化学反应动力学第二章习题参考答案

化学反应动力学 第二章习题1、The first-order gas reaction SO 2Cl 2 → SO 2 + Cl 2 has k = 2.20 ⨯ 10-5 s -1 at 593K,(1) What percent of a sample of SO 2Cl 2 would be decomposed by heating at 593K for 1 hour?(2) How long will it take for half the SO 2Cl 2 to decompose? 解：一级反应动力学方程为：t k e Cl SO Cl SO ⋅-⋅=ο][][2222 ⇒t k e Cl SO Cl SO ⋅-=ο][][2222(1) 反应达1小时时：60601020.222225][][⨯⨯⨯--=e Cl SO Cl SO ο=0.924=92.4%已分解的百分数为：100%-92.4%=7.6% (2) 当21][][2222=οCl SO Cl SO 时，7.3150621ln 1=-=k t s521102.2693.0-⨯=t = 31500 s = 8.75 hour2、T-butyl bromide is converted into t-butyl alcohol in a solvent containing 90 percent acetone and 10 percent water. The reaction is given by(CH 3)3CBr + H 2O → (CH 3)3COH + HBrThe following table gives the data for the concentration of t-utyl bromide versus time:T(min) 0 9 18 24 40 54 72 105(CH 3)CBr (mol/L) 0.1056 0.0961 0.0856 0.0767 0.0645 0.0536 0.0432 0.0270(1) What is the order of the reaction?(2) What is the rate constant of the reaction? (3) What is the half-life of the reaction? 解： (1) 设反应级数为 n ，则 n A k dtA d ][][=-⇒kt A A n n =---11][1][1ο 若 n=1，则 ][][ln 1A A t k ο=t = 9 01047.00961.01056.0ln91==k ， t = 18 01167.00856.01056.0ln 181==k t = 24 01332.00767.01056.0ln 241==k ， t = 40 01232.00645.01056.0ln 401==k t = 54 01256.0=k ， t = 72 01241.0=k ， t = 105 01299.0=k若 n=2，则 )][1][1(1οA A t k -= t ： 9 18 24 40 54 k ： 0.1040 0.1229 0.1487 0.1509 0.1701 若 n=1.5t ： 9 18 24 k ： 0.0165 0.0189 0.0222 若 n=3t ： 9 18 24 k ： 2.067 2.60 3.46反应为一级。

(2) k = 0.0123 min -1= 2.05×10-4 s -1(3)0123.0693.021=t = 56.3 min = 3378 s3、已知复杂反应：的速率方程为]][[][][321111A A k A k dt A d --=-，推导其动力学方程。

要求写出详细的推导过程。

解：设 0=t 时， ο][][11A A = ，ο][][22A A = ，ο][][33A A =t t = 时， x A A -=ο][][11 ，x A A +=ο][][22 ，x A A +=ο][][33 代入 ]][[][][321111A A k A k dtA d --=- 得：)])([]([)]([32111x A x A k x A k dtdx++--=-οοο 212131321111][][][][][x k x A k x A k A A k x k A k ---------=οοοοο 212131132111)][][(][][][x k x A k A k k A A k A k -----++--=οοοοο 令 α = οοο][][][32111A A k A k -- ， β = οο][][21311A k A k k --++ ， γ = 1--k则2x x dtdxγβα++= ， 移项积分： ⎰⎰=++x t dt x x dx002γβα⎰=-----+--xt x x dx22)24)(24(γαγββγαγββA 1A 2 + A 3k k -1令 αγβ42-=q ，⎰=++--xt q x q x dx)2)(2(γβγβqt q x q x x =++--22lnγβγβ得动力学方程：qt q q q x qx =+--++-+}{ln }22{lnββγβγβ 4、已知复杂反应由下列两个基元反应组成：求反应进行过程中，A 1物种浓度与A 3物种浓度间的关系。

