In [13]:

```
import math
T =273.15+25 #[K] given temperature
R = 8.314 #[J/(mol*K)] universal gas consmath.tant
g_0_H = 0 #[kJ/mol]
g_0_OH = -157.29 #[kJ/mol]
g_0_H2O = -237.1 #[kJ/mol]
delta_g_0 = g_0_H + g_0_OH - g_0_H2O #[kJ/mol]
delta_g_1 = delta_g_0*1000 #[J/mol]
K = math.exp((-delta_g_1)/(R*T))
K_w = K
print "At the equilibrium the product of the hydrogen ion and hydroxil ion is %0.1e"%(K_w)
```

In [14]:

```
from scipy.optimize import fsolve
import math
n_H2SO4 = 1. #[mol] mole of the sulphuric acid
w_water = 1000. #[g] weight of the water
T =273.15+25 #[K] temperature
R = 8.314 #[J/(mol*K)]
g_0_H = 0 #[J/mol] free energy of the hydrogen ion
g_0_HSO4 = -756.01*1000 #[J/mol] free energy of the bisulphate ion
g_0_H2SO4 = -744.50*1000 #[J/mol] free enery of sulphuric acid
delta_g_0 = g_0_H + g_0_HSO4 - g_0_H2SO4 #[J/mol]
K_1 = math.exp((-delta_g_0)/(R*T))
g_0_H = 0 #[J/mol] free energy of the hydrogen ion
g_0_SO4 = -744.62*1000 #[J/mol] free energy of sulphate ion
g_0_HSO4 = -756.01*1000 #[J/mol] free energy of the bisulphate ion
delta_g_1 = g_0_H + g_0_SO4 - g_0_HSO4 #[J/mol]
K_2 = math.exp((-delta_g_1)/(R*T))
def F(e):
f = [0,0]
f[0] = ((e[0]-e[1])*(e[0]+e[1]))/(1-e[0]) - K_1
f[1] = ((e[1])*(e[0]+e[1]))/(e[0]-e[1]) - K_2
return f
e = [0.8,0.1]
y = fsolve(F,e)
e_1 = y[0]
e_2 = y[1]
m_H2SO4 = 1-e_1 # [molal]
m_HSO4 = e_1 - e_2 #[molal]
m_SO4 = e_2 #[molal]
m_H = e_1 + e_2 #[molal]
print " The equilibrium concentration of H2SO4 in terms of molality is %f molal"%(m_H2SO4)
print " The equilibrium concentration of HSO4- in terms of molality is %f molal"%(m_HSO4)
print " The equilibrium concentration of SO4-- in terms of molality is %f molal"%(m_SO4)
print " The equilibrium concentration of H+ in terms of molality is %f molal"%(m_H)
```

In [15]:

```
from scipy.optimize import fsolve
import math
P = 10. #[MPa] given pressure
T = 250. #[C] Temperature
n_T_0 = 1. #[mol]
n_CO = 0.15 #[mol]
n_CO2 = 0.08 #[mol]
n_H2 = 0.74 #[mol]
n_CH4 = 0.03 #[mol]
V_1 = -2
V_2 = 0
K_1 = 49.9 # For the first reaction
K_2 = 0.032 # For the second reaction
v_CO_1 = -1
v_H2_1 = -2
v_CH3OH_1 = +1
v_CO2_2 = -1
v_H2_2 = -1
v_CO_2 = +1
v_H2O_2 = +1
def F(e):
f = [0,0]
f[0] = ((0 + e[0])/(1 - 2*e[0]))/(((0.15 - e[0] + e[1])/(1 - 2*e[0]))*((0.74 - 2*e[0] - e[1])/(1 - 2*e[0]))**(2)) - K_1
f[1] = (((0.15 - e[0] + e[1])/(1 - 2*e[0]))*((0 + e[1])/(1 - 2*e[0])))/(((0.08 - e[1])/(1 - 2*e[0]))*((0.74 - 2*e[0] - e[1])/(1 - 2*e[0]))) - K_2
return f
e = [0.109, 0]
y = fsolve(F,e)
e_1 = y[0]
e_2 = y[1]
c_CO2 = e_2/(n_CO2)*100
c_CO = e_1/(n_CO + 0.032)*100
print " Percent conversion of CO is %f%%"%(c_CO)
print " Percent conversion of CO2 is %f%%"%(c_CO2)
```

In [16]:

```
import math
T = 273.15+25 #[K] Temperature
R = 8.314 #[J/(mol*K)] universal gas consmath.tant
g_0_Ag = 77.12*1000 #[J/mol]
g_0_Cl = -131.26*1000 #[J/mol]
g_0_AgCl = -109.8*1000 #[J/mol]
delta_g_0 = g_0_Ag + g_0_Cl - g_0_AgCl #[J/mol]
K = math.exp((-delta_g_0)/(R*T))
a_AgCl = 1.00
print "The amount of solid dissolved in terms of solubility product is %0.2e"%(K)
```

