Consider the \"steam-methane\" reforming reaction Part A: Use published thermody

Question

Consider the "steam-methane" reforming reaction Part A: Use published thermodynamic data to calculate the heat of reaction (DeltaH_g) at 25degreeC Part B: Use published thermodynamic data to calculate the Gibbs free energy of reaction (DeltaG_R) at 25 degreeC Part C: Calculate the equilibrium constant at a temperature of 650degreeC. It may be assumed that the heat or reaction (DeltaH_R) is constant. Part D: A mixture of reactant gases consisting 1.5 mol of CH_4 and 2.5 mol of H_2O is allowed to equilibrate at a constant temperature of 650degreeC and a constant pressure of 3.5 atm. Calculate the equilibrium composition of the gas in mol fraction of each component.

Explanation / Answer

Part-A :

delta H rxn = Heat of formation of products - Heat of formation of reactants

= [1Hf(CO (g)) + 3Hf(H2 (g))] - [1Hf(CH4 (g methane)) + 1Hf(H2O (g))]
= [1(-110.54) + 3(0)] - [1(-74.85) + 1(-241.82)]

= 206.13 kJ
delta G rxn = [1Gf(CO (g)) + 3Gf(H2 (g))] - [1Gf(CH4 (g methane)) + 1Gf(H2O (g))]
= [1(-137.28) + 3(0)] - [1(-50.84) + 1(-228.59)]

= 142.15 kJ

K298 = exp (-delta G /RT) == 1.244 x10-25

ln ( K923.15 / K298 ) = -206.13 kJ /8.314 ( 1/923.15 - 1/298 )

K923.15   = 0.366

Now. initail CH4 = 1.5 and H2O = 2.5 mol

K =3.66 = (x) (3x)3 / (1.5-x) (2.5-x)

on solving , x =0.824 mole

mol CH4 = 1.5-0.824 = 0.676 mol

mol H2O = 2.5-x = 1.676 mol

mol CO = 0.824 mol

mol H2 = 3x = 2.472 mol

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