Chapter 13:Chemical Kinetics

Example no:13.2,Page no:564

In [2]:
#Variable declaration
dO2=-0.024 #rate of reaction of O2, M/s

#Calculation
#(a)
dN2O5=-2*dO2 #rate of formation of N2O5, M/s
#(b)
dNO2=4*dO2 #rate of reaction of NO2, M/s

#Result
print"(a).The rate of formation of N2O5 is :",dN2O5,"M/s"
print"(b).The rate of reaction of NO2 is :",dNO2,"M/s"

(a).The rate of formation of N2O5 is : 0.048 M/s
(b).The rate of reaction of NO2 is : -0.096 M/s

Example no:13.3,Page no:568

In [72]:
#Variable declaration
NO1=5*10**-3 #conc of NO from 1st experiment, M
H21=2*10**-3 #conc of H2 from 1st experiment, M
r1=1.3*10**-5 #initial rate from 1st experiment, M/s
NO2=10*10**-3 #conc of NO from 2nd experiment, M
H22=2*10**-3 #conc of H2 from 2nd experiment, M
r2=5*10**-5 #initial rate from 1st experiment, M/s
NO3=10*10**-3 #conc of NO from 3rd experiment, M
H23=4*10**-3 #conc of H2 from 3rd experiment, M
r3=10*10**-5 #initial rate from 3rd experiment, M/s

#Calculation
import math
#(a)
#r=k*NO**x*H2**y, dividing r2/r1 and r3/r2
x=math.log(r2/r1)/math.log(NO2/NO1) #since H21=H22
y=math.log(r3/r2)/math.log(H23/H22) #since NO3=NO2
x=round(x) 
y=round(y) 
#(b)
k=r2/((NO2)**x*H22**y) #rate constant, /M**2 s
#(c)
NO=12*10**-3 #conc of NO, M
H2=6*10**-3 #conc of H2, M
rate=k*(NO**x)*H2**y #rate, M/s

#Result
print"(a) the rate of reaction is : r=k[NO]**",x,"*[H2]**",y
print"(b) the rate constant of the reaction is :%.1e"%k," /M**2 s" 
print"(c) the rate of reaction at given concentration is :%.1e"%rate," M/s"

(a) the rate of reaction is : r=k[NO]** 2.0 *[H2]** 1.0
(b) the rate constant of the reaction is :2.5e+02  /M**2 s
(c) the rate of reaction at given concentration is :2.2e-04  M/s

Example no:13.4,Page no:571

In [75]:
#Variable declaration
t=8.8*60 #time, s
k=6.7*10**-4 #rate constant, s-1

#Calculation
import math
#(a)
Ao1=0.25 #initial conc, M
A1=math.exp(-k*t+math.log(Ao1)) #final conc, M
#(b)
A2=0.15 #initial conc, M
Ao2=0.25 #final conc, M
t2=-math.log(A2/Ao2)/(60*k) #time, min
#(c)
percent=74.0 
#let initial conc be 1
Ao3=1.0 #initial conc, M
A3=1.0-percent/100.0 #final conc, M
t3=-math.log(A3/Ao3)/(k*60.0) #time, min
print"(a) the concentration of cyclopropane at given time is :",round(A1,2),"M"
print"(b) the time required is :",round(t2),"min"
print"(c) the time required for required conversion is :%.1f"%t3,"min(approx)"

(a) the concentration of cyclopropane at given time is : 0.18 M
(b) the time required is : 13.0 min
(c) the time required for required conversion is :33.5 min(approx)

Example no:13.5,Page no:573

In [2]:
#Variable declaration
t=[0,100,150,200,250,300] #time(data given), s
P=[284.0,220.0,193.0,170.0,150.0,132.0] #pressure(data given) in mmHg corresponding to time values
lnP=[0,0,0,0,0,0]

