# Chapter 5:The second Law of Thermodynamics¶

## Example 5.1,Page no:59¶

In [1]:
T1=373.0 				#initial temperature [K]
T2=573.0 				#final temperature [K]
Q2=750.0 				#Heat absorbed by carnot engine[J]

e=(T2-T1)/T2 			#efficiency of the engine
W=e*Q2 				#Workdone by the engine[J]
Q1=T1*Q2/T2 			#Heat rejected by the engine[J]
print"Efficiency of the engine =",round(e,3)
print"\n Workdone by the engine =",round(W),"J"
print"\n Heat rejected by the engine =",round(Q1),"J"

Efficiency of the engine = 0.349

Workdone by the engine = 262.0 J

Heat rejected by the engine = 488.0 J


## Example 5.2,Page no:61¶

In [2]:
T1=250.0 #temperature of heat rejection[K]
T2=1000.0 #temperature of heat absorption[K]
e=1-(T1/T2)
print"Efficiency of the corresponding carnot engine =",e,"or",e*100,"%"
print" Therefore , the inventors claim of 80% efficiency is absurd.The patent application should be rejected"

Efficiency of the corresponding carnot engine = 0.75 or 75.0 %
Therefore , the inventors claim of 80% efficiency is absurd.The patent application should be rejected


## Example 5.3,Page no:62¶

In [46]:
T1=323.0 				#temperature [K]
T2=423.0 				#temperature [K]
W=1.3 				#work [KJ]
e=(T2-T1)/T2 			#efficiency
Q2=W/e 				#minimum heat withdrawal from heat source[KJ]
print"Minimum heat withdrawal from heat source=",round(Q2,2),"kJ"

Minimum heat withdrawal from heat source= 5.5 kJ


## Example 5.5,Page no:64¶

In [4]:
T=298 			#Temperature [K]
n=1 			#no. of moles
V1=500 			#initial volume [cm3]
V2=1000 		#final volume [cm3]
R=8.314 		#Universal gas constant [J/mol/K]
import math
S=R*math.log(V2/V1)		#molar entropy change at constant temperature[J/K]
print"Molar entropy change of argon =",round(S,1),"J/K"

Molar entropy change of argon = 5.8 J/K


## Example 5.6,Page no:64¶

In [47]:
W=1728.0 			#Isothermal and reversible work done[J/mol]
T=298.0 			#Isothermal temperature[K]

S=W/T 			#change in molar entropy for isothermal and reversible process
print"The change in molar entropy =",round(S,1),"JK^-1mol^-1"

The change in molar entropy = 5.8 JK^-1mol^-1


## Example 5.7,Page no:68¶

In [6]:
H=-92.22 			#Standard reaction enthalpy[KJ]
T=298 				#Temperature [K]

#standard reaction enthalpy is H.Therefore, heat gained by the surroundings at 298K is -H
S=-H*1000/T 			#Change in entropy[J/K]
print"Change in entropy of the surroundings at 298k =",round(S,1),"J/K"

Change in entropy of the surroundings at 298k = 309.5 J/K


## Example 5.8,Page no:69¶

In [7]:
T1=298.0 				#Initial Temperature[K]
T2=573.0 				#Final Temperature[K]
Cv=29.1 			#Specific Heat capacity of argon gas [J/K/mol]
n=1 				#no. of moles

import math
S=n*Cv*math.log(T2/T1) 		#Change in entropy [J/K]
print"The change in entropy of the argon gas is",round(S,2),"J/K"

The change in entropy of the argon gas is 19.03 J/K


### Example 5.9,Page no:69¶

In [8]:
T1=276.0 				#Initial temperature[K]
Tf=278.7 			#Freezing point temperature[K]
Tb=353.3 			#Boiling point temperature[K]
T2=373.0 				#Final temperature[K]
Hf=9870.0 			#Standard enthalpy of fusion[J/mol]
Hv=30800.0 			#Standard enthalpy of vaporization[J/mol]
Cp=136.1 			#Specific heat capacity of benzene[J/K/mol]
mol_wt=78.0 			#molecular weight of benzene[g/mol]
mass=200.0			#weight of solid benzene[g]
print"Cp doesnot change within this temp limit"
import math
n=mass/mol_wt 			#no. of moles

