Acetaldehyde, CH3CHO, may be produced by the reaction of ethylene with oxygen in

Question

Acetaldehyde, CH3CHO, may be produced by the reaction of ethylene with oxygen in

the presence of a tetrachloropalladate catalyst according to the reaction:

2 CH2CH2+ O2? 2 CH3CHO

In an undesired side reaction some of the ethylene is oxidized to produce carbon

dioxide and water,

CH2CH2+ 3 O2? 2 CO2 + 2 H2O

A catalytic reactor is fed with a mixture of ethylene and oxygen only, and only the two

above reactions occur. The exit stream contains gas with the following composition:

CH2CH2 18.5 mol %

O2 unknown

CH3CHO 47.0 mol %

CO2 6.1 mol %

H2O unknown

Calculate:

a)The composition of the feed mixture.

b)The excess O2(expressed as a percentage).

c) The conversion of ethylene (expressed as a percentage).

d) The yield of acetaldehyde (expressed as a percentage).

e) The selectivity (expressed as a ratio).

Explanation / Answer

In the side reaction, no of moles of CO2 produce = no. of moles of H2O produce

=> H2O mole % in outlet = CO2 mole% in outlet=6.1%

Sum of Mole % =100

=> O2 mole% in outlet = 100-6.1-6.1-18.5-47=22.3%

a) One mole CH3CHO requires one mole CH2CH2, two moles CO2 requires one mole CH2CH2

Assuming 100 moles in outlet

=> Total CH2CH2 moles in inlet=Moles of CH3CHO in outlet +Outlet Moles of CO2/2+Outlet moles of CH2CH2 = 47+6.1/2+18.5=68.55 moles

two mole CH3CHO requires one mole O2, two moles CO2 requires three mole O2

=>

Total O2 moles in inlet=Moles of CH3CHO in outlet/2 +Outlet Moles of CO2/2*3+Outlet moles of O2= 47/2+6.1/2*3+22.3=54.95 moles

b) Excess O2=(O2 in inlet - O2 requires to convert complete inlet CH2CH2 in desired reaction)/O2 requires to convert complete inlet CH2CH2 in desired reaction=(54.95-68.55/2)/(68.55/2)*100=60.32%

c) Yield of Acetaldehyde = Acetene converted in main reaction/total Acetene inlet =Moles of Acetaldehyde form/Mole of Acetene in Inlet = 47/68.55*100=68.5%

d) Selectivity = Acetene converted in main reaction/Acetene converted in undesired reaction=Mole of Acetaldehyde form/(Mole of CO2 form/2)=47/(6.1/2)=15.4

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