: 2m = 21 + 22 2 1 2 cos 1200 =21 + 22 + 1 2 = 7,35 m =7,35 = 2,71DDn xut para: 2p = 1 + 2=1,60 + 1,53 = 3,13D9. Tnh pH ca dung dch benzoatnatri C6H5COONa nng 2,0 105 M. Bit hng s axit ca axit benzoic bng 6,29 105. C6H5COONaNa+ + C6H5COO C6H5COO +H+</p> <p>C6H5COOHKa1 H2O H++OH KwT hp 2 phng trnh cho: C6H5COO +H2OC6H5COOH+OH KtpKtp = waKK = 145106,29 10= 1,59 1010Donng u ca C6H5COO nh; mt khc hng s ca qu trnh khng ln hn nhiu so vi 1014 nn phi tnh n s in li ca nc.C6H5COO +H2OC6H5COOH+OH Ktp(1) 2,0 105 [OH] H2O H++OH Kw(2)Theo nh lut bo ton in tch: [OH] = [C6H5COOH] + [H+] hay[C6H5COOH] =[OH] [H+] = [OH] 1410OH 1 ]thay vo biu thc hng s cn bng ca (1): K =[ ]6 56 5C H COOHOHC H COO 1 ] 1 ] = 146 510OH OHOHC H COO _ 11 ] ] 1 ] , 1 ] = 1,59 1010 2145OH 102 10 OH 1 ] 1 ] = 1,59 1010 [OH]2 + 1,59 1010[OH] 13,18 1015 = 0CTDB[OH] =1,148 107 pOH = lg(1,148 107) = 6,94 pH = 7,069. 200C ha tan vo dung dch NaOH nng 0,016 g/lt mt lng iot phn ng sau xy ra hon ton:2NaOH +I2 NaI+ NaIO+ H2O Tnh pH ca dung dch thu c. Bit hng s axit ca HIO = 2,0 1011 Nng u ca OH = 0,01640 = 4,0 104 mol/lt Phn ng 2OH +I2I + IO + H2O 4,0 104 2,0 104 IO + H2OHIO+ OH [ ] 2,0 104 xx x [HIO] =[OH] HIO H++IO</p> <p> Ka =2,0 1011 Ta c:Ka = [ ]IO HHIO + 11 ] ] = 2,0 1011 IO HOH + 11 ] ] 1 ] = 2,0 1011 4(2,0 10 OH) HOH + 11 ] ] 1 ] = 1441410(2,0 10 ) HH10H +++ 1 ] 1 ] 1 ] = 2,0 1011 2,0 1014[H+]2 1,0 1014[H+] 2,0 1025 = 0 [H+] = 6,53 1011 pH = lg[H+] = lg(6,53 1011)= 10,18510.Hn hp A gm 3 este n chc, mch thng, to thnh t cng mt ru B vi 3 axit hu c, trong c hai axit no l ng ng k tip nhau v mt axit khng no cha mt lin kt i. X phng ho hon ton 14,7 gam A bng dung dch NaOH, thu c hn hp mui v p gam ru B. Cho p gam ru B vo bnh ng natri d, sau phn ng c 2,24 lt kh thot ra v khi lng bnh ng natri tng 6,2 gam. Mt khc t chy hon ton 14,7 gam A, thu c 13,44 lt CO2 v 9,9 gam H2O. Xc nh cng thc cu to ca tng este trong A. (Cc th tch kh o iu kin tiu chun). Xc nh ru B:v este n chc nn ru B n chcR OH+NaR ONa+12H2</p> <p> 0,20,1 mol tng KL = KL (R O)= 6,2 gKL mol (R O) = 6,20,2 = 31R + 16 = 31R = 15 l CH3 Ru B:CH3OHCng thc ca 2 este no l: CnH2n 1 +COOCH3 s mol = xCng thc ca este cha no l CmH2m1COOCH3s mol = yCnH2n 1 +COOCH3 + 3n42+O2 (n + 2) CO2 +(n + 2) H2O x (n + 2) x (n + 2) x CTCmH2m1COOCH3+3m 32+O2 (m + 2) CO2 + (m + 1) H2O y (m + 2) y(m + 1) yta c h pt: x+y= 0,2 (1) (n + 2) x+ (m + 2) y= 0,6 (2)(n + 2) x+ (m + 1) y= 0,55 (3)Gii h pt cho x =0,15 ;y =0,05v3n + m = 4Do n 0 v m 2 nn 2 m 3bi ton c 2 nghim m = 2 v m = 3Vi m = 2n = 23 ng vi nghim HCOOCH3 ;CH3COOCH3 v CH2=CH-COOCH3Vi m = 3n = 13 ng vi nghim HCOOCH3 ;CH3COOCH3 v C3H5-COOCH310.Nitrosyl clorua l mt cht rt c, khi un nng s phn hu thnh nit monoxit v clo.a) Hy vit phng trnh cho phn ng nyb) Tnh Kp ca phn ng 298K(theo atm v theo Pa). Cho:Nitrosyl clorua Nit monoxit Cl2 Ho298(kJ/mol)51,71 90,25 ?S0298 (J/K.mol)264 211 223c) Tnh gn ng Kp ca phn ng 475K a) 2NOCl 2NO+ Cl2.b) Hng s cn bng nhit ng lc hc c tnh theo phng trnh G = RTlnKTrong G = H T. S H = [(2 90,25. 103) + 0 (2 51,71. 103 ) = 77080 J/mol S= [(2 211) + 233 (2 264) = 117 J/mol G = 77080 298 117 = 42214 J/mol vln K = 422148, 314 298 = 17 Kp = 3,98. 108 atm vKp = 4,04. 103 Pa c) Tnh gn ng: ln 21( )( )Kp TKp T= 1 21 1 HR T T _ , lnKp(475K) = 77080 1 18, 314 298 475 _ ,+ lnKp(298) ln Kp (475) = 5,545 Kp = 4,32. 10 3 atm hay Kp = 437PaCT</p>