1、化學反應動力學第二章習題-5-11、The first-order gas reaction SOCb SO2+ CI2has k = 2.20 10 s at 593K,(1) What percent of a sample of SO2CI2would be decomposed by heating at 593K for 1hour?How long will it take for half the SO2CI2to decompose?解：一級反應動力學方程為:0.693t1亍=31500 s = 8.75 hour22.2 102、T-butyl bromide is conv

2、 erted into t-butyl alcohol in a solve nt containing 90 perce nt acetone and 10 perce nt water. The reacti on is give n by(CH3)3CBr + H2O (CH3)3COH + HBrThe following table gives the data for the concentration of t-utyl bromide versus time:若 I 則k10.105610.1056t =9kIn0.01047， t = 18kIn0.0116790.09611

3、80.085610.105610.1056t =24kIn-0.01332， t = 40kIn0.01232240.0767400.0645t=54k= 0.01256，t = 72k =0.01241t :=105k= 0.01299SO2CI2eSO2CI2(1)反應達 1 小時時：SO2Cl220宀60=0.924=92.4%SO2CI2 *已分解的百分數為:100%-92.4%=7.6%SO2CI21SQCI2 - 2T(min)1824405472105(CH3)CBr (mol/L) 0.10560.09610.08560.07670.0270(1) What is the or

4、der of the reaction?(2) What is the rate con sta nt of the react ion?(3) What is the half-life of the reaction?解：(1)設反應級數為 n,則0.06450.05360.0432dAdtAAn4=ktSO2Cl2 =SO2CI2 e當t一n丄= 31506.7sk 2二kAn01 1 1若n=2,則両）t918244054k0.10400.12290.14870.15090.1701若n=1.5t91824k0.01650.01890.0222若n=3t91824k2.0672.603

5、.46反應為一級。-1-4-1(2)k = 0.0123 min = 2.05 10 s(3)110.693= 56.3 min = 3378 s10.01233、已知復雜反應：,k1A1A2+ A3的速率方程為-dA1】k1A-kA2A3，推導其動力學方程。要求寫出詳細的推dt導過程。解：設t =0時，A =A , A2，A3= A3t=t 時，AJXA -X，陽二地x , A3】=A3 x代入dA1kdA -kA2A3得：dx k1(A -x)-k(A2x)(A3x)dt二 kA -kX -kA2 A _ kA3 x _ k JA2 x _ k二 kJA -kJ A A3 - (kk A3

6、九 kA2 )x - k令a= k - k/A? A， B=匕kjAsk/A?，丫= - kdx則x2，移項積分:dx0_dx_(x丄)(x一 _ “q - B q + P(xx xq2了In x 2+P-q x 2Y得動力學方程：In上x +-L24、已知復雜反應由下列兩個基元反應組成:A2+ A1求反應進行過程中，A1物種濃度與 A3物種濃度間的關系。要求寫出詳細的推導過 程。解：速率方程：ki考慮物料平衡：A A?AJ -AJ-2A3，代入上式,得AiA3關系式為:嚴 1 J1_k2(A2kmAJ-QA3) 一(【A? . A-A, -2/3 -地)=A3 k2k,-k2A2+nk1八A

7、1A2kdA2*內-k2A.A2dt(1)dAdtWA1A2(1)得:得:(2)設t =0dA人譏內dgk2A2時，A2 =A2 ,乓=0,移項積分:門k2 A2dA2 =A2&七血【A】、dA3kiA2A2(AT1)dA2=A3k1_k2 A2-料 nkl-熾屮2!/_知一2陽)_(陽.A._AI_2A3-A2) = A3 k2ki- k2【A2即：kiK k2(A2h +Ah A2民)A3丄1 n-1 22113AiAk2ki-k2【A2k若t =0時，A “A?,陽二入。，則:ki|nK kzUAJr+IAbA2民)k2k1_k2A2；5、Consider the reactio

