化学动力学第二章习题和答案

1、化学反应动力学第二章习题-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：一级反应动力学方程为: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 则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)反应达 1 小时时：SO2Cl220宀60=0.924=92.4%SO2CI2 *已分解的百分数为: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)设反应级数为 n,则0.06450.05360.0432dAdtAAn4=ktSO2Cl2 =SO2CI2 e当t一n丄= 31506.7sk 2二kAn01 1 1若n=2,则両）t918244054k0.10400.12290.14870.15090.1701若n=1.5t91824k0.01650.01890.0222若n=3t91824k2.0672.603

5、.46反应为一级。-1-4-1(2)k = 0.0123 min = 2.05 10 s(3)110.693= 56.3 min = 3378 s10.01233、已知复杂反应：,k1A1A2+ A3的速率方程为-dA1】k1A-kA2A3，推导其动力学方程。要求写出详细的推dt导过程。解：设t =0时，A =A , A2，A3= A3t=t 时，AJXA -X，阳二地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则x2，移项积分:dx0_dx_(x丄)(x一 _ “q - B q + P(xx xq2了In x 2+P-q x 2Y得动力学方程：In上x +-L24、已知复杂反应由下列两个基元反应组成:A2+ A1求反应进行过程中，A1物种浓度与 A3物种浓度间的关系。要求写出详细的推导过 程。解：速率方程：ki考虑物料平衡：A A?AJ -AJ-2A3，代入上式,得AiA3关系式为:严 1 J1_k2(A2kmAJ-QA3) 一(【A? . A-A, -2/3 -地)=A3 k2k,-k2A2+nk1八A

7、1A2kdA2*内-k2A.A2dt(1)dAdtWA1A2(1)得:得:(2)设t =0dA人讥內dgk2A2时，A2 =A2 ,乓=0,移项积分:門k2 A2dA2 =A2&七血【A】、dA3kiA2A2(AT1)dA2=A3k1_k2 A2-料 nkl-炽屮2!/_知一2阳)_(阳.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时，A “A?,阳二入。，则: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.对 X 进行稳态近似，则 鱼

9、卷二0dt即: XHkiABkB + k2(ki_ kB k2)AB_k4Bk2AB即：业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机理为：k1NO+O NO3快速平衡k-1k2NO3+ NO2NO2决速步据快速平衡：NO3k1NOgJNO3叮NOO2k_LdNO2=2k2NO3】NO二2k1k2NO2dtkj(b)(1)NO NO、N2O2k1(2) N2O2 2NOk2(3)N2O2O2 2NO2k3对N2O2进行稳态近似dN22二kNO2-k2N2O2 -k3N2O2O2 =0 dt若只有很少量的 N2O2转变为 NO2,而绝大部分转变为 NO,即 k2 k3O2,贝 y：N2O2k/NO2k2k3O2dNO2dt=

13、 2k3N2O2O22k1k3NO2O2k2k3O2-iEs(overall) = Eai+ Ea3- Ea2=79.5 + 84 -205 = - 41.5 kJ- mol(c)(1)检测中间体 N2O2或 NO3(2)大大增加 O2的浓度第一历程为： WfNOgdtk第二历程为：dt测定速率常数大小是否与 O2浓度有关(3) 作 r O2(固定NO,测不同O2下的反应速率)第一历程为直线， 第二历程不为线性。常用来描述酶催化反应和热活化单分子反应，若其总包反应的反应速率方程为:(2)当 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总包反应产生 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拟定该反应的反应机理。要求写出详细推导过程解：写出反应体系可能存在的各基元反应：根据活化能大小，拟定可能的反应机理:(kJ/mol)20.90.00.00.0= EalEa2Rdln(k-12)E ERd(k k2)=Ea1Ea2代入（6）得：Hss曲4)也k3Cl212对活性中间体 H、Cl 应用稳态近似：=2k1Cl2Hk2ClH2 k3H Cl2 - 2k4Cl 2= 0 dt響十0屮2十屮灿=0dt(5)+(6)得：k1Cl2-k4【CI2=0 二Clss二Cl?12k4(5)(6)则有：dtk4k3Cl2/2= 2k2（A）i2H2Cl2i2k4则：2dH5C=