CH4-----> 2H2 (s) +C
delG= Change in Gibbs free energy= delH- TdelS= -74.85*1000
J/mol- 298*(-80.67)=50810 J/mol= -50.810 Kj/mol
delG= -RT ln K
lnK= -delG/RT= 50.810*1000 /(8.314*298)=20.51
K= 8.06*108
KP= [PH2]2/ P[CH4]
let x= degree of dissociation of CH4
                           Â
CH4--------------> C+2H2
Initial                    Â
1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â 0
dissociation          Â
-x                 Â
x      2x
Equilibrium        Â
1-x                  Â
x       2x
total moles of gas = 1-x+2x= 1+x
Mole fraction :Â Â CH4= (1-x)/
(1+x)Â Â Â Â H2= 2x/(1+x)
Partial pressures : CH4 =0.01*(1-x)/
(1+x)Â Â Â Â Â Â Â Â Â Â Â Â
H2= 2x*0.01/(1+x)
Kp = {2x*0.01)2/(1+x)}2/
(0.01*(1-x)/(1+x)= 4*0.01*x2/(1-x2)=
8.06*108
               Â
x2/(1-x)2= 201.5*108
x/(1-x)= 141951
x= 141951- 141951x
141952x= 141951
x= 141951/141952 =0.999993
=