S1=n*Cp*math.log(Tf/T1) 	#entropy change in heating [J/K]
S2=n*Hf/Tf 			#entropy change in melting[J/K]
S3=n*Cp*math.log(Tb/Tf) 	#entropy change in heating[J/K]
S4=n*Hv/Tb 			#entropy change in vaporization[J/K]
S5=n*Cp*math.log(T2/Tb) 	#entropy change in heating[J/K]
S=S1+S2+S3+S4+S5 		#total entropy change in heating from 276 to 373K
print"Total entropy change in heating 200g benzene from 3 to 100C is",round(S,1),"J/K or",round(S/1000,3),"KJ/K"
print"\nNOTE:In textbook the value of 'n' is wrongly calculated as 25.64 instead of 2.564,SO there is a error in answer shown in book"

Cp doesnot change within this temp limit
Total entropy change in heating 200g benzene from 3 to 100C is 419.4 J/K or 0.419 KJ/K

NOTE:In textbook the value of 'n' is wrongly calculated as 25.64 instead of 2.564,SO there is a error in answer shown in book


## Example 5.10,Page no:71¶

In [9]:
mass=32 			#weight of methane gas[gm]
P1=6*10**5 			#Initial temperature[N/m2]
P2=3*10**5 			#Final pressure[N/m2]
mol_wt=16 			#molecular weight of methane gas[g/mol]
T=298 				#Temperature[K]
R=8.314 			#Universal gas constant[J/K/mol]
import math
n=mass/mol_wt 			#no. of moles
S=n*R*math.log(P1/P2) 		#change in entropy of gas[J/K]

print"The change in entropy of the gas is",round(S,2),"J/K"

The change in entropy of the gas is 11.53 J/K


## Example 5.11,Page no:75¶

In [4]:
import math
black=2              #No. of black balls
white=1              #No. of white ball

W=math.factorial(black+white)/(math.factorial(black)*math.factorial(white))
P=1.0/W
E1=0+1+2
E2=E1
E3=E2
E=E1+E2+E3
E_av=E/3
E1_dash=1+2+3
E2_dash=E1_dash
E3_dash=E2_dash
E_dash=E1_dash+E2_dash+E3_dash
change=E_dash-E

print"1.Total No. of possible configuration:",W
print"2.Probability of getting a configuration=",P,"or 1/3"
print"3.Total energy of system=",E
print"  Therefore,Average energy=",E_av
print"4.In this case,Total energy=",E_dash
print"  Change in total energy of system=",change

1.Total No. of possible configuration: 3
2.Probability of getting a configuration= 0.333333333333 or 1/3
3.Total energy of system= 9
Therefore,Average energy= 3
4.In this case,Total energy= 18
Change in total energy of system= 9


## Example 5.12,Page no:77¶

In [50]:
n=1.0 			#no. of moles
T=273.0 			#temperature [K]
Hf=6000.0 		#enthalpy of fusion at 273K [J/mol]
k=1.38*(10**-23) 	#boltzmann constant[J/K]

p=Hf/(k*T)/2.303
print"\nTHE RESULT IS 10^24,which is too large to be displayed by ipython "
print"This value of w is very large to calculate for python,because it's in the range of 10^24"
print"The relative no. of distinguishable quantum states in 1 mole of water and ice at 273K is 10^24"
print"\nTHE RESULT IS 10^24,which is too large to be displayed by ipython "

THE RESULT IS 10^24,which is too large to be displayed by ipython
This value of w is very large to calculate for python,because it's in the range of 10^24
The relative no. of distinguishable quantum states in 1 mole of water and ice at 273K is 10^24

THE RESULT IS 10^24,which is too large to be displayed by ipython


## Example 5.13,Page no:86¶

In [51]:
T=300 			#temperature[K]
n=4 			#no. of moles of an ideal gas
P1=2.02*10**5 		#initial pressure[N/m2]
P2=4.04*10**5 		#final pressure[N/m2]
R=8.314 		#Universal gas constant[J/K/mol]
import math
G=n*R*T*2.303*math.log10(P2/P1) 	#[J]
print" The change in Gibbs free energy is",round(G,1),"J"

