# Chapter 12 : Chemical Equilibrium¶

### Example 12.1 Page: 311¶

In [14]:
import math

T = 298.15      #[K] temperature
P = 1.          #[atm] pressure
R = 8.314*10**(-3)      #[kJ/(mol*K)]

g_a_0 = -20.9       #[kJ/mol]
g_b_0 = -17.2       #[kJ/mol]

x_a = math.exp((g_b_0-g_a_0)/(R*T))/(1+math.exp((g_b_0-g_a_0)/(R*T)))
x_b = 1-x_a

print " The chemical equilibrium composition of the gaseous mixture contains %f mol fraction isobutane \t\t\t\t\t\t\t\tan %f mol fraction n-bumath.tane"%(x_a,x_b)

 The chemical equilibrium composition of the gaseous mixture contains 0.816475 mol fraction isobutane 								an 0.183525 mol fraction n-bumath.tane


### Example 12.2 Page: 319¶

In [8]:
import math

T = 298.15              #[K] temperature
P = 0.987               #[atm] pressure
g_0_NO = 86.6           #[kJ/mol] Free energy of formation the NO from elements
R = 8.314               #[J/(mol*K)]

g_0_O2 = 0.00
g_0_N2 = 0.00

delta_g_0 = 2*g_0_NO - g_0_O2 - g_0_N2          #[kJ/mol]
delta_g_01 = delta_g_0*1000                     #[J/mol]

K_298 = math.exp((-delta_g_01)/(R*T))

f_0_N2 = 1.             #[bar]
f_0_O2 = 1.             #[bar]
f_0_NO = 1.             #[bar]

y_N2 = 0.78
y_O2 = 0.21

y_NO_298 = math.sqrt(K_298*y_N2*y_O2)

T_1 = 2000                  #[K]
K_2000 = 4.0*10**-4

y_NO_2000 = math.sqrt(K_2000*y_N2*y_O2)*10**(6)         #[ppm]

print " The equilibrium constant for the reaction at 298.15 K is \t\t\t %.1e"%(K_298)
print " The concentration of NO at equilibrium at temperature 298.15 K is \t\t %.1e"%(y_NO_298)
print " The equilibrium consmath.tant for the reaction at 2000 K is \t\t\t %.1e"%(K_2000)
print " The concentration of NO at equilibrium at temperature 2000 K is \t\t %.0f ppm"%(round(y_NO_2000,-2))

 The equilibrium constant for the reaction at 298.15 K is 			 4.5e-31
The concentration of NO at equilibrium at temperature 298.15 K is 		 2.7e-16
The equilibrium consmath.tant for the reaction at 2000 K is 			 4.0e-04
The concentration of NO at equilibrium at temperature 2000 K is 		 8100 ppm


### Example 12.3 Page: 321¶

In [16]:
from scipy.optimize import fsolve
import math

Temp = 2000.                #[K]
n_air = 1.                  #[mol] no of moles of the air

K_2000 = 4*10**(-4)

def f(x):
return  (2*x)**(2) - K_2000*(0.78-x)*(0.21-x)
#return  (K_2000-2)*x**(2)-K_2000*(0.78+0.21)*x+K_2000*0.78*0.21
x = fsolve(f,0)
c_NO = 2*x*10**(6)              #[ppm]
p = c_NO/8100.*100

print " The calculated NO cocentration is %f ppm, which %f%% of the value computed in example 12.1"%(c_NO,p)

 The calculated NO cocentration is 7996.442873 ppm, which 98.721517% of the value computed in example 12.1


### Example 12.5 Page: 324¶

In [18]:
from scipy.optimize import fsolve
import math

Temp = 298.             #[K]
K = 29.6                # equilibrium consmath.tant at 298 K
P = 1.                  #[bar]
n_water_0 = 0.833       #[mol]
n_ethylene_0 = 1.       #[mol]
n_ethanol_0 = 0.        #[mol]

n_T_0 = (n_water_0+n_ethylene_0+n_ethanol_0)        #[mol]

def f(e):
return  ((0+e)/(1.833-e))/(((1-e)/(1.833-e))*((0.833-e)/(1.833-e)))-K*P/(1)
e_1 = fsolve(f,0)
e_2 = fsolve(f,0.5)

y_ethanol = ((0+e_2)/(1.833-e_2))
y_ethylene = ((1-e_2)/(1.833-e_2))
y_water = ((0.833-e_2)/(1.833-e_2))

print "Concentration of the ethylene at the equilibrium is %f"%(y_ethylene)
print " Concentration of the water at the equilibrium is    %f"%(y_water)
print " Concentration of the ethanol at the equilibrium is  %f"%(y_ethanol)