In [17]:

```
import math
T = 273.15+25 #[K] Given temperature of air
P = 1. #[atm] Pressure of the air
y_CO2 = 350.*10**(-6) # Amount of CO2 present in air at the given condition
R = 8.314 #[J/(mol*K)]
g_0_H2CO3 = -623.1*1000 #[J/mol]
g_0_H = 0. #[J/mol]
g_0_HCO3 = -586.85*1000 #[J/mol]
delta_g_0 = g_0_H + g_0_HCO3 - g_0_H2CO3 #[J/mol]
K_1 = math.exp((-delta_g_0)/(R*T))
g_0_CO3 = -527.89*1000 #[J/mol]
delta_g_1 = g_0_H + g_0_CO3 - g_0_HCO3 #[J/mol]
K_2 = math.exp((-delta_g_1)/(R*T))
H = 1480. #[atm]
x_CO2 = P*y_CO2/H
n_water = 1000/18. #[mol]
m_CO2 = x_CO2*n_water #[molal]
m_HCO3 = math.sqrt(K_1*m_CO2) #[molal]
m_H = m_HCO3 #[molal]
m_CO3 = K_2*(m_HCO3/m_H) #[molal]
print " Amount of the CO2 dissolved in water in equilibrium with air is \t\t\t%0.2e molal"%(m_CO2)
print " Conentration of HCO3 ion and hydrogen ion H- in solution in equilibrium with air is %0.2e molal"%(m_HCO3)
print " And concentration of CO3 ion in the solution in equilibrium with air is\t\t%0.2e molal"%(m_CO3)
```

In [18]:

```
import math
m_H = 10**(-10) #[molal] molality of hydrogen ion
K_1 = 4.5*10**(-7)
K_2 = 4.7*10**(-11)
m_CO2 = 1.32*10**(-5) #[molal] from previous example
m_HCO3 = K_1*(m_CO2/m_H) #[molal]
m_CO3 = K_2*(m_HCO3/m_H) #[molal]
print " Amount of the CO2 dissolved in water in equilibrium with air is \t%0.2e molal"%(m_CO2)
print " Conentration of HCO3 ion in solution in equilibrium with air is \t %0.2e molal"%(m_HCO3)
print " And concentration of CO3 ion in the solution in equilibrium with air is %0.2e molal"%(m_CO3)
```

In [19]:

```
import math
T = 298.15 #[K] Temperature
F = 96500. #[(coulomb)/(mole*electrons)] faraday consmath.tant
n_e = 6. #[electron]
g_0_CO2 = -394.4*1000 #[J/mol]
g_0_Al = 0 #[J/mol]
g_0_C = 0 #[J/mol]
g_0_Al2O3 = -1582.3*1000 #[J/mol]
delta_g_0 = 1.5*g_0_CO2 + 2*g_0_Al - 1.5*g_0_C - g_0_Al2O3 #[J/mol]
E_0 = (-delta_g_0)/(n_e*F) #[V]
print "Standard state cell voltage for the production of aluminium is %f Volt"%(E_0)
```

In [20]:

```
import math
T = 298.15 #[K] Temperature
F = 96500. #[(coulomb)/(mole*electrons)] faraday consmath.tant
delta_g_0 = -587.7*1000 #[J/mol]
n_e = 1 #[electron] no of electron transferred
E_298_0 = (-delta_g_0)/(n_e*F) #[V]
print "The reversible voltage for given electrochemical device is %f Volt"%(E_298_0)
```

In [21]:

```
import math
T = 298.15 #[K] Temperature
P_0 = 1. #[atm]
P = 100. #[atm]
E_0 = -1.229 #[V]
F = 96500. #[(coulomb)/(mole*electrons)] faraday consmath.tant
R = 8.314 #[J/(mol*K)] universal gas consmath.tant
n_e = 2. #[(mole electrons)/mole]
E = E_0 - 1.5*(R*T)*math.log(P/P_0)/(n_e*F)
print "The equilibrium cell voltage of electrolytic cell if feed and product are at the pressure 100 atm is %f Volt"%(E)
```

In [22]:

```
import math
T = 273.15+25 #[K] Temperature
P = 11.38/760 #[atm] Pressure
R = 0.08206 #[(L*atm)/(mol*K)] Gas consmath.tant
v = 0.6525/0.04346 #[L/g] Specific volume
M = 60.05 #[g/mol] Molecular weight of HAc in the monomer form
V = v*M #[L/mol]
z = (P*V)/(R*T)
print "The value of the compressibility factor for HAc at given condition is %f"%(z)
```

In [23]:

```
from scipy.optimize import fsolve
import math
T = 273.15+25 #[K] Temperature
P = 11.38 #[torr] Pressure
K = 10**(-10.4184 + 3164/T) #[1/torr]
def f(y_HAc_2):
return K*(P*(1-y_HAc_2))**(2)/P-y_HAc_2
y_HAc_2 = fsolve(f,0)
y_HAc = 1-y_HAc_2
print "Mole fraction of the monomer in the vapour phase is %f"%(y_HAc)
print "Mole fraction of the dimer in the vapour phase is %f"%(y_HAc_2)
```

In [25]:

```
import math
T = 273.15+25 #[K] Temperature
P = 11.38/760 #[atm] Pressure
R = 0.08206 #[(L*atm)/(mol*K)] Gas consmath.tant
v = 0.6525/0.04346 #[L/g] Specific volume
y_HAc = 0.211 # monomer
y_HAc_2 = 0.789 # dimer
M_HAc = 60.05 #[g/mol] monomer
M_HAc_2 = 120.10 #[g/mol] dimer
M_avg = M_HAc*y_HAc + M_HAc_2*y_HAc_2 #[g/mol]
V = v*M_avg #[L/mol]
z = (P*V)/(R*T)
print "The compressibility factor z for the gaseous mixture is %f"%(z)
```

In [ ]:

```
```