# Chapter11-Free energy and chemical equillibrium¶

## Ex1-pg327¶

In [1]:
import math
##Intitalisation of variables

p1= 1. ##atm
p2= 0.1 ##atm
p3= 0.1 ##atm
R= 1.987 ##cal mole^-1 K^-1
T= 2000. ##K
Kp= 1.55*10**7
##CALCULATIONS
Qp= p1/(p2**2*p3)
dF= 2.303*R*T*math.log10(Qp/Kp)/1000.
dF1= -2.303*R*T*math.log10(Kp)/1000.
##RESULTS
print'%s %.2f %s'% ('free energy change = ',dF,' kcal ')
print'%s %.2f %s'% ('\n free energy change = ',dF1,' kcal ')

free energy change =  -38.35  kcal

free energy change =  -65.81  kcal


## Ex2-pg333¶

In [2]:
import math
##Intitalisation of variables
T= 500. ##C
Kp= 1.43*10**-5 ##atm
R= 1.987 ##cal
##CALCULATIONS
dF= -2.303*R*(273.+T)*math.log10(Kp)
##RESULTS
print'%s %.2f %s'% ('dF = ',dF+3,' cal ')

dF =  17140.01  cal


## Ex3-pg335¶

In [3]:
import math
##Intitalisation of variables
n1= 2. ##moles
n2= 2. ##moles
n3= 1. ##mole
h1= 54.6 ##cal
h2= 7.8 ##cal
h3= -69.6 ##cal
R= 1.987 ##cal
T= 25. ##C
##CALCULATIONS
dF= -n1*h1-(-n2*h2+n3*h3)
Kp= 10**(-dF*1000./(2.303*R*(273.2+T)))
##RESULTS
print'%s %.2f %s'% ('dF = ',dF,' kcal ')
print'%s %.2e %s'% ('\n equillibrium constant =  ',Kp,'')

dF =  -24.00  kcal

equillibrium constant =   3.87e+17


## Ex4-pg340¶

In [4]:
import math
##Intitalisation of variables
dH= 12300. ##cal
T= 25. ##C
dS= -60.1 ##cal deg^-1 mole^-1
##CALCULATIONS
dF= dH-dS*(273.+T)
##RESULTS
print'%s %.2f %s'% ('Standard free energy of formation = ',dF-10,' cal mole^-1 ')

Standard free energy of formation =  30199.80  cal mole^-1