#Calculation
import matplotlib.pyplot as plt
import numpy as np
from pylab import *
%matplotlib inline
lnP=np.log(P) #lnP values corresponding to P
A=plot(t,lnP)
plt.xlim(0,350)
plt.ylim(4.8,5.8)
plt.ylabel('$ln Pt$')
plt.xlabel('$t(s)$')
plt.title('Plot of ln Pt versus time for the decomposition of azomethane\n')
slope=np.polyfit(t,lnP,1)
k=-slope[0]

#Result
show(A)
print"The rate constant for the decomposition is :%.2e"%k,"s**-1"

The rate constant for the decomposition is :2.55e-03 s**-1

Example no:13.6,Page no:576

In [77]:
#Variable declaration
k=5.36*10**-4 #rate constant, s-1

#Calculation
t_half=0.693/(60*k) #half life of the reaction, min

#Result
print"The half life for the decomposition of ethane is :",round(t_half,1),"min"

The half life for the decomposition of ethane is : 21.5 min

Example no:13.7,Page no:578

In [79]:
#Variable declaration
k=7*10**9 #rate constant, M s

#Calculation
#(a)
t=2*60 #half life of the reaction, s
Ao1=0.086 
A1=(k*t+1/Ao1)**-1 
#(b)
Ao2=0.6 
t_half2=1/(Ao2*k) #half life of the reaction, s
Ao3=0.42 
t_half3=1/(Ao3*k) #half life of the reaction, s

#Result
print"(a.)The concentration of I is :%.1e"%A1,"M"
print"(b.)The half life for Io=0.6 is :%.1e"%t_half2,"s"
print"    The half life for Io=0.42 is :%.1e"%t_half3,"s"

(a.)The concentration of I is :1.2e-12 M
(b.)The half life for Io=0.6 is :2.4e-10 s
    The half life for Io=0.42 is :3.4e-10 s

Example no:13.8,Page no:584

In [4]:
#Variable declaration
T=[700.0,730.0,760.0,790.0,810.0] #temperature(data given), K
R=8.314 #gas constant, kJ/mol
k=[0.011,0.035,0.105,0.343,0.789] #rate constant(data given) in 1/M**1/2 s corresponding to temperature values
import numpy
#Calculation
x=numpy.reciprocal(T) #1/T values corresponding to Temp values above, K-1
x=numpy.round(x,5)
import math

from pylab import *
%matplotlib inline
lnk=numpy.log(k) #lnk values corresponding to k
lnk[0]=numpy.round(lnk[0],2)
lnk[1]=numpy.round(lnk[1],2)
lnk[2]=numpy.round(lnk[2],3)
lnk[3]=numpy.round(lnk[3],3)
lnk[4]=numpy.round(lnk[4],3)
A=plot(x,lnk,marker='o', linestyle='-')
plt.xlim(1.2*10**-3,1.45*10**-3)
plt.ylim(-5.0,0.0)
plt.ylabel('$ln k$')
plt.xlabel('$1/T(K^-1)$')
plt.title('Plot of ln k versus 1/T\n')
slope=np.polyfit(x,lnk,1)
Ea=-slope[0]*R #activation energy, kJ/mol

#Result
show(A)
print"The activation energy for the decomposition is :%.3e"%(Ea/1000),"kJ/mol"
print"NOTE:Slope is approximately calculated in book,that's why wrong answer"

The activation energy for the decomposition is :1.797e+02 kJ/mol
NOTE:Slope is approximately calculated in book,that's why wrong answer

Example no:13.9,Page no:586

In [81]:
import math
#Variable declaration
k1=3.46*10**-2 #rate constant at T1
T1=298 #temp K
T2=350 #temp K
R=8.314 #gas constant,J/K mol
Ea=50.2*1000 #activation energy, J/mol

#Calculation
k2=k1/math.exp(Ea/R*(T1-T2)/(T1*T2)) #from equation ln(k1/k2)=Ea*(T1-T2)/(T1*T2*R), S-1

#Result
print"The rate constant at temp 350 is :",round(k2,3),"s**-1"

The rate constant at temp 350 is : 0.702 s**-1