文档介绍：《化学反应动力学》
第六、七、九章习题
Calculate k (T) given the following forms for the reactive cross section:
(a)бR(v) =πd2, ., 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 these 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:

其中::Maxwell-Boltzmann分布函数:折合质量。
则有:k(T) =бR<u R>
解法2:

(b)
Arrhenius公式为:,k与T呈指数关系,随T的增加,k呈指数增长。
而(a) k(T)∝T1/2 ,
(b) ,
与Arrhenius公式中k与T的关系不同。
For a Lennard-Jones “ 6 - 12 ” potential determine the internuclear distance at the potential energy minimum.
解:Lennard-Jones 势能函数为:

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:
当时,势能V为最小值,

得: 即,势能最小时,核间距
The following reaction was done experimentally at various temperatures in order to obtain rate constants:
O + HO2 → OH + O2
T1 = 298 K k1 = x 10-11 cm3 / molecule·s
T1 = 229 K k2 = x 10-11 cm3 / molecule·s
Using the experimental data, find the activation energy Eact.
Calculate the following thermodynamic properties at T = 298 K using transition state theory: ΔE≠, ΔH≠, ΔS≠, ΔG≠.

解:(a) 由Arrhenius公式
得: (1)
(2)
:
则:
(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:

据,得:
或:
由过渡态理论速率常数表达式:
得:
又:,因其为气相反应,标准态通常选为