Question

Using The Degree of Advancement Variable

Methane and oxygen react in the presence of a catalyst to give formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water:

CH4 + O2 = HCHO + H2O

CH4 + O2 = CO2 + H2O

The reactor feed contains equimolar amounts of methane and oxygen. Assume as a base a feed of 100 mol / s.

The methane conversion fraction is 0.90 and the formaldehyde yield fraction is 0.855. Use the reaction advance degree method to determine the molar composition of the reactor outlet stream and the selectivity of formaldehyde production relative to carbon dioxide production.

Answer 1

Oxidization reactions are -

CH4 + O2 -----> HCHO + H2O ,

CH4 + 2O2 -----> CO2 + 2H2O

inlet feed is 50 mol CH4 and 50 mol O2.

conversion fraction of CH4 = 0.9 = CH4 reacted / CH4 entered CH4 reacted / 50 mol.

so, CH4 reacted = 0.9*50 mol = 45 mol

HCHO yeild = 0.855 = moles of HCHO formed / maximum mole of HCHO would be produced

= moles of HCHO formed / maximum mole of CH4 would be reacted

= moles of HCHO formed / moles of CH4 entered = moles of HCHO formed / 50 mol

so, HCHO formed = 0.855 * 50mol = 42.75 mol.

from 1 mol of CH4 and 1 mol of O2, 1 mol of HCHO and 1 mol of H2O is formed.

so, 42.75 mol of HCHO is formed with 42.75 mol of H2O from 42.75 mol of CH4 and 42.75 mol of O2.

CH4 reacted = 45 mol.

So CH4 reacted to form CO2 = 45 mol - 42.75mol = 2.25 mol.

1 mol of CH4 reacts with 2 mol of O2 to produce 1 mol of CO2 and 2 mol of H2O.

so, 2.25 mol of CH4 reacts with 2*2.25 = 4.5 mol of O2 to produce 2.25 mol of CO2 and 4.5 mol of H2O.

So, selectivity of HCHO production relative to CO2 production

= moles of HCHO produced / moles of CO2 prodeced = 42.75 mol / 2.25 mol = 19.

In the outlet stream, HCHO present = 42.75 mol, CO2 present = 2.25 mol, H2O present = (4.5 + 42.75) = 47.25mol,

unreacted CH4 present = (50 - 42.75 - 2.25) = 5mol, unreacted O2 present = (50 - 42.75 - 4.5) = 2.75 mol.

Total outlet stream flow rate = (42.75 + 2.25 + 47.25 + 5 + 2.75) = 100 mol.

so, composition of outlet stream -

CH4 = 5mol /100mol = 5%, O2 = 2.75%, H2O = 47.25%, HCHO = 42.75%, CO2 = 2.25%.