Question

2. Ethylene is produced by the dehydrogenation of ethane. If the feed includes 0.65 mol of steam (an inert diluent) per mole of ethane and if the reaction reaches equilibrium at 1200 K and 1 bar, what is the composition of the product gas on a water-free basis?

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Answer 1

Ethylene is the product

The reaction occuring is

$C_{2}H_{6}(g) \rightarrow C_{2}H_{4}(g) + H_{2}(g)$

Th reaction occurs at P = 1 bar and T = 1200 K

From handbook

∆H(ethane) (g) (298 K) = -84 KJ/mol

∆H(ethylene) (g) (298 K) = +52.4 KJ/mol

∆H(hydrogen) = 0

∆H_r = ∆H_products - ∆H_reactants

∆H_r = 52.4-(-84) = 136.4 KJ/mol

According to kirchoffs law

∆H(1200 K) = ∆H(298 K) + ∆C_p(1200-298)

∆S(ethane) (298 K) = 229.6 J/mol K

∆S(ethylene) (298 K) = 219.32 J/mol K

∆S(hydrogen gas) (298 K) = 130.68 J/mol K

∆S(298 K) - ∆S_products - ∆S_reactants

∆S(298 K) = (130.68) +(219.32) -(229.6) =

120.4 J/mol K

Cp is taken at average temperature of (1200+298) /2 = 749 K

C_p(ethane) (g) = 103.452 J/mol K

C_p(ethylene) (g) = 80.698 J/mol K

C_p(hydrogen) (g) = 29.5282 J/mol K

∆C_p= (nC_p) _products - (nC_p) _reactants

∆C_p = (29.5282+80.698) -(103.452) = 6.7742 J/mol K

∆H(1200 K) = 136.4(1000) +(6.7742) (1200-298)

∆H(1200 K) = 142.510 KJ/mol

∆S (1200 K) = C_p ln(T/298)

C_pavg according to stiochiometry

103.452(0.33) +(80.698(0.33) +(29.528(0.33) )= 70.5137 J/mol K

∆S(1200 K) = 70.5137 ln(1200/298)

∆S(1200 K) = 98.224 J/mol k

∆S(total) = ∆S(298 K) + ∆S(1200 K)

∆S(total) = 120.4+98.224 = 218.624 J/mol K

∆G = ∆H -T∆S

∆G(1200 K) = 142.510(1000) -(218.624×1200)

∆G(1200 K) = -119838. 8 J/mol

$\frac{\Delta G}{RT}= -lnK$

-119838.8 (8.314) (1200) -Ink

K = 164677.987

Product gas should be water free basis

Basis

1 mole ethane

$K = \frac{\epsilon ^{2}}{(1-\epsilon )}$

$\epsilon$- extent of reaction

$164677.987= \frac{\epsilon ^{2}}{(1-\epsilon )}$

$\epsilon$ = 0.999

Equilibrium composition of product on steam(water free basis)

Moles of ethylene = 0.999

Moles of ethane = 0.001

Moles of hydrogen = 0.999

Mole fraction of hydrogen = 0.999/(0.999+0.999+0.001) = 0.999/1.999 =0.4997

Mole fraction of ethylene = 0.4997

Mole fraction of ethane = 0.0006

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