 The change in Gibbs free energy is 6916.6 J


## Example 5.14,Page no:86¶

In [52]:
n=1 			#no. of moles
T=300 			#temperature[K]
V1=2 			#initial volume[m3]
V2=20 			#final volume[m3]
R=8.314 		#Universal gas constant[J/K/mol]
import math

A=-n*R*T*2.303*math.log10(V2/V1) 	#Change in work function[J/mol]
print"The change in Helmholts free energy is",round(A),"J/mol"

The change in Helmholts free energy is -5744.0 J/mol


## Example 5.15,Page no:87¶

In [14]:
print"C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)"
T=298 				#Temperature[k]
R=8.314 			#Universal gas constant[J/K/mol]
S=182.45 			#standard entropy change at 298K [J/K]
U=-2808 			#change in internal energy at 298K[KJ/mol]
#reaction is taking place in bomb calorimeter so no volume change
#therefore U=Q at constant volume

A=U-T*S*0.001 			#Energy extracted as heat[KJ/mol]
Wmax=A 				#work done [KJ/mol]
dn=6-6 				#change in no. of moles
H=U+dn*R*T 			#Change in enthalpy of the bomb calorimeter[KJ]
print"The energy change that can be extracted as heat is",round(A),"KJ/mol"
print"\nThe energy change that can be extracted as work is",round(-A),"KJ/mol"
print"\nThe change in enthalpy of bomb calorimeter is",round(H),"KJ/mol"

C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)
The energy change that can be extracted as heat is -2862.0 KJ/mol

The energy change that can be extracted as work is 2862.0 KJ/mol

The change in enthalpy of bomb calorimeter is -2808.0 KJ/mol


## Example 5.16,Page no:87¶

In [55]:
print"C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(l)"

T=298.0 				#temperature[K]
S=421.5 			#change in entropy[J/K]
H=-5109000.0 			#Heat of reaction[J]
R=8.314 			#Universal gas constant[J/K/mol]
dn=8-(1+12.5) 			#change in no. of moles

U=H 				#[J]
A=U-T*S 			#Change in helmholts free energy[J]
G=A+dn*R*T 			#Change in Gibbs free energy[J]
print"The change in Helmholts free energy is",round(A),"J"
print"\nThe change in Gibbs free energy is",round(G),"J"
print"The calculation is not precise in book,that's why a slight change in answer"

C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(l)
The change in Helmholts free energy is -5234607.0 J

The change in Gibbs free energy is -5248234.0 J
The calculation is not precise in book,that's why a slight change in answer


## Example 5.17,Page no:88¶

In [56]:
print"C3H6(g)+4.5O2(g)-->3CO2(g)+3H2O(l)"
S=-339.23 			#standard change in entropy [J/K]
T=298 				#temperature[K]
Hf1=20.42 			#enthalpy of formation of C3H6(g)[J]
Hf2=-393.51 			#enthalpy of formation of CO2(g)[J]
Hf3=-285.83 			#enthalpy of formation of H2O(l)[J]
dn=3-4.5-1 			#change in no. of moles
R=8.314 			#Universal gas constant[J/K/mol]

H=3*Hf2+3*Hf3-Hf1 		#Enthalpy of the reaction[J]
U=H-dn*R*0.001*T 		#Change in internal energy of the reaction[J]
A=U-T*S*0.001 			#Helmholts free energy change[J]
G=A+dn*R*0.001*T 		#Gibbs free energy change[J]
print"The change in Helmholts free energy is",round(A,2),"kJ"
print"\nThe change in Gibbs free energy is",round(G,2),"kJ"

C3H6(g)+4.5O2(g)-->3CO2(g)+3H2O(l)
The change in Helmholts free energy is -1951.16 kJ

The change in Gibbs free energy is -1957.35 kJ


## Example 5.19,Page no:92¶

In [17]:
print"CH4(g)+2O2(g)-->CO2(g)+2H2O(l)"

S1=-242.98 				#standard entropy change for the combustion reaction[J/K]
Hf1=-74.81 				#Enthalpy of formation of CH4(g)[KJ/mol]
Hf2=-393.51 				#Enthalpy of formation of CO2(g)[KJ/mol]
Hf3=-285.83 				#Enthalpy of formation of H2O(l)[KJ/mol]
T=298 					#temperature[K]