8、n mechanism kiA + BX + Bk-ik2X -C + Di.Write chemical rate equatio ns for A and X.ii.Employi ng the steady-state approximatio n, show that an effective rate equation for A isdA/dt = -keffABiii.Give an expressi on forlefi n terms of ki, k-i, k2, and B.解：i宀二kiAB -kXBdt警二kiAB-kXB-k2Xdtii.對 X 進行穩態近似，則 魚

9、卷二0dt即: XHkiABkB + k2(ki_ kB k2)AB_k4Bk2AB即：業AJ二_keffABdtASAI；-A AS。dAdt= kiAB-kBkiABk4B k2MBkB k26、(a) The reaction 2 NO + O2 2 NO2is third order. Assuming that a small amount of NQexists in rapid reversible equilibrium with NO and 02 and that the rate-determining stepis the slow bimolecular reacti

10、on NO + NO2 NO2, derive the rate equation for this themecha ni sm.(b) Ano ther possible mecha nism for the react ion 2 NO +O2 NO2is(1) NO + NO N2O2k1N2O2 2 NOk2(3) N2O2+ O2 2 NO2k3Apply the steady state approximati on to N?O2 to obtai n the rate law for dNO2/dt.If only a very small fraction of the N

11、2O2formed in (1) gose to form products in reaction (3), while most of the N2O2reverts to NO in react ion (2), and if the activati on energies are E = 79.5 kJ/mol, E2= 205 kJ/mol, and 巳=84 kJ/mol, what is the overallactivatio n en ergy?(c) How would you disti nguish experime ntally betwee n the mecha

12、 nism suggested inpart (a) and (b)?解：(a)2NO O2 2NO2機理為：k1NO+O NO3快速平衡k-1k2NO3+ NO2NO2決速步據快速平衡：NO3k1NOgJNO3叮NOO2k_LdNO2=2k2NO3】NO二2k1k2NO2dtkj(b)(1)NO NO、N2O2k1(2) N2O2 2NOk2(3)N2O2O2 2NO2k3對N2O2進行穩態近似dN22二kNO2-k2N2O2 -k3N2O2O2 =0 dt若只有很少量的 N2O2轉變為 NO2,而絕大部分轉變為 NO,即 k2 k3O2,貝 y：N2O2k/NO2k2k3O2dNO2dt=

13、 2k3N2O2O22k1k3NO2O2k2k3O2-iEs(overall) = Eai+ Ea3- Ea2=79.5 + 84 -205 = - 41.5 kJ- mol(c)(1)檢測中間體 N2O2或 NO3(2)大大增加 O2的濃度第一歷程為： WfNOgdtk第二歷程為：dt測定速率常數大小是否與 O2濃度有關(3) 作 r O2(固定NO,測不同O2下的反應速率)第一歷程為直線， 第二歷程不為線性。常用來描述酶催化反應和熱活化單分子反應，若其總包反應的反應速率方程為:(2)當 k-1 Cl Cl117.2(1)k1Cl H2 H HCl25.1(2)k2H Cl2 HCl Cl8

14、.415.1(3)k3Cl Cl M Cl2M0.0k4總包反應產生 HCI 的生成速率：=k2【ClH2 k3【HCl2 dtEa( kJ/mol)EaCI2 Cl Cl117.2H HCl Cl H2H2H H436Cl Cl M Cl2MCl H2 H HCl25.1H Cl M HOLMH Cl2 HCl Cl8.415.1H H M H2M擬定該反應的反應機理。要求寫出詳細推導過程解：寫出反應體系可能存在的各基元反應：根據活化能大小，擬定可能的反應機理:(kJ/mol)20.90.00.00.0= EalEa2Rdln(k-12)E ERd(k k2)=Ea1Ea2代入（6）得：Hss曲4)也k3Cl212對活性中間體 H、Cl 應用穩態近似：=2k1Cl2Hk2ClH2 k3H Cl2 - 2k4Cl 2= 0 dt響十0屮2十屮燦=0dt(5)+(6)得：k1Cl2-k4【CI2=0 二Clss二Cl?12k4(5)(6)則有：dtk4k3Cl2/2= 2k2（A）i2H2Cl2i2k4則：2dH5C=