《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) =πd2, 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 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 求k(T) 根据气体分子运动论：
?u???uf(u)du??0??0?2??uu()e2?kBT2322kBT4?u2du
3??2?2?()?e??u02?kBT2kBTu2du2
32kBT?2?2?()?(2?kBT?2?)e0???u22kBT?(ud?)(u2kBT2kTB?2?2)?(8kBT??)12
其中：f(u)：Maxwell-Boltzmann分布函数 ?：折合质量。 则有：k(T) =бR??d2(解法2：
k(T)??0?Ruf(u)du??0??8kBT??)12?d2(8?kBT?)2
1?3?du()2e??u2?kBT2222kBT4?u2du
???03?4?2d2()2e??u2?kBT22kBTu3du
??d(8kBT??)2
1 (b)
k(T)???Ruf(u)du???03??0(1?e?a?)u()2e??u2?kBT22kBT4?u2du181/2??()??0(1?e?a?)e??kBT?d?
0kBT??kBT?81/2???kBT?(){?e?d???e?(a?1kBT)??d?}
00kBT??kBT?0????081/2?12??()?(kBT)-12? kBT??kBT?(a+)?kBT????12??()?0?1-()?
???akBT+1?8kBT1/2
?EArrhenius公式为：k?Aexp(ak呈指数增长。 而(a) k(T)∝T1/2 , (b) k(T)?(RT)，k与T呈指数关系，随T的增加，
8kBT??1)1/2?0?1-()2?, ???akBT+1?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：
?? V(r)?4?[()12?()6]
rr 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。
dV?12?6?4?[?1213?67] 则：drrr 当
dV?0时，势能V为最小值， drdV?12?6?4?[?1213?67]?0 drrr 得：r?2? 即，势能最小时，核间距 r?2?
3. The following reaction was done experimentally at various temperatures in order to obtain rate constants:
O + HO2 → OH + O2
T1 = 298 K k1 = 6.1 x 10-11 cm3 / molecule·s
T1 = 229 K k2 = 7.57 x 10-11 cm3 / molecule·s
(a) Using the experimental data, find the activation energy Eact.
(b) Calculate the following thermodynamic properties at T = 298 K using transition state theory: ΔE≠, ΔH≠, ΔS≠, ΔG≠.
?E 解：(a) 由Arrhenius公式 k?Aexp(a)
RT1616?E 得：k1?Aexp(a) (1)
RT1?E k2?Aexp(a) (2)
RT2)RT211(2)k2??exp{(?)Ea} ：
?Eak1RTRT(1)12Aexp()RT1?EAexp(a 则
lnk2k1：
RTTk28.314?298?2297.57?10?1112Ea??ln?ln??1.775kJ?mol?1?1111T2?T1k1229?2986.1?10?RT1RT2
（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： kp(298K)kc6.1?10?11?10-3?6.023?1023???1.504?109 R'T0.082?298kBT??G?exp()，得： 据k?hRTk298h1.504?109?6.626?10-34?1 ?G??RTln()??8.314?298ln?20.63kJ?mol-23kBT1.38?10?298?或：
由过渡态理论速率常数表达式： k?kBT?hkK 得：K?? hkBT 又：?G???RTlnK?(?)，因其为气相反应，标准态通常选为1个大气压，
K?(?)hkp?K?p?p?/p?[c]?RT? ????p???kBTRT(pO/p)(pHO2/p)(cORT)(cHO2RT)RT?hkp?故：?G??RTln(?)
kBTRT6.626?10?34?6.023?1023?6.1?10?11?10?61.013?105??8.314?298ln(?) ?231.38?10?2988.314?298?20.63kJ?mol?1
?H??Ea?RT??1.775?8.314?10?3?298??4.25kJ/mol 由?G???H??T?S?得：
?H???G??4.25?20.63?S????83.49J?K?1?mol?1
T298??E???H???nRT?Ea?RT?(1?n)RT?Ea?1.775kJ?mol?1
4. The reaction of CH with N2 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 is
CH + N2 → ( CHN2 )≠ → HCN + N
Using 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 kexp = 7.1 x 10-14 cm3 / molecule·s.
Properties of the Transition State + Molecules CH N2 HCN2 ( assume linear adduct ) M (amu = g / mol) 13.019 28.013 41.032 I ( 10 -40 g cm2 ) 1.935 13.998 73.2
Eo ≈221 J/ mol
ν(cm-1) 2733 2330 3130; 2102; 1252; 1170; 564; 401 gelec 2 1 2 解：
EkBTQ? k?exp?(0 )hQAQBRTEkBTQ?Q?Q?Q??()t?()v?()r?()e?exp(?0) hQAQBQAQBQAQBQAQBRT12?mCHN2kBT?2kBT?h??h?2?mCHkBT2?mN2kBT??2hh2???????3/2h?i)]kTi?1B?)v?11?Qh?CHh?N2B1?exp(?)1?exp(?)kBTkBT?[1?exp(?6
8?2IHCN2kBT?HCNh2E02??exp(?) 22RT8?IHCkBT8?INkBT2?1??HCh2?Nh2222?2.04?10?13cm3?molec?1?s?1