Concentration of the ethylene at the equilibrium is 0.243702
Concentration of the water at the equilibrium is    0.092079
Concentration of the ethanol at the equilibrium is  0.664219


### Example 12.6 Page: 324¶

In [19]:
from scipy.optimize import fsolve
import math

Temp = 273.15+25            #[C]
P = 1.                      #[bar]
R = 8.314                   #[J/(mol*K)]

g_H2O_0 = -237.1            #[kJ/mol]
g_O2_0 = 0                  #[kJ/mol]
g_H2_0 = 0                  #[kJ/mol]

delta_g_0 = g_H2O_0 - 0.5*g_O2_0-g_H2_0         #[kJ/mol]
delta_g_1 = delta_g_0*1000                      #[J/mol]

K = math.exp((-delta_g_1)/(R*Temp))

n_T_0 = 1.5#[mol]

def f(e):
return (e/(1.5-0.5*e))/(((1-e)/(1.5-0.5*e))*((0.5-0.5*e)/(1.5-0.5*e))**(0.5))

y_H2 = 2.4e-28
y_O2 = 0.5*y_H2

print " The equilibrium mol fraction of the hydrogen is   %0.3e"%(y_H2)
print " And the equilibrium mol fraction of the oxygen is %e "%(y_O2)

 The equilibrium mol fraction of the hydrogen is   2.400e-28
And the equilibrium mol fraction of the oxygen is 1.200000e-28


### Example 12.7 Page: 327¶

In [20]:
import math

Temp = 298.15           #[K]
Press = 1*10**(5)       #[Pa]
R = 8.314               #[J/(mol*K)]

v_liquid = 1.805*10**(-5)           #[m**(3)/mol] this liquid specific volume and we will treat is as a consmath.tant

P_vapour_25 = 0.0317*10**(5)            #[Pa]

def f0(P):
return v_liquid*P**(0)

delta_g_0_2 = 0             #[J/mol]

def f1(P):
return 1./P

delta_g_0 = delta_g_0_1+delta_g_0_2+delta_g_0_3         #[J/mol]
delta_g_1 = delta_g_0/1000                              #[kJ/mol]

print " Total change in the free energy of water going under given conditions is %0.2f kJ/mol"%(delta_g_1)

delta_g_0_ideal_gas = -228.6                            #[kJ/mol]
delta_g_0_liquid = -237.1                               #[kJ/mol]
delta_g_o = delta_g_0_ideal_gas-delta_g_0_liquid        #[kJ/mol]

print " From the values of Table A.8 given in the book the free energy change is %0.2f kJ/mol"%(delta_g_o)

 Total change in the free energy of water going under given conditions is 8.55 kJ/mol
From the values of Table A.8 given in the book the free energy change is 8.50 kJ/mol


### Example 12.8 Page: 330¶

In [21]:
import math

T1 = 273.15+25              #[K]
T2 = 273.15+400             #[K]
R = 8.314                   #[J/(mol*K)]

g0_NH3 = -16.5              #[kJ/mol]
g0_N2 = 0                   #[kJ/mol]
g0_H2 = 0                   #[kJ/mol]

delta_g_0 = g0_NH3 - 0.5*g0_N2 - 1.5*g0_H2          #[kJ/mol]

K_1 = math.exp(-delta_g_0*1000/(R*T1))          # Equilibrium consmath.tant of the reaction at temperature 298.15 K

h0_NH3 = -46.1          #[kJ/mol]
h0_N2 = 0               #[kJ/mol]
h0_H2 = 0               #[kJ/mol]

del_h_1 = h0_NH3 - 0.5*h0_N2 - 1.5*h0_H2        #[kJ/mol]

a_NH3 = 3.578
a_H2 = 3.249
a_N2 = 3.280
b_NH3 = 3.020*10**(-3)      #[1/K]
b_H2 = 0.422*10**(-3)
b_N2 = 0.593*10**(-3)
c_NH3 = 0                   #[1/K**(2)]
c_H2 = 0                    #[1/K**(2)]
c_N2 = 0                    #[1/K**(2)]
d_NH3 = -0.186*10**(5)      #[K**(2)]
d_H2 = 0.083*10**(5)        #[K**(2)]
d_N2 = 0.040*10**(5)        #[K**(2)]

del_a = a_NH3 - 0.5*a_N2 - 1.5*a_H2
del_b = b_NH3 - 0.5*b_N2 - 1.5*b_H2
del_c = c_NH3 - 0.5*c_N2 - 1.5*c_H2
del_d = d_NH3 - 0.5*d_N2 - 1.5*d_H2

I = R*( del_a*T1 + del_b*T1**(2)/2 + del_c*T1**(3)/3 - del_d/T1)        #[J/mol]

def f5(T):
return (del_h_1*1000 - I + R*(del_a*T + del_b*T**(2)/2 + del_c*T**(3)/3 - del_d/T))/T**(2)

