Question

Consider the dissociaton of methane, into the elements H2g and C(s) given

a) the change of enthalpy of reaction is -74.85 kJ/mol and change of entropyof reaction is -80.67 J/mol K at 298K. Calculate the Keq at 298K.

b) Assuming that change in enthalpy of formation is independent of temperature, calculate K at 323K.

c) Calculate the degree of dissociation of methane at 298K and Ptotal at 0.01 bar.

Answer 1

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

                             CH4--------------> C+2H2

Initial                      1                    0      0

dissociation            -x                   x       2x

Equilibrium          1-x                    x        2x

total moles of gas = 1-x+2x= 1+x

Mole fraction :   CH4= (1-x)/ (1+x)     H2= 2x/(1+x)

Partial pressures : CH4 =0.01*(1-x)/ (1+x)             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

                 x²/(1-x)²= 201.5*10⁸

x/(1-x)= 141951

x= 141951- 141951x

141952x= 141951

x= 141951/141952 =0.999993

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