Question

Vanadium oxide was used as a catalyst in the oxidation reaction of methane to produce formaldehyde in a reactor operatin...


Vanadium oxide was used as a catalyst in the oxidation reaction of methane to produce formaldehyde in a reactor operating at 973 K. Carbon dioxide was also produced in an undesired side reaction. The feed to the reactor contains 60 mole% methane and 40 mole% oxygen. The reactor is designed on the methane fractional conversion of 0.622 and a selectivity of 4.1 mol HCHO / mol CO2. CH4 + O2 → HCHO + H2O CH4 + 2O2 → CO2 + 2H2O By using these information, Calculate the fractional yield of formaldehyde production.

Answer

Given data:

Feed to the reactor is,

Methane = 60 mole %

Oxygen = 40 mole %

Fractional conversion of methane = 0.622

Selectivity = 4.1 mol HCHO/ mol CO2

Formaldehyde is produced by the oxidation of methane in presence of vanadium oxide. Carbon dioxide also produced in an undesired side reaction.

The reactions are given as follows:

CH4 + O2  HCHO + H2O        desired reaction

CH4 +2O2 CO2 + 2H2O         undesired reaction

 

Basis: 100 mol fed to the reactor.

Therefore, the number of moles of methane and oxygen fed to the reactor becomes,

CH4 = 60 mol

Oxygen = 40 mol

The fraction la conversion of methane = 0.662

Therefore, the total number of moles of methane reacted becomes,

CH4reacted = 60 mol × 0.662 =39.72 mol

The number of moles of methane required producing formaldehyde and carbon oxide is not given but selectivity of formaldehyde per carbon dioxide is given.

Selectivity of formaldehyde per mol of carbon dioxide = 4.1 mol HCHO / mol CO2

 Assume x mole of methane reacts in desired reaction and 39.72 – x mol of methane reacts in undesired reaction.

The whole scenario can be represented on a diagram as follows:

Chemical Engineering homework question answer, step 4, image 1

From the above diagram,

mol HCHO / mol CO2  = x/39.72-x         ………. (1)

From the selectivity data,

mol HCHO / mol CO2  = 4.1 mol HCHO /mol CO2        …………. (2)

Equate equation (1) and (2)

x39.72-x = 4.11x = 4.1 × 39.72-xx = 162.852 - 4.1 x5.1 x = 162.852x =31.93 mol

The number of moles of methane reacts in desires reaction = 31.93 mol

Therefore, the number of moles of methane reacts in undesired reaction becomes = 39.72 – 31.93

                                                                                                                                     = 7.79 mol                                                            

 

Fractional yield of a chemical compound is the ratio of moles of that compound formed to the number of moles of reactant consumed.

For the production of HCHO, the formula can be written as follows:

Yield =  mol of HCHO formedmol of CH4 consumed                ............. (3)

 mol of HCHO formed = 31.93 mol

mol of CH4 reacted = 39.72 mol

Plugin the values in equation (3)

Yield = 31.93 mol39.72 mol=0.8030.8

 

Fraction yield of formaldehyde production is 0.8.

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