Chapter 17 Homogeneous Chemical Reactions

Example 17_1_1 pgno:485

In [1]:
#initialization of variables
K = 1.46*10**-4 # lit/mol-sec (rate constant)
cpyridine = 0.1 # mol/lit
K1 = 2.0*10**-5 # cm**2/sec
#Calculations
D = K*cpyridine # sec**-1
k0 = ((D*K1)**0.5)*10**5#in x*10**-5 cm/sec
#Results
print"The diffusion co efficient of methyl iodide in benzene is  x10**-5 cm/sec",round(k0,1)

The diffusion co efficient of methyl iodide in benzene is  x10**-5 cm/sec 1.7

Example 17_1_2 pgno:486

In [3]:
#initialization of variables
R = 0.3 # cm
K1 = 18.6 # sec**-1
D = 0.027 # cm**2/sec
from sympy import coth
#Calculations
l = R/3 # cm
n = ((D/(K1*(l**2)))**0.5)*coth((K1*(l**2)/D)**0.5)
#Results
print"The value of reduction in reaction rate due to diffusion is ",round(n,3)

The value of reduction in reaction rate due to diffusion is  0.385

Example 17_1_3 pgno:486

In [4]:
#initialization of variables
k = 16*10**-3 # m.t.c in cm/sec
D  = 1.25*10**-5 # Diffusion co efficient in cm**2/sec
#Calculations 
K1 = (k**2)/D
#Results
print"The rate constant is  sec**-1",round(K1)

The rate constant is  sec**-1 20.0

Example 17_2_1 pgno:490

In [5]:
#initialization of variables
D2 = 5*10**-6 # The diffusion co efficient of the new compound in cm**2/sec
Nu = 3 # The factor
D1 = 0.7*10**-5 # The diffusion co efficient of the original compound in cm**2/sec
c2l = 1.5*10**-5 # the new solubility in mol/cc
c1l = 3*10**-7 # The old solubility in mol/cc
#Calculations
k = 1 + ((D2*c2l)/(Nu*D1*c1l))# The number of times the rate has increased to the previous rate
#Results
print"There is about a  fold increase in rate",round(k)

There is about a  fold increase in rate 13.0

Example 17_4_1 pgno:503

In [6]:
#initialization of variables
from math import pi
#For first reaction
D1 = 9.3*10**-5 # cm**2/sec
D2 = 5.3*10**-5 # cm**2/sec
K1exp = 1.4*10**11 # litre/mol-sec
sigma12 = 2.8*10**-8 # cm
N = (6.02*10**23)/10**3# liter/cc-mol
K1 = 4*pi*(D1+D2)*sigma12*N/10**10 # Rate constant for first reaction in litre/mol-sec
print"The rate constant for this reaction is litre/mol-sec",round(K1)
if K1>K1exp:
     	 print"This reaction is controlled more by chemical factors"
else:
         print"This reaction is diffusion controlled"

#Second reaction
D1 = 5.3*10**-5 # cm**2/sec
D2 = 0.8*10**-5 # cm**2/sec
sigma12 = 5*10**-8 # cm
K1exp = 3.8*10**7 # litre/mol-sec
K1 = 4*pi*(D1+D2)*sigma12*N/10**10 # Rate constant for second reaction in litre/mol-sec
print"The rate constant for this reaction is 10**10 litre/mol-sec",round(K1)
if K1>K1exp: 
    print"This reaction is controlled more by chemical factors"
else: 
    print"The reaction is diffusion controlled"

The rate constant for this reaction is litre/mol-sec 3.0
This reaction is diffusion controlled
The rate constant for this reaction is 10**10 litre/mol-sec 2.0
The reaction is diffusion controlled

Example 17_5_1 pgno:506

In [7]:
#intitialization of variables
d = 5# cm
v = 200 # cm/sec
nu = 0.01 # cm**2/sec
D = 3.2*10**-5 # cm**2/sec
l = 30*10**-4 # cm
#Calculations
Re = d*v/nu # Flow is turbulent
E = d*v/2 # cm**2/sec
tou1 = (d**2)/(4*E)# sec
tou2 = (l**2)/(4*D)
tou = tou1 + tou2 # sec
#Results
print"The relaxation time is sec",round(tou,3)

The relaxation time is sec 0.07