Question

Formaldehyde is manufactured by the catalytic oxidation of methanol, using excess air: CH_3OH + 1/2 O_2 rightarrow HCHO + H_2O If the reactor conditions are not properly controlled, a secondary oxidation occurs: HCHO + 1/2 O_2 rightarrow HCOOH The measured product gas composition from a test run is given in table Q_2. Determine the percentage conversion of methanol to formaldehyde (methanal). Find the percentage conversion of methanol to formic acid (methanoic acid). Calculate the molar ratio of air to methanol in the feed. Determine the yield and selectivity of the formaldehyde under the test conditions. Table Q2: Product gas composition in Question 2

Answer 1

First consider the reaction (ii)

HCHO + 1/2 O₂ = HCOOH ...... (ii)

