Chapter 18:Entropy, Free Energy,and Equilibrium¶

Example no:18.2,Page no:809¶

In [22]:
#Variable declaration
#(a)
SCaO=39.8 #standard entropy of CaO, J/K mol
SCO2=213.6 #standard entropy of CO2, J/K mol
SCaCO3=92.9 #standard entropy of CaCO3, J/K mol
#(b)
SNH3=193 #standard entropy of NH3, J/K mol
SN2=192 #standard entropy of N2, J/K mol
SH2=131 #standard entropy of H2, J/K mol
#(c)
SHCl=187 #standard entropy of HCl, J/K mol
SH2=131 #standard entropy of H2, J/K mol
SCl2=223 #standard entropy of Cl2, J/K mol

#Calculation
#(a)
deltaSrxn1=SCaO+SCO2-SCaCO3 #standard entropy change of the reaction, J/K mol
#(b)
deltaSrxn2=2*SNH3-(SN2+3*SH2) #standard entropy change of the reaction, J/K mol
#(c)
deltaSrxn3=2*SHCl-SH2-SCl2 #standard entropy change of the reaction, J/K mol

#Result
print"(a) the standard entropy of reaction is :",deltaSrxn1,"J/K mol"
print"(b) the standard entropy of reaction is :",deltaSrxn2,"J/K mol"
print"(c) the standard entropy of reaction is :",deltaSrxn3,"J/K mol"

(a) the standard entropy of reaction is : 160.5 J/K mol
(b) the standard entropy of reaction is : -199 J/K mol
(c) the standard entropy of reaction is : 20 J/K mol


Example no:18.4,Page no:817¶

In [2]:
#Variable declaration
#(a)
GCO2=-394.4 #free energy of formation of CO2, kJ/mol
GH2O=-237.2 #free energy of formation of H2O, kJ/mol
GCH4=-50.8 #free energy of formation of CH4, kJ/mol
GO2=0 #free energy of formation of O2, kJ/mol
#(b)
GMg=0 #free energy of formation of Mg, kJ/mol
GMgO=-569.6 #free energy of formation of MgO, kJ/mol
GO2=0 #free energy of formation of O2, kJ/mol

#Calculation
deltaGrxn1=(GCO2+GH2O*2)-(GCH4+2*GO2) #standard free energy change of the reaction, kJ/mol
deltaGrxn=(GO2+GMg*2)-(2*GMgO) #standard free energy change of the reaction, kJ/mol

#Result
print"(a) The standard free energy change of reaction is :",deltaGrxn1,"kJ/mol"
print"(b) The standard free energy change of reaction is :",round(deltaGrxn),"kJ/mol"

(a) The standard free energy change of reaction is : -818.0 kJ/mol
(b) The standard free energy change of reaction is : 1139.0 kJ/mol


Example no:18.5,Page no:820¶

In [24]:
#Variable declaration
#for fusion
T1=5.5+273 #temperature of fusion, K
deltaH1=10.9*1000 #change in enthalpy, J/mol
#for vaporisation
T2=80.1+273 #temperature of vaporisation, K
deltaH2=31*1000 #change in enthalpy, J/mol

#Calculation
deltaSf=deltaH1/T1 #since in fusion deltaG=0, J/ K mol
deltaSv=deltaH2/T2 #since in vaporisation deltaG=0, J/ K mol

#Result
print"The entropy change for fusion is",round(deltaSf,1),"J/K.mol"
print"The entropy change for condensation is : ",round(deltaSv,1),"J/K mol"

The entropy change for fusion is 39.1 J/K.mol
The entropy change for condensation is :  87.8 J/K mol


Example no:18.6,Page no:823¶

In [25]:
T=298 #temperature, K
R=8.314 #gas constant, J/K mol
GH2=0 #free energy of formation of H2, kJ/mol
GH2O=-237.2 #free energy of formation of H2O, kJ/mol
GO2=0 #free energy of formation of O2, kJ/mol

#Calculation
import math
deltaG=1000*(2*GH2+GO2-2*GH2O) #free energy of rxn, J/mol
Kp=math.exp(-deltaG/(R*T)) #equilibrium constant for rxn

#Result
print"deltaG_rxn=",deltaG/1000,"kJ/mol"
print"The equilibrium constant for the given reaction is :%.e"%Kp

deltaG_rxn= 474.4 kJ/mol
The equilibrium constant for the given reaction is :7e-84


Example no:18.7,Page no:824¶

In [26]:
#Variable declaration
T=298 #temperature, K
R=8.314 #gas constant, J/K mol
Ksp=1.6*10**-10 #solubility constant

#Calculation
import math
deltaG=-R*T*math.log(Ksp) #here solubility product is equal to equilibrium constant

#Result
print"The free energy for the given reaction is :",round(deltaG*10**-3),"kJ/mol"

The free energy for the given reaction is : 56.0 kJ/mol


Example no:18.8,Page no:825¶

In [27]:
#Variable declaration
T=298 #temperature, K
R=8.314 #gas constant, J/K mol
deltaG0=5.4*10**3 #standard free energy, kJ/mol
pNO2=0.122 #pressure of NO2, atm
pN2O4=0.453 #pressure of N2O4, atm

#Calculation
import math
deltaG=deltaG0+R*T*math.log(pNO2**2/pN2O4) #here solubility product is equal to equilibrium constant

#Result
if(deltaG<0):#equilibrium determination
d="net reaction proceeds from left to right to reach equilibrium"
else:
d="net reaction proceeds from right to left to reach equilibrium"
print"The free energy for the given reaction is :",round(deltaG*10**-3,2),"kJ/mol and",d

The free energy for the given reaction is : -3.06 kJ/mol and net reaction proceeds from left to right to reach equilibrium