H=Hf2+2*Hf3-Hf1 			#Change in enthalpy of reaction[KJ]
S2=-H*1000/T 				#Change in entropy of the surrounding[J/K]
Stotal=(S1+S2)*0.001 			#Total entropy change
print"The total change in entropy is",round(Stotal,2),"KJ/K"

CH4(g)+2O2(g)-->CO2(g)+2H2O(l)
The total change in entropy is 2.74 KJ/K


## Example 5.20,Page no:93¶

In [18]:
print"2H2(g)+O2(g)-->2H2O(l)"
Hf1=-285.83 				#standard enthalpy of formation of H2O(l)[KJ/mol]
S=-327 					#Standard entropy change for the same reaction[J/K]
T=298 					#temperature[K]

H=2*Hf1-0-0 				#Enthalpy of the reaction[KJ/mol]
G=H-T*S*0.001 				#Change in Gibbs free energy[KJ]
print"The change in Gibbs free energy is",round(G,2),"KJ\n "
print"As change in Gibbs free energy is negative.Therefore,the reaction is spontaneous"

2H2(g)+O2(g)-->2H2O(l)
The change in Gibbs free energy is -474.21 KJ

As change in Gibbs free energy is negative.Therefore,the reaction is spontaneous


## Example 5.21,Page no:94¶

In [57]:
print"CH4(g)+2O2(g)-->CO2(g)+2H2O(l)"
S=-242.98 			#standard entropy change for reaction [J/K]
T=298 				#temperature[K]
Gf1=-50.72 			#standard Gibbs free energy of formation for CH4(g)[KJ/mol]
Gf2=-394.36 			#standard Gibbs free energy of formation for CO2(g)[KJ/mol]
Gf3=-237.13 			#standard Gibbs free energy of formation for H2O(l)[KJ/mol]

G=Gf2+2*Gf3-Gf1 		#Standard Gibbs free energy for reaction[KJ/mol]
H=G+T*S*0.001 			#Standard enthalpy of reaction [KJ]
print"The standard enthalpy of reaction is",round(H,2),"kJ"

CH4(g)+2O2(g)-->CO2(g)+2H2O(l)
The standard enthalpy of reaction is -890.31 kJ


## Example 5.22,Page no:94¶

In [21]:
print"C6H12O6(s)+6O2(g)-->6CO2(g)+6H2O(l)"
mass=25.0 			#mass of glucose for combustion under standard condition[gm]
T=298 				#temperature[K]
Gf1=-910 			#Standard Gibbs free energy of formation for C6H12O6[KJ/mol]
Gf2=-394.4 			#Standard Gibbs free energy of formation for CO2(g)[KJ/mol]
Gf3=-237.13 			#Standard Gibbs free energy of formation for H2O(l)[KJ/mol]
mol_wt=180.0 			#molecular weight of glucose[gm/mol]

G=6*Gf2+6*Gf3-Gf1
n=mass/mol_wt 			#no. of moles
Gactual=G*n 			#Gibbs free energy for the combustion of 0.139mol of glucose
print"The energy that can be extracted as non-expansion work is",round(-Gactual),"KJ"

C6H12O6(s)+6O2(g)-->6CO2(g)+6H2O(l)
The energy that can be extracted as non-expansion work is 400.0 KJ


## Example 5.23,Page no:97¶

In [58]:
a=1.39*10**-2 		#constant for a vanderwaal's gas[lit2.atm/mol2]
b=3.92*10**-2 		#constant for a vanderwaal's gas[lit2.atm/mol2]
R=0.082 		#Universal gas constant[lit.atm/deg/mol]

Ti=(2*a)/(R*b) 		#inversion temperature [K]
print"The inversion temperature for the gas is",round(Ti,3)," K"

The inversion temperature for the gas is 8.649  K


## Example 5.26,Page no:100¶

In [59]:
T=169.25 			#Boiling point[K]
R=8.314 			#Universal gas constant[J/K/mol]
print"dlnP/dT=He/R*T**2"
print"dlnP/dT=(2.303*834.13/T**2)+(1.75/T)-(2.30*8.375*10**-3)"
print"Therefore using these two equations we calculate the He(enthalpy) of ethylene"

x=(2.303*834.13/T**2)+(1.75/T)-(2.30*8.375*10**-3) #it is dlnP/dT
He=R*0.001*T**2*x #Enthalpy of vaporization[J/mol]
print"\n\nThe Enthalpy of vaporization of ethylene at its boiling point is",round(He,3),"KJ/mol"

dlnP/dT=He/R*T**2
dlnP/dT=(2.303*834.13/T**2)+(1.75/T)-(2.30*8.375*10**-3)
Therefore using these two equations we calculate the He(enthalpy) of ethylene