要求写出详细的推导过程。

解：速率方程： ]][[][][212112A A k A k dtA d -= （1） ]][[][2123A A k dtA d = （2）)2()1(，得：][][][][2222132A k A k k A d A d -= 设 0=t 时，ο][][22A A = ，0][3=A ， 移项积分：232[][]2223[]0122[][][][]A A A k A d A d A k k A =-⎰⎰o⎰=--][][32221122][][)1][(A A A A d A k k k ο][)][]([][][ln 32222122121A A A A k k A k k k k =-----οο考虑物料平衡： ][2][][][][31122A A A A A --+=οο，代入上式， 得[A 1]~[A 3]关系式为：][)][][2][][]([][])[2][][]([ln 32311222131122121A A A A A A A k k A A A A k k k k=---+----+--οοοοοο A 1A 321A 2A 1A 2 +][)][][2][][]([][])[2][][]([ln 32311222131122121A A A A A A A k k A A A A k k k k =---+----+--οοοοοο 即：12211313112122([][][]2[])[]ln [][][]k k A A A A k A A A k k k A -+--=+--o o o o若0=t 时，ο][][22A A = ，330[][]A A =，则：1221131311302122([][][]2[])[]ln [][][][]k k A A A A k A A A A k k k A -+--=+-+-o o o o5、Consider the reaction mechanismk -1k 1k X C + D A + B X + Bi. Write chemical rate equations for [A] and [X].ii.Employing the steady-state approximation, show that an effective rate equation for [A] isd[A]/dt = -k eff [A][B]iii. Give an expression for k eff in terms of k 1, k -1, k 2, and [B].解：ⅰ. ]][[]][[][11B X k B A k dt A d --=-][]][[]][[][211X k B X k B A k dtX d --=-ⅱ. 对X 进行稳态近似，则 0][=dt X d即：211][]][[][k B k B A k X +=-21111][]][[][]][[][k B k B A k B k B A k dt A d +-=--- ]][[][]][)[][][(212121111B A k B k k k B A k B k B k k k +=+-=---即：]][[][B A k dtA d eff -= ⅲ. 2121][k B k k k k eff +=-6、(a) The reaction 2 NO + O 2 → 2 NO 2 is third order. Assuming that a small amount of NO 3exists in rapid reversible equilibrium with NO and O2 and that the rate-determining step is the slow bimolecular reaction NO 3 + NO → 2 NO 2, derive the rate equation for this the mechanism.(b) Another possible mechanism for the reaction 2 NO + O 2 → 2 NO 2 is (1) NO + NO → N 2O 2 k 1 (2) N 2O 2 → 2 NO k 2(3) N 2O 2 + O 2 → 2 NO 2 k 3Apply the steady state approximation to [N 2O 2] to obtain the rate law for d[NO 2]/dt. If only a very small fraction of the N 2O 2 formed in (1) gose to form products in reaction (3), while most of the N 2O 2 reverts to NO in reaction (2), and if the activation energies are E 1 = 79.5 kJ/mol, E 2 = 205 kJ/mol, and E 3 = 84 kJ/mol, what is the overall activation energy?(c) How would you distinguish experimentally between the mechanism suggested in part (a) and (b)?解：（a ） 2222NO O NO →+ 机理为：NO+O 2NO 31-1快速平衡NO 3+ NO2NO 2k 2决速步据快速平衡：1123]][[][-=k kO NO NO]][[][2113O NO k k NO -=][][2]][[2][22121322O NO k kk NO NO k dt NO d -== (b) （1）22O N NO NO →+ k 1 （2）NO O N 222→ k 2 （3）22222NO O O N →+ k 3 对 [N 2O 2] 进行稳态近似0]][[][][][22232222122=--=O O N k O N k NO k dtO N d ][][][2322122O k k NO k O N +=][][][2]][[2][232223122232O k k O NO k k O O N k dt NO d +== 若只有很少量的N 2O 2转变为 NO 2，而绝大部分转变为 NO ，即k 2 >> k 3 [O 2], 则：][][2][222312O NO k k k dt NO d = E a (overall) = E a1+ E a3 - E a2 =79.5 + 84 -205 = - 41.5 kJ ·mol -1(c) (1) 检测中间体 N 2O 2 或 NO 3（2）大大增加O 2的浓度第一历程为：][][2][221212O NO k k k dt NO d -= 第二历程为：212][2][NO k dtNO d = 测定速率常数大小是否与O 2浓度有关。