Gaseous hydrogen can be produced by the steam cracking of methane in a catalytic
ID: 1074159 • Letter: G
Question
Gaseous hydrogen can be produced by the steam cracking of methane in a catalytic reactor at 500 degree C and 1 bar according to the following reaction: 4+22 leftrightarrow 2+42 If five moles of steam are fed into the reactor for every mole of methane, at equilibrium, how many moles of hydrogen are produced. Determine the equilibrium conversion of CH4. Determine the equilibrium conversion of CH4 if methane and steam are fed in stoichiometric proportion. Will it make sense to increase the pressure in order to increase the equilibrium conversion? Show with the help simulations. Compute the equilibrium conversion of CH4 over a range of temperatures from 700 K to 900K at 1 bar. Explain your results with the help of a plot.Explanation / Answer
The overall reaction is CH4+2H2O---àCO2+4H2 deltaH298= 165 Kj/mole
Standard entropy change of reaction= 1* entropy of CO2+4* entropy of H2- { 1* entropy of CH4+2* entropy of H2O}
1, 4, 1 and 2 are coefficients of given reaction
1*213.6+4*130.6- { 1*186.2+2*188.7} =172.4 J/K
At 500deg.c =773K, deltaG= deltaH-TdeltaS = 165*1000-172.4*773 =31735Joules= 31.735KJ
deltaG= -RTlnK, K is the equilibrium constant
lnK= -deltaG/RT = -31735/(8.314*773), K= 0.0072
CH4 fed = 1 moles, Steam = 5 moles
Let x= moles of CH4 consumed
At equilibrium moles are , CH4= 1-x, Steam= 5-2x, H2=4x and CO2=x
Total moles at equilibrium = 1-x+5-2x+4x+x= 6+2x
Since the pressure is one atmosphere and partial pressure = mole fraction* total pressure, mole fractions are same as partial pressures)
Mole fractions : CH4= (1-x)/(6+2x), H2O= (5-2x)/ (6+2x), H2= 4x/(6+2x) and CO2= x/ (6+2x)
K= equilibrium constant = [H2]4[CO2]/[CH4][H2O]2 = 256x5/{(6+2x)2*(1-x)*(5-2x)}=0.0072
x5/{(6+2x)2*(1-x)*(5-2x)2}=0.0072/256=2.813*10-5
when solved using excel, x= 0.421
So at 500 deg.c, equilibrium conversion of CH4 =0.421
2. When the reactants are fed in stoichiometric ratio, CH4=1 and H2O= 2
, At equilibrium moles are , CH4= 1-x, Steam= 2-2x, H2=4x and CO2=x
Total moles at equilibrium , 1-x+2-2x+4x+x= 3+2x
Compositions at equilibrium, CH4= (1-x)/(3+2x), Steam = (2-2x)/ (3+2x), H2= 4x/(3+2x), CO2= x/(3+2x)
K= x5/{(3+2x)2*(1-x)*(2-2x)2}=2.813*10-5
when solved using excel, x =0.2282, at equilibrium, CH4= 0.2233, conversion is 0.2233
3. The reaction is gas phase reaction where there is an increase in number of moles as the reaction proceeds ( there are 3 moles of gases on reactant side and 5 mole of gases on product side). As the reaction proceeds, the number of moles are increasing and according to Lechatlier principle, as the pressure is increased, the reaction shifts in a direction so as to nullify the effect of pressure, that is where there is a decrease in number of moles. So back ward reaction is favoured. So there is no point in increasing the pressure .
When pressure is increased, let us say from 1 bar to 2 bar, the partial pressures becomes
CH4= 2*(1-x)/(3+2x), Steam = 2*(2-2x)/ (3+2x), H2= 2*4x/(3+2x), CO2= 2x/(3+2x)
K= x5/{(3+2x)2*(1-x)*(2-2x)2}=2.813*10-5/4 =0.70325*10-5
when solved using excel, at 500K, x= 0.1775 which is less than the equilibrium conversion at 1 bar
4. The sample calculations along with conversion are shown ( for stoichiometric ratio of CH4 and H2O) below.
x
0.7697
0.2303
x^5
0.270151531
4.3492
(3+2x)2
20.60615236
1.3016
1-x
0.2303
-0.050733577
(2-2x)^2
0.21215236
0.268329685
0.2684
-7.0315E-05
Temperature (K)
deltaG
K( Equilibrium constant)
Temperature (K)
Eq.conversion
700
44320
0.000492812
700
0.37
750
35700
0.003262431
750
0.487
800
27080
0.017052717
800
0.5992
850
18460
0.073374567
850
0.694
900
9840
0.268460653
900
0.7697
x
0.7697
0.2303
x^5
0.270151531
4.3492
(3+2x)2
20.60615236
1.3016
1-x
0.2303
-0.050733577
(2-2x)^2
0.21215236
0.268329685
0.2684
-7.0315E-05
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