The Enthalpy of vaporization of ethylene at its boiling point is 13.846 KJ/mol


## Example 5.27,Page no:101¶

In [24]:
P1=101.3 			#Initial Pressure[KPa]
P2=60 				#Final Pressure[KPa]
He=31.8 			#Enthalpy of vaporization[KJ/mol]
R=8.314 			#Universal gas constant[J/K/mol]
T1=353.2 			#boiling point of benzene at 101.3KPa[K]
import math

x=(T1**-1)-(R*0.001*math.log(P2/P1)/He)
T2=x**-1 			#Boiling point of benzene at 60KPa
print"The boiling point of benzene at 60KPa is",round(T2,1),"K"

The boiling point of benzene at 60KPa is 336.9 K


## Example 5.28,Page no:101¶

In [60]:
P1=0.016 			#Vapour pressure of pure ethanol at 273K[bar]
P2=0.470 			#Vapour pressure of pure ethanol at 333K[bar]
T1=273 				#initial temperature [K]
T2=333 				#final temperature[K]
R=8.314 			#Universal gas constant[J/K/mol]
P=1.01 				#vapour pressure at normal boiling point[bar]
import math

x=(T2**-1)-(T1**-1)
He=-R*0.001*math.log(P2/P1)/x 	#molar enthalpy of vaporization[J/mol]
t=(T2**-1)-(R*0.001*math.log(P/P2)/He)
T=(t**-1)-273 			#normal boiling point [C]
print"\n\nThe normal boiling point for pure ethanol is ",round(T,1),"C"
print"The molar enthalpy of vapourization is",round(He,2),"J/mol"


The normal boiling point for pure ethanol is  77.4 C
The molar enthalpy of vapourization is 42.58 J/mol


## Example 5.29,Page no:102¶

In [5]:
T2=353.2 			#normal boiling point of benzene at 1.01325bar[K]
T1=298	 			#temperature [K]
R=8.314 			#Universal gas constant[J/K/mol]
P2=1.01325 			#Vapour pressure of benzene[bar]
import math
#benzene obey's Trouton's rule
print" from Troutons rule , "
print" He/Tb=85J/K/mol"

He=85*T2 			#molar enthalpy of vapourization[J/K/mol]
x=(T2**-1)-(T1**-1)
t=-He*x/R
P1=P2/math.exp(t)
print"\nThe vapour pressure of benzene at 298K is",round(P1,3)," bar"

 from Troutons rule ,
He/Tb=85J/K/mol

The vapour pressure of benzene at 298K is 0.152  bar


## Example 5.30,Page no:111¶

In [27]:
c=1 			#no. of components(only CO2)
p=2 			#no. of phases(liquid + gas)

F=c-p+2 		#degree of freedom
print"Degrees of freedom is",F
print"Degrees of freedom 1 means that either pressure or temperature can be varied independently,i.e.when temperature is fixed,pressure is automatically fixed"

Degrees of freedom is 1
Degrees of freedom 1 means that either pressure or temperature can be varied independently,i.e.when temperature is fixed,pressure is automatically fixed


## Example 5.31,Page no:111¶

In [61]:
c=1 			#no. of components
p=1 			#no. of phases

F=c-p+2 		#Degrees of freedom
print"Degrees of freedom,F is",F
print"Degrees of freedom 2 means both the pressure and temperature can be varied independently"

Degrees of freedom,F is 2
Degrees of freedom 2 means both the pressure and temperature can be varied independently


## Example 5.32,Page no:113¶

In [30]:
P=1.75*10**-5 			#Vapour pressure of pure water at 293K[torr]
dP=1.1*10**-7 			#Lowering in vapour pressure of water

x=dP/P 				#mole fraction of sucrose
print"The mole fraction of sucrose is",round(x,6)

