# Chapter 8 Chemical Equlibrium¶

## Example 8.1 , Page no:164¶

In [1]:
import math
from __future__ import division

#initialisation of variables
x= 3.33
n= 5 #moles

#CALCULATIONS
N= x**2/(n-x)**2

#RESULTS
print"moles of water and ester formed=",round(N);

moles of water and ester formed= 4.0


## Example 8.2 , Page no:165¶

In [2]:
import math
from __future__ import division

#initialisation of variables
n= 1 #mole
x= 3
y= 4

#CALCULATIONS
r= x**2/n**2
z= n/x
n= n+z
n1= x-z

#RESULTS
print"moles of acid and alcohol=",round(n,2),"moles";
print"moles of ester and water=",round(n1,2),"moles";

moles of acid and alcohol= 1.33 moles
moles of ester and water= 2.67 moles


## Example 8.3 , Page no:165¶

In [3]:
import math
from __future__ import division

#initialisation of variables
k= 1.1*10**-5
V= 600 #ml
n= 0.4 #mole

#CALCULATIONS
m= n*1000/V
x= (-k+math.sqrt(k**2+4*4*0.67*k))/(2*4)
M= 2*x
P= x*100/m

#RESULTS
print"molar concentration of NO2=",'%.2E'%M,"mol per litre";
print"per cent dissociation=",round(P,2),"per cent";

molar concentration of NO2= 2.71E-03 mol per litre
per cent dissociation= 0.2 per cent


## Example 8.4 , Page no:167¶

In [4]:
import math
from __future__ import division

#initialisation of variables
pno2= 0.31 #atm
pn2o2= 0.69 #atm
p= 10 #atm

#CALCULATIONS
Kp= pno2**2/pn2o2
x= (-Kp+math.sqrt(Kp**2+4*4*p*Kp))/(2*4)
p1= p-x
p2= 2*x

#RESULTS
print"Kp=",round(Kp,2);
print"N2O4=",round(p1,2);
print"NO2=",round(p2,2);

Kp= 0.14
N2O4= 9.43
NO2= 1.15


## Example 8.5 , Page no:172¶

In [5]:
import math
from __future__ import division

#initialisation of variables
T= 65 #C
R= 1.98 #cal/mol K
kp= 2.8
kp1= 0.141
T1= 25 #C

#CALCULATIONS
H= math.log10(kp/kp1)*2.303*R*(273+T1)*(273+T)/(T-T1)

#RESULTS
print"average heat of reaction=",round(H+62),"cal";

average heat of reaction= 14965.0 cal