K_2 = K_1*math.exp(X)

print " Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are "
print " K = %0.0f at temperature 25 deg C"%(K_1)
print " K = %f at temperature 400 deg C"%(K_2)

 Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are
K = 778 at temperature 25 deg C
K = 0.013588 at temperature 400 deg C


### Example 12.9 Page: 335¶

In [22]:
from scipy.optimize import fsolve
import math

n_H2_0 = 1.5            #[mol]
n_N2_0 = 0.5            #[mol]
n_NH3_0 = 0             #[mol]
T_1 = 298.15            #[K]
T_2 = 673.15            #[K]
P = 1.                  #[bar]

K_298 = 778.            # at temperature 298.15K
K_673 = 0.013           # at temperature 673.15K

def g(e_1):
return  ((0+e_1)/(2-e_1))/(((0.5-0.5*e_1)/(2-e_1))**(0.5)*((1.5-1.5*e_1)/(2-e_1))**(1.5))-K_298

e_1 = fsolve(g,0.97)
y_NH3_298 = e_1/(2-e_1)

def h(e_2):
return  ((0+e_2)/(2-e_2))/(((0.5-0.5*e_2)/(2-e_2))**(0.5)*((1.5-1.5*e_2)/(2-e_2))**(1.5))-K_673

e_2 = fsolve(h,0)
y_NH3_673 = e_2/(2-e_2)

print " The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is %f"%(y_NH3_298)
print " The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is %f"%(y_NH3_673)

 The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is 0.939036
The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is 0.004187


### Example 12.10 Page: 337¶

In [23]:
from scipy.optimize import fsolve
import math

Temp = 273.15+400           #[K]
P = 150*1.01325             #[bar]

K_673 = 0.013

K = K_673*(P/1)**(1.5+0.5-1)

def f(e):
return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K
e=fsolve(f,0.5)

y_NH3 = (0+e)/(2-e)

print "The mole fraction of the ammonia in the equilibrium is %0.2f"%(y_NH3)

The mole fraction of the ammonia in the equilibrium is 0.31


### Example 12.11 Page: 338¶

In [24]:
from scipy.optimize import fsolve
import math

T = 273.15+400              #[K] given temperature
P = 150*1.01325             #[bar] given pressure

K_673 = 0.013

K_v = (10**((0.1191849/T + 91.87212/T**(2) + 25122730/T**(4))*P))**(-1)

K = (K_673/K_v)*(P/1)**(1.5+0.5-1)

def f(e):
return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K
e = fsolve(f,0.2)

y_NH3 = (0+e)/(2-e)

print " The mole fraction of the ammonia in the equilibrium is %0.2f"%(y_NH3)

 The mole fraction of the ammonia in the equilibrium is 0.34


### Example 12.12 Page: 340¶

In [25]:
from scipy.optimize import fsolve
import math

p_i = 1.                #[atm] initial pressure
P = 150.                #[atm] final pressure
T = 273+25.             #[K] Given temperature
R = 8.314               #[J/(mol*K)]

delta_g_0 = 10.54*1000          #[J/mol]

K = math.exp((-delta_g_0)/(R*T))

def f(e):
return  (e*e)/((1-e)*(1-e))-K
e = fsolve(f,0)

v_C2H5OOC2H5 = 97.67            #[ml/mol]
v_H2O = 18.03                   #[ml/mol]
v_C2H5OH = 58.30                #[ml/mol]
v_CH3COOH = 57.20               #[ml/mol]

delta_v = v_C2H5OOC2H5 + v_H2O - v_C2H5OH - v_CH3COOH       #[ml/mol]

def g(e_1):
return  (e_1*e_1)/((1-e_1)*(1-e_1))*math.exp((delta_v)/(R*T)*(P-p_i))-K
e_1 = fsolve(g,0.2)

a = e_1/e

print "On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar \
reacmath.tants at equilibrium becomes %f times of initial"%(a)

On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar reacmath.tants at equilibrium becomes 0.994639 times of initial


### Example 12.13 Page: 342¶

In [26]:
import math

P = 150.            #[atm] given pressure
T = 400.            #[C] temperature
K = 0.013
K_v = 0.84
delta_v = 1.5+0.5-1

K_p = (K/K_v)*(1/1)**(delta_v)              #[1/bar]

print " Value of the K_p at the given condition is %f 1/bar)"%(K_p)
print  " The basic K is dimensionless%( but K_p has the dimensions of pressure to the power."

 Value of the K_p at the given condition is 0.015476 1/bar)
The basic K is dimensionless%( but K_p has the dimensions of pressure to the power.

In [ ]: