Chapter9 Chemical Equilibria

Example 9.1, Page no.35

In [2]:
import math
#initialisation of variables
T=400        #in Celsius
R=0.08205    #in l−atm moleˆ−1 degˆ−1
Kp=1.64*10**-4
n=2.0
P=10         #in atm
#CALCULATIONS
Kc=Kp*(R*(273.1+T))**n
Kx=Kp*P**n
#RESULTS
Kc=round(Kc,4)
print 'Kc=',Kc,'lˆ2 moleˆ−2'
print 'Kx=',Kx,'e'

Kc= 0.5002 lˆ2 moleˆ−2
Kx= 0.0164 e

Example 9.2, Page no.35

In [3]:
import math
#initialisation of variables
R=0.08205     #in l−atm moleˆ−1 degˆ−1
T=25          #in Celsius
g=1.588       #in gms
P=1           #in atm
V=0.5         #litres
M1=92.02      #g moleˆ−1 
#CALCULATIONS
M2=R*(273.1+T)*g/(P*V)
a=(M1-M2)/M2 
#RESULTS
a=round(a,4)
print 'degree of dissociation=',a

degree of dissociation= 0.1846

Example 9.3, Page no.36

In [4]:
import math
#initialisation of variables
P=1        #atm
a=18.46    #in percentage
P1=0.5     #atm
#CALCULATIONS
Kp=P*4*(a/100)**2/(1-(a/100)**2)
#RESULTS
Kp=round(Kp,2)
print 'Kp=',Kp

Kp= 0.14

Example 9.4, Page no.36

In [5]:
import math
#initialisation of variables
M1=208.3     #gms
g=2.69       #gms
R=0.08205    #l−atm moleˆ−1 degˆ−1
T=250        #Celsius
P=1          #atm
V=1          #lit
#CALCULATIONS
M2=g*R*(273.1+T)/(P*V)
a=(M1-M2)/M2
Kp=a**2*P/(1-a**2)
#RESULTS
Kp=round(Kp,3)
print 'Kp=',Kp

Kp= 1.83

Example 9.5, Page no.37

In [6]:
import math
#initialisation of variables
x=0.0574           #mole
n=0.1              #mole
#CALCULATIONS
a=x/n
#RESULTS
print 'degree of dissociation=',a

degree of dissociation= 0.574

Example 9.6, Page no.37

In [7]:
import math
#initialisation of variables
R=0.08205       #l−atm moleˆ−1 degˆ−1
T=250           #Celsius
n=0.1           #mole
Kp=1.78
#CALCULATIONS
x=n+(n**2*R*(273.1+T)/Kp)
#RESULTS
x=round(x,3)
print 'x=',x,'mole'

x= 0.341 mole

Example 9.7, Page no.38

In [8]:
import math
from math import sqrt
#initialisation of variables
Ppcl5=1    #atm
Kp=1.78 
#CALCULATIONS
Ppcl2=sqrt(Kp)
P=2*Ppcl2+Ppcl5
#RESULTS
P=round(P,3)
print 'P=',P,'atm'

P= 3.668 atm

Example 9.8, Page no.38

In [9]:
import math
#initialisation of variables
Kp=1.78
a=0.2
#CALCULATIONS
P=Kp*(1-a**2)/a**2
#RESULTS
print 'Kp=',P,'atm'

Kp= 42.72 atm

Example 9.10, Page no.38

In [10]:
import math
#initialisation of variables
n=0.6667       #mole
#CALCULATIONS
K=n**2/((1-n)**2)
#RESULTS
K=round(K,3)
print 'K=',K

K= 4.001

Example 9.11, Page no.39

In [11]:
import math
#initialisation of variables
pN2O4=0.141          #atm
pNO2=1               #atm
R=1.987              #cal moleˆ−1 degˆ−1
T=25                 #Celsius
#CALCULATIONS
dG=-R*2.303*(273.1+T)*math.log10(pN2O4/pNO2**2)
#RESULTS
dG=round(dG,3)
print 'dG=',dG,'cal'

dG= 1160.57 cal

Example 9.12, Page no.40

In [12]:
import math
#initialisation of variables
pN2O4=1 #atm 
pNO2=0.141 #atm 
R=1.987 #cal moleˆ−1 degˆ−1 
T=25    #C 
#CALCULATIONS 
dG=-R*2.303*(273.1+T)*math.log10(pN2O4/pNO2) 
#RESULTS 
dG=round(dG,3)
print 'dG=',dG,'cal'

dG= -1160.57 cal

Example 9.13, Page no.40

In [13]:
import math
#initialisation of variables
Kc=2.7*10**2 
R= 1.987  #cal moleˆ−1 degˆ−1 
T= 43.9   #c 
#CALCULATIONS 
dG=-R*(273.1+T)*2.303*math.log10(Kc) 
#RESULTS
dG=round(dG,3)
print 'dG=',dG,'cal'

dG= -3526.964 cal

Example 9.14, Page no.40

In [14]:
import math
#initialisation of variables
dH=-17.889    # cal degˆ−1 
T=25          #C 
dS=-19.28     # cal degˆ−1 
R=1.987       # cal moleˆ−1 degˆ−1 
#CALCULATIONS 
dG=dH-dS*(273.1+T) 
Kp=10**(dG/(-R*(273.1+T)*2.303)) 
#RESULTS 
print 'Kp=',Kp,'e'
print 'ANSWER IN THE TEXTBOOK IS WRONG'

