Hydrogen is to be produced via the steam reforming of methane according to the f

Question

Hydrogen is to be produced via the steam reforming of methane according to the following reaction:

CH4 + 2H2O CO2 + 4H2

An initial mixture at 1 atm and 800K consists of 50% excess steam. Determine the equilibrium composition of the products at 800 K and I atm.

Take the values of the equilibrium constants of formation for CH4, H2O and CO2 at 800K to be 0.124, 5.08 and 13.91, respectively. (5 points)

[Hint: Use the equation relating the value of Kp of the reaction to the (Kp)CH4, (Kp)CO2, and (Kp)H2O]

Explanation / Answer

for the given reaction, Kp = [CO2] [H2]4 / [ CH4][H2O]2 =

for CH4, C(s) + 2H2--->CH4, KpCH4 = [CH4]/ [H2]2

for H2O, H2(g) +0.5O2---------> H2O , KpH2O= [H2O]/ [H2][O2]0.5

[KPH2O]2= [H2O]2/ [H2]2 [O2]

[KPH2O]2* KPCH4 = [H2O2[CH4] /[H2]4 [O2]

for CO2 , C+O2---------->CO2, KpCO2 = [CO2]/ [O2]

[KPH2O]2* [KPCH4]/ [KPCO2] = [H2O]2 [CH4]/ [H2]4[CO2] =1/KP

hence 1/KP= [5.08)2*0.124/13.91 = 0.23

Kp = 1/0.23= 4.35

Basis : 1 mole of CH4, Steam required as per the reaction = 2 moles

Steeam supplied = 2*1.5= 3 moles

Partial pressures : Mole fraction* total pressure , CH4= (1/4)*1= 0.25 atm and Steam = 3/4=0.75 atm

let P= drop in pressure of CH4 to reach equilibirum

So at equilibrium [CH4]= 0.25-P, steam = 0.75-2P, CO2= P and H2= 4P

Kp = 4.35 = P*(4P)4/ (0.25-P)*(0.75-2P)2

P5/ (0.25-P)* (0.75-2P)4= 4.35/64= 0.068

when solved using excel, P= 0.157 atm

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