The mole fraction of sucrose is 0.006286


## Example 5.33,Page no:114¶

In [31]:
P=94.6 				#The vapour pressure of pure benzene at 298K[torr]
n1=20.0 				#no. of moles of pure benzene
n2=5.0 				#no. of moles of pure naphthalene

x=n1/(n1+n2) 			#(mole fraction of benzene)
p=x*P 				#the partial vapour pressure of benzene[torr]
print"The partial vapour pressure of benzene is",p,"torr"

The partial vapour pressure of benzene is 75.68 torr


## Example 5.34,Page no:114¶

In [32]:
x=0.28 				#mole fraction of solute
R=8.314 			#Universal gas constant[J/K/mol]
T=298 				#temperature[K]
import math

du=R*T*math.log(1-x) 		#reduction in chemical potential[J/mol]
print"The reduction in chemical potential is",round(-du,1),"J/mol"

The reduction in chemical potential is 813.9 J/mol


## Example 5.35,Page no:116¶

In [62]:
Kb=0.51 			#ebullioscopic constant of water [K*Kg/mol]
n=155.0/180.0 			#no. of moles of glucose
m=n/1 				#[mol/Kg]
Ti=373.0 				#Boiling point temperature of water[K]

Tf=(Ti+Kb*m)-273 		#boiling point temperature of the solution[C]
print"The boiling point of the solution is",round(Tf,2),"degree C"

The boiling point of the solution is 100.44 degree C


## Example 5.36,Page no:117¶

In [65]:
Ti=5.44 #freezing point of pure benzene[K]
Tf=4.63 #freezing point of solution[K]
m1=2.12 #mass of the solute[gm]
m2=125.0 #mass of the benzene[gm]
Kf=5.12 #cryoscopic constant of pure benzene[K*Kg/mol]

dTf=Ti-Tf 	#depression in freezing point[K]
M2=(m1*1000*Kf)/(m2*dTf) #molar mass of solute
print"The molar mass of solute is",round(M2),"(approx)"

The molar mass of solute is 107.0 (approx)


## Example 5.38,Page no:124¶

In [66]:
print"N2(g)+3H2(g)<=>2NH3(g)"
T=298 			#Temperature[K]
Gf1=-16450 		#Gibb's free energy of formation for NH3(g)[J/mol]
R=8.314 		#Universal gas constant[J/K/mol]
import math

Gf=2*Gf1			#Gibb's free energy for the reaction[KJ]
x=Gf/R/T
Kp=math.exp(-x)
print"The Kp for above reaction is",round(Kp),"or 5.85*10^5,in scientific notation(APPROX)"

N2(g)+3H2(g)<=>2NH3(g)
The Kp for above reaction is 584861.0 or 5.85*10^5,in scientific notation(APPROX)


## Example 5.39,Page no:124¶

In [37]:
print"0.5N2(g)+1.5H2(g)<=>NH3(g)"
T=298 #Temperature[K]
Kp=900 #Equilibrium constant for above reaction
P1=0.32 #partial pressure of N2(g)[bar]
P2=0.73 #partial pressure of H2(g)[bar]
P3=0.98 #partial pressure of NH3(g)[bar]
R=8.314 #Universal gas constant[J/K/mol]
import math

G=-R*T*math.log(Kp)
x=(P1**0.5)*(P2**1.5)
p=P3/x
Gr=(G+R*T*math.log(p))*0.001
print"The reaction Gibbs free energy is",round(Gr*1000),"J/mol "

0.5N2(g)+1.5H2(g)<=>NH3(g)
The reaction Gibbs free energy is -14322.0 J/mol


## Example 5.40,Page no:125¶

In [38]:
print"N2(g)+3H2(g)<=>2NH3(g)"

Kp1=5.85*10**5 #equilibrium constant at 298K
H1=-46.11 #standard enthalpy of formation of NH3(g)[KJ/mol]
T1=298 #Initial temperature[K]
T2=423 #Final temperature[K]
R=8.314 #Universal gas constant[J/K/mol]
import math

H=2*H1 #enthalpy for reaction [KJ]
t=(T1**-1)-(T2**-1)
x=-H*t/(R*0.001)
Kp2=Kp1*math.exp(x)
print"The Equilibrium constant for reaction  at 423K is",round(Kp2)

N2(g)+3H2(g)<=>2NH3(g)
The Equilibrium constant for reaction  at 423K is 35004905509.0


