# Chapter 10 - Chemical Kinetics¶

## Example 1 - pg 543¶

In [1]:
#calculate the Pressure
#initialisation of variables
t= 3 #sec
P0= 200 #mm
k= 17.3 #mm/sec
P1= 104 #mm
#CALCULATIONS
P= P0-k*t
P2= P+P1
#RESULTS
print '%s %d %s' % (' Pressure=',P2,' mm of Hg')

 Pressure= 252  mm of Hg


## Example 2 - pg 545¶

In [2]:
#calculate the Half time
#initialisation of variables
k= 2.63*10**-3 #min^-1
#CALCULATIONS
t1= 0.693/k
#RESULTS
print '%s %.1f %s' % (' Half time=',t1,'min')

 Half time= 263.5 min


## Example 3 - pg 546¶

In [4]:
#calculate the Partial Pressure of the reactant
#initialisation of variables
P= 200. #mm
t= 30. #min
k= 2.5*10**-4 #sec^-1
#CALCULATIONS
P0= P/(10**(k*t*60/2.303))
P1= P-P0
#RESULTS
print '%s %d %s' % (' Partial Pressure of reactant=',P1,'mm of Hg')

 Partial Pressure of reactant= 72 mm of Hg


## Example 4 - pg 548¶

In [6]:
#calculate the no of atoms
#initialisation of variables
t= 5600*365*24*60.
x= 5 #atoms
#CALCULATIONS
k= 0.693/t
N= x/k
#RESULTS
print '%s %.2e %s' % (' No of atoms=',N, 'atoms')

 No of atoms= 2.12e+10 atoms


## Example 5 - pg 548¶

In [7]:
#calculate the time passed
#initialisation of variables
import math
t= 5600 #sec
r= 0.256
#CALCULATIONS
t1= (t/0.693)*2.303*math.log10(1/r)
#RESULTS
print '%s %d %s' % (' Time=',t1,'years ago')
print 'The answer is a bit different due to rounding off error in textbook'

 Time= 11012 years ago
The answer is a bit different due to rounding off error in textbook


## Example 6 - pg 549¶

In [8]:
#calculate the first order rate constant and half life
#initialisation of variables
import math
t= 25.1 #hr
C= 0.004366
C1= 0.002192
C2= 0.006649
#CALCULATIONS
r= (C-C1)/(C2-C1)
k= 2.303*math.log10(1/r)/t
t1= 0.693/k
#RESULTS
print '%s %.1f %s' %(' Time=',t1,' hr')
print '%s %.2e %s' %(' Time=',k,' hr')

 Time= 24.2  hr
Time= 2.86e-02  hr


## Example 7 - pg 552¶

In [9]:
#calculate the Rate constant
#initialisation of variables
s= 18.6*10**4 #mm of hg
#CALCULATIONS
k= 1./s
#RESULTS
print '%s %.2e %s' % (' Rate constant=',k,' (mm Hg)^-1 sec^-1')

 Rate constant= 5.38e-06  (mm Hg)^-1 sec^-1


## Example 8 - pg 552¶

In [10]:
#calculate the requried Pressure
#initialisation of variables
k= 1.14*10**-4 #sec^-1
k1= 5.38*10**-6 #sec^-1
#CALCULATIONS
P= k/k1
P2=0.01*P
#RESULTS
print '%s %.3f %s' % (' Pressure=',P2,'mm of Hg')

 Pressure= 0.212 mm of Hg


## Example 9 - pg 555¶

In [11]:
#calculate the no of molecules
#initialisation of variables
T= 600 #K
P= 1 #atm
R= 0.082 #atm lit/mol K
#CALCULATIONS
C= P/(R*T)
r= C**2*4*10**-6
r1= 6*10**23*r
#RESULTS
print '%s %.1e %s' % (' No of molecules=',r1,'molecules l^-1 sec^-1')

