化学反应动力学第六、七、九章习题和答案

《化学反应动力学》 第六、七、九章习题1. Calculate k (T) given the following forms for the reactive cross section:(a)бR (v) =πd 2, i.e., a constant. If d is the molecular diameter, this represents an “encounter limited rate coefficient”. (b)бR (E) =б0 ( 1 - e - a E )Do any of t hese calculated k’s have an Arrhenius -like temperature dependence?解：（a ）反应截面与反应速率常数的关系为： k(u) =бR u R 其中，u R ：相对平动速率 解法1：因反应截面为常数，故可先求出平均速率，然后据k(T) =бR <u R >求k(T) 根据气体分子运动论：⎰⎰∞∞-=>=<022234)2()(2du u e Tk u du u uf u Tk uB B ππμμ⎰∞-=0222232)2(2du u e Tk Tk uB B μπμπ232222222()()()()222B u k TB B B Bk T eu d uk T k T k T μμμμππμ∞-=---⎰21)8(πμTk B =其中：)(u f ：Maxwell-Boltzmann 分布函数 μ：折合质量。

则有：k(T) =бR <u R >212212)8()8(μππμπTk d Tk d B B ==解法2：232222R 00k(T)()d ()42B uk TB uf u du u e u du k Tμμσπππ∞∞-==⎰⎰232223204d ()2B uk TB e u du k Tμμππ∞-=⎰1228()B k Td ππμ=(b)232a 22R 00k(T)()(1e )()42B uk TB uf u du u e u duk Tμεμσσππ∞--==-⎰⎰1/2a 00B 18()(1e )k T B k T B e d k Tεεσεεπμ∞--=-⎰(a 1)1/2000B 8(){}k T B B k Tk T B e d e d k Tεεσεεεεπμ∞∞--+=-⎰⎰1/220B 2B B 81()(k T)-1k T (a+)k T B k T σπμ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦1/220B 81()1-()ak T+1B k Tσπμ⎡⎤=⎢⎥⎣⎦Arrhenius 公式为：)exp(RTE A k a -=，k 与T 呈指数关系，随T 的增加，k 呈指数增长。

而(a) k(T)∝T 1/2 , (b) 1/220B 81k(T)()1-()ak T+1B k Tσπμ⎡⎤=⎢⎥⎣⎦, 与Arrhenius 公式中k 与T 的关系不同。

2. For a Lennard-Jones “ 6 - 12 ” potential determine the internuclear distance at the potential energy minimum.解：Lennard-Jones 势能函数为：])()[(4)(612rr r V σσε-=其中：ε：Lennard-Jones 势能函数的势阱深度 r ：分子间距离σ：V(r)=0, r ≠∞时的r 值。

则：]612[4761312rr dr dV σσε+-=当0=drdV时，势能V 为最小值， 0]612[4761312=+-=rr dr dV σσε 得：σ612=r 即，势能最小时，核间距 σ612=r3. The following reaction was done experimentally at various temperatures in order to obtain rate constants:O + HO 2 → OH + O 2T 1 = 298 K k 1 = 6.1 x 10-11 cm 3 / molecule ·sT 1 = 229 K k 2 = 7.57 x 10-11 cm 3 / molecule ·s(a) Using the experimental data, find the activation energy E act .(b) Calculate the following thermodynamic properties at T = 298 K using transition state theory: ΔE ≠, ΔH ≠, ΔS ≠, ΔG ≠.解：(a) 由Arrhenius 公式 )exp(RTEA k a -=得：)exp(11RT EA k a -= (1))exp(22RT EA k a -= (2))1()2(：})11exp{()exp()exp(211212a a aE RT RT RT E A RT E A k k -=--= 则：211111221121112ln 8.3142982297.5710ln ln 1.775k mol 229298 6.110a k k RTT k E J T T k RT RT ---⨯⨯⨯====-⋅--⨯-（b ）考虑恒温恒压过程:因其为气相反应，标准态通常选为1个大气压，故需据kc 计算出kp ： 11-3239c p(298K)k 6.11010 6.02310k 1.50410R 'T 0.082298-⨯⨯⨯⨯===⨯⨯ 据B k T G exp()h k RT≠-∆=，得：9-341298-23B k h 1.50410 6.62610G RTln()8.314298ln 20.63kJ mol T 1.3810298k ≠-⨯⨯⨯∆=-=-⨯=⋅⨯⨯ 或：由过渡态理论速率常数表达式： B k T k K h ≠=得：Tk hk K B =≠ 又：)(ln θ≠≠-=∆K RT G ，因其为气相反应，标准态通常选为1个大气压，22()/[](/)(/)()()O HO O HO K p p p c RT Kp p p p p c RT c RT RTθθθθθθ≠≠≠≠===B hk p k T RT θ=⨯ 故：ln()B hk p G RT k T RTθ≠∆=-⨯34231165236.62610 6.02310 6.11010 1.013108.314298ln()1.38102988.314298----⨯⨯⨯⨯⨯⨯⨯=-⨯⨯⨯⨯⨯ 120.63kJ mol -=⋅31.7758.31410298 4.25/a H E RT kJ mol ≠-∆=-=--⨯⨯=- 由≠≠≠∆-∆=∆S T H G 得：114.2520.6383.49298H G S J K mol T ≠≠≠--∆-∆--∆===-⋅⋅1(1) 1.775a a E H nRT E RT n RT E kJ mol ≠≠-∆=∆-∆=---==⋅4. The reaction of CH with N 2 is of importance in the chemistry of planetary atmospheres and hydrocarbon flames. By understanding the basic reaction, other complex systems such as combustion in the environment can be studied. The reaction isCH + N 2 → ( CHN 2 )≠ → HCN + NUsing transition state theory, find the partition functions for each molecule and then the rate constant at 298 K. Also, specify all the degrees of freedom and the temperature dependence of the A factor. From experiment, the second-order rate constant at 298 K has been found to be k exp = 7.1 x 10-14 cm 3 / molecule ·s.Properties of the Transition State + MoleculesCH N 2 HCN 2 ( assume linear adduct )M (amu = g / mol) 13.019 28.013 41.032 I ( 10 -40 g cm 2 ) 1.935 13.998 73.2Eo ≈221 J/ molν(cm -1) 2733 2330 3130; 2102; 1252; 1170; 564; 401 g elec 2 1 2 解：0e x p ()B A B E k T Q k h Q Q RT≠=- 0()()()()exp()B t v r e A B A B A B A B E k T Q Q Q Q h Q Q Q Q Q Q Q Q RT≠≠≠≠=⋅⋅⋅⋅- 2223/2212212[1exp()])21121exp()1exp()iCHN B i B Bv N B CH B BCH N B B h m k T k T k T h m k Th m k T Q h h h h k T k Tπππνν=⎛⎫--⎪=⋅⨯⎪ ⎪⋅⋅ ⎪⎝⎭----∏2222220222282exp()8218HCN B HCN N B HC B HC N I k Th E I k T RT I k T h h πσππσσ⋅⋅-⨯⋅133112.0410cm molec s ---=⨯⋅⋅。