Kp= 6.30787913713e-05 e
ANSWER IN THE TEXTBOOK IS WRONG

Example 9.15, Page no.40

In [15]:
import math
#initialisation of variables
HCO2=-94.2598 # kcal
HH2=0 # kcal
HCO=-32.8079 # kcal
HH2O=-54.6357 # kcal
R=1.987 # cal degˆ−1 moleˆ−1
T=25    #C
#CALCULATIONS 
Kp=10**(-(HCO2 -HCO -HH2O)/(R*2.303*(273.1+T)))
#RESULTS 
print 'Kp=',Kp,'e' 
print 'ANSWER IN THE TEXTBOOK IS WRONG'

Kp= 1.01157190896 e
ANSWER IN THE TEXTBOOK IS WRONG

Example 9.16, Page no.41

In [16]:
import math
#initialisation of variables
G0=1161.0 # cal
R=1.987 # cal moleˆ−1 degˆ−1
T=25.0 #C
P=1.0 #atm
P1=10.0 #atm
#CALCULATIONS
dG=G0+R*(273.0+T)*2.303*math.log10(P**2/P1) 
#RESULTS 
dG=round(dG,3)
print 'dG=',dG,'cal'

dG= -202.666 cal

Example 9.17, Page no.41

In [17]:
import math
#initialisation of variables
K2500=3.6*10**-3
K2000=4.08*10** -4 
R=1.987 # cal moleˆ−1 Kˆ−1
T1=2500 #K
T2=2000 #K
#CALCULATIONS
dH=math.log10(K2500/K2000)*2.303*R*T1*T2/(T1-T2) 
#RESULTS 
dH=round(dH,3)
print 'enthalpy change=',dH,'cal'

enthalpy change= 43273.17 cal

Example 9.18, Page no.42

In [18]:
import math
#initialisation of variables
dH=-10200    #cal
R=1.987     # cal degˆ−1 moleˆ−1
T1=690      #K
T2=800      #K
KT1=10
#CALCULATIONS
KT2=KT1*10**(dH*(T2-T1)/(2.303*R*T1*T2)) 
#RESULTS 
KT2=round(KT2,3)
print 'K800=',KT2

K800= 3.596

Example 9.19, Page no.42

In [19]:
import math
#initialisation of variables
T=1000.0 #K 
R=1.987 # cal moleˆ−1 Kˆ−1 
G=-1330.0 # cal moleˆ−1 
#CALCULATIONS 
Kp=10.0**(G/(-R*T*2.303)) 
#RESULTS
Kp=round(Kp,3)
print 'Kp=',Kp

Kp= 1.953

Example 9.20, Page no.43

In [20]:
import math
#initialisation of variables
from math import sqrt
Kp=1.78 
P=0.1 #atm 
#CALCULATIONS 
a=sqrt(Kp/(Kp+P))*100 
#RESULTS 
a=round(a,3)
print 'percent dissaciated=',a,'percent'

percent dissaciated= 97.304 percent

Example 9.21, Page no.43

In [21]:
import math
#initialisation of variables
R=1.987 # cal moleˆ−1 Kˆ−1 
T=2000 #K 
dH= 117172 # cal moleˆ−1 
H=-43 # cal moleˆ−1 
n=2 
H1=-56.12 # cal moleˆ−1 
#CALCULATIONS 1
K=10**( -(1/(2.303*R))*((dH/T)+n*H-H1))
#RESULTS
print 'equilibrium constant=',format(K, '.8f')

equilibrium constant= 0.00000072

Example 9.22, Page no.43

In [22]:
import math
#initialisation of variables
T=25.0 #C 
R=1.987 # cal moleˆ−1 Kˆ−1 
n=2.0 
dH=-21.840 # cal moleˆ−1 
HHCl=-37.73 # cal moleˆ−1 
HH2=-24.44 # cal moleˆ−1 
HCl=-45.95 # cal moleˆ−1 1
#CALCULATIONS
K=10**(( -1/(2.303*R))*((dH*n/(273.15+T))+n*HHCl -HH2 - HCl))
print 'ANSWER IN TEXTBOOK IS WRONG'

ANSWER IN TEXTBOOK IS WRONG