 No of molecules= 9.9e+14 molecules l^-1 sec^-1


## Example 10 - pg 555¶

In [12]:
#calculate the time required
#initialisation of variables
k= 6.3*10**2 #ml mole^-1 sec^-1
P= 400. #mm
T= 600. #K
R= 82.06
#CALCULATIONS
C= (P/760.)/(R*T)
t= 1/(9.*C*k)
#RESULTS
print '%s %.1f %s' % (' time=',t,' sec')

 time= 16.5  sec


## Example 11 - pg 556¶

In [13]:
#calculate the pressure of No2 in both cases
#initialisation of variables
pf2= 2.00 #mm Hg
y= 0.96 #mm Hg
Pn= 5 #mm Hg
#CALCULATIONS
pF2= pf2-y
pNO2= Pn-2*y
pNO2F= 2*y
#RESULTS
print '%s %.2f %s' % (' pressure of NO2=',pNO2,'mm of Hg')
print '%s %.2f %s' % (' \n pressure of NO2 after 30 sec=',pNO2F,'mm of Hg')

 pressure of NO2= 3.08 mm of Hg

pressure of NO2 after 30 sec= 1.92 mm of Hg


## Example 13 - pg 561¶

In [14]:
#calculate the Rate constant
#initialisation of variables
k= 4*10**-6 #mol^-1 sec^-1
Kc= 73
#CALCULATIONS
K1= k*Kc/2
#RESULTS
print '%s %.2e %s' % (' Rate constant=',K1,'l mol^-1 sec^-1')

 Rate constant= 1.46e-04 l mol^-1 sec^-1


## Example 14 - pg 568¶

In [15]:
#calculate the activation energy
#initialisation of variables
import math
R= 1.987 #atm lit/mol K
T= 573.2 #K
T1= 594.6 #K
k= 3.95*10**-6 #mol^-1 sec^-1
k1= 1.07*10**-6 #mol^-1 sec^-1
#CALCULATIONS
H= R*T*T1*2.303*math.log10((k/k1))/(T1-T)
#RESULTS
print '%s %d %s' %(' activation energy=',H,'calmol^-1')
print 'The answers in the texbook are a bit different due to rounding off error'

 activation energy= 41338 calmol^-1
The answers in the texbook are a bit different due to rounding off error


## Example 15 - pg 568¶

In [16]:
#calculate the time required
#initialisation of variables
import math
H= 41300. #cal
T= 673. #K
T1= 595. #K
R= 1.987 #cal/mol K
K= 3.95*10**-6
P= 1 #atm
R1= 0.08205 #j/mol K
#CALCULATIONS
k2= math.e**(H*(T-T1)/(R*T*T1))*K
C= P/(R1*T)
t= 44.8/C
t2=R1*T*10**-2 /k2
#RESULTS
print '%s %d %s' %(' time =',t,'sec')
print '%s %d %s' %('Time required in case 2 = ',t2,'sec')
print 'The answers in the texbook are a bit different due to rounding off error'

 time = 2473 sec
Time required in case 2 =  2438 sec
The answers in the texbook are a bit different due to rounding off error


## Example 16 - pg 569¶

In [17]:
#calculate the collision diameter
#initialisation of variables
import math
H= 41300.
R= 1.987 #atm lit/mol K
T= 595. #K
M= 128. #gm
R1= 8.314*10**7 #atm lit/mol K
N= 6.02*10**23 #moleccules
k= 3.95*10**-6 #sec**-1
#CALCULATIONS
s= math.sqrt((k*10**3/(4*N))*(128/(math.pi*R1*T))**0.5*math.e**(H/(R*T)))
#RESULTS
print '%s %.3e %s' % (' collision diameter=',s,' cm')

 collision diameter= 8.356e-09  cm


## Example 18 - pg 577¶

In [18]:
#calculate the Concentration of A and B
#initialisation of variables
import math
import numpy
from numpy import linalg
p= 20.3 #percent
p1= 1.77 #percent
I= 100.
n= 2.
l= 300. #l mol^-1 cm^-1
l1= 30. #l mol^-1 cm^-1
l2= 10. #l mol^-1 cm^-1
l3= 200. #l mol^-1 cm^-1
#CALCULATIONS
A= ([[n*l, n*l1],[n*l2, n*l3]])
b= ([[math.log10(I/p1)],[math.log10(I/p)]])
c= numpy.dot(numpy.linalg.inv(A),b)
R1=c[0]
R2=c[1]
#RESULTS
print '%s %.2e %s' % (' Concentration of A =',R1,' mole l^-1')
print '%s %.2e %s' % (' \n Concentration of B =',R2,' mole l^-1')

 Concentration of A = 2.76e-03  mole l^-1

Concentration of B = 1.59e-03  mole l^-1