Chapter 19 - Photochemistry

Example 1 - pg 488

In [1]:
#calculate the Overall transmittance
#Initialization of variables
r1=0.727
r2=0.407
#calculations
r3=r1*r2
#results
print '%s %.3f' %("Overall transmittance = ",r3)

Overall transmittance =  0.296

Example 2 - pg 488

In [2]:
#calculate the Extinction coefficient
import math
#Initialization of variables
r=0.450
c=0.02 #M
l=4 #cm
#calculations
e=-math.log10(r) /(c*l)
#results
print '%s %.2f %s' %("Extinction coefficient =",e,"litres mole^-1 cm^-1")

Extinction coefficient = 4.33 litres mole^-1 cm^-1

Example 3 - pg 488

In [3]:
#calculate the Transmittance of the solution
#Initialization of variables
import math
from math import log10
r1=0.850
r2=0.50
#calculations
Da=-log10(r1)
Db=-log10(r2)
D=Da+Db
r3=10**(-D)
#results
print '%s %.3f' %("Transmittance of solution =",r3)

Transmittance of solution = 0.425

Example 4 - pg 491

In [4]:
#calculate the Extinction coefficient
#Initialization of variables
c=0.000025 #M
l=2 #cm
D=0.417
#calculations
e=D/(c*l)
#result
print '%s %d %s' %("Extinction coefficient =",e," liters mole^-1 cm^-1")

Extinction coefficient = 8340  liters mole^-1 cm^-1

Example 5 - pg 491

In [5]:
#calculate the Kc for dissociation
#Initialization of variables
c=0.5 #M
c1=0.000025 #M
D2=0.280
D1=0.417
#calculations
c2=D2*c1/(D1)
dC=c1-c2
SCN=c- 6*c2 -4*dC
K=dC*SCN**2 /c2
#results
print '%s %.2f %s' %("Kc for dissociation =",K," M^2")

Kc for dissociation = 0.12  M^2

Example 6 - pg 492

In [6]:
#calculate the value of m
#Initialization of variables
import math
from math import log
D2=0.249
D1=0.172
a2=0.00752
a1=0.00527
#calculations
m=(log(D2) -log(D1))/(log(a2) - log(a1))
#results
print '%s %.2f' %("m = ",m)

m =  1.04

Example 7 - pg 495

In [8]:
#calculate the Increase in optical density
#Initialization of variables
c=0.1 #M
V=100 #ml
v1=25 #ml
D=0.980
d1=0.090
d2=0.150
#calculations
a=v1*c/V
b=(V-v1)*c/V
Da=a*d1/c
Db=b*d2/c
Ddash=Da+Db
dD=D-Ddash
#results
print '%s %.3f' %("Increase in optical density =",dD)

Increase in optical density = 0.845

Example 8 - pg 496

In [9]:
#calculate the value of lambda for the reaction to occur
#Initialization of variables
E=50000. #cal/mol
#calculations
lam=2.8593/E
#results
print '%s %d %s' %("For the reaction to occur lambda <",lam*10**8,"A")

For the reaction to occur lambda < 5718 A

Example 9 - pg 497

In [10]:
#calculate the Amount of reactant disappeared
#Initialization of variables
lam=3000*10**-8 #cm
ield=0.420
Et=70000 #cal
#calculations
E=2.8593/lam
n=ield*Et/E
#results
print '%s %.3f %s' %("Amount of reactant disappeared =",n," mol")

Amount of reactant disappeared = 0.308  mol