Question

Please answer as soon as possible (Chemical Engineering)

Given The reactions

C₃H₈ → C₃H₆ + H₂

C₃H₈ + 2H₂ → 3CH₄

The feed contains 90.0 mole% propane(C₃H₈) and the balance inert. The fractional conversion of propane is 0.603 and the fractional yield of propylene is 0.541.

Calculate the molar composition of the product gases and selectivity of ethylene to methane production.

Answer 1

basis : 1 mole of mixture of propane and inerts. Propane in the feed =0.9 moles.

Propane converted = 0.9*0.603=0.543

let x mole got converted to propylene = x

0.543-x = moles got converted to methane

given x= 0.541 = moles of propylene formed

moles of propane got converted to methane =0.543-0.541 =0.02

methane formed =32 times that of propane = 3*0.02= 0.06

Hydrogen produced= 0.541 hydrogen produced = 2*0.541=1.082 moles

Hydrogen consumed due to formation of CH4= 0.04

hydrogen remaining= 1.082-0.04=1.042

Products : CH4= 0.06 moles Propane = proane supplied - propane reacted = 0.9-0.543=0.357

Propylene =0.541 hydrogen =1.042 inerts =0.1

total mole =0.06+0.357+0.541+1.042 +0.1= 2.1

Molal comosition : CH4= 100*0.06/2.1=2.86% Propane= 100*0.357/2.1=17% Propylene= 100*0.541/2.1=25.76%

hydrogen =100*1.042/2.1=49.62 inerts = 100*0.1/2.1=4.76%

selectivity= moles of propylene formed/ moles of CH4 formed= 0.541/0.06=9.02