## Example 5.41,Page no:128¶

In [39]:
print"Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)"
T=298.0 			#Temperature[K]
R=8.314 		#Universal gas constant[J/K/mol]
E1=-0.7618 		#Standard electrode potential for Zn2+/Zn [volts]
E2=-0.403 		#Standard electrode potential for Cd2+/Cd [volts]
n=2.0 			#no. of electrons balancing

Ei=E2-E1 		#Standard potential for the reaction[volts]
import math
Gi=-n*F*Ei 		#Standard Gibb's Free Energy [KJ/mol]
Ki=math.exp(-Gi/R/T) 	#Equilibrium constant
print"The Free energy for the rection is",Gi*0.001,"KJ/mol"
print"The value of equilibrium constant is",Ki

print"Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)"
E3=0.6141 		#Standard electrode potential for Hg2SO4(s),SO4^2-/Hg(l) [volts]

Eii=E3-E2 		#Standard potantial for the reaction[volts]
Gii=-n*F*Eii 		#Standard Gibb's free energy[KJ/mol]
Kii=math.exp(-Gii/R/T) 	#Equilibrium constant
print"The Free energy for the rection is",round(Gii*0.001,1),"KJ/mol"
print"The value of equilibrium constant is",Kii
print"PLEASE REDO the last line calculation,It is showing wrong result in my PC"

Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)
The Free energy for the rection is -69.2484 KJ/mol
The value of equilibrium constant is 1.37586809667e+12
Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)
The Free energy for the rection is -196.3 KJ/mol
The value of equilibrium constant is 2.56773255559e+34
PLEASE REDO the last line calculation,It is showing wrong result in my PC


## Example 5.42,Page no:130¶

In [67]:
print"Zn(s)|ZnCl2(soln)||AgCl(s)|Ag-Ag|AgCl(s)|ZnCl2(soln)|Zn(s)"

m1=0.02 			#concentration[M]
Y1=0.65 			#mean ionic activity coefficient
m2=1.5 				#concentration[M]
Y2=0.29 			#mean ionic activity coefficient
R=8.314 			#Universal gas constant[J/K/mol]
T=298 				#Temperature [K]
import math

E=R*T*(math.log(m2*Y2/m1/Y1))*3/2/F 	#[volts]
print"The overall e.m.f of the cell is",round(E,4),"volt"

Zn(s)|ZnCl2(soln)||AgCl(s)|Ag-Ag|AgCl(s)|ZnCl2(soln)|Zn(s)
The overall e.m.f of the cell is 0.1352 volt


## Example 5.43,Page no:131¶

In [41]:
print"H2(g,1atm)|HCl(aq)|HCl(aq)|H2(g,1atm)"
m1=0.02 			#concentration[M]
Y1=0.88 			#mean ionic activity coefficient
m2=1 				#concentration[M]
Y2=0.81 			#mean ionic activity coefficient
R=8.314 			#universal gas constant[J/K/mol]
T=298 				#Temperature[K]
t=0.178 			#Tranference number of Cl-1
import math

E=-2*t*R*T*(math.log(m1*Y1/m2/Y2))/F 	#e.m.f of the cell[volts]
print"The e.m.f of the cell is",round(E,3)," volts"
print"\nWrongly calculated in book as 0.351 volt"

H2(g,1atm)|HCl(aq)|HCl(aq)|H2(g,1atm)
The e.m.f of the cell is 0.035  volts

Wrongly calculated in book as 0.351 volt


## Example 5.44,Page no:133¶

In [68]:
print"The values for reaction that goes on within the cadmium cell"
n=2 			#no. of moles
d=-5*10**-5 		#i.e d=dE/dT[V/K]
F=96500 		#[coulomb/mol]
T=298 			#Temperature [K]

dG=-n*E*F 		#Change in Gibb's free energy[J]
dS=n*F*d 		#Change in entropy [J/K]
dH=dG+T*dS 		#change in enthalpy[J]
print" dG=",dG,"J\nWrongly calculated in book as -195815 J"
print"\n dS=",dS,"J/K"
print"\n dH=",dH,"J"

The values for reaction that goes on within the cadmium cell
dG= -195823.59 J
Wrongly calculated in book as -195815 J

dS= -9.65 J/K

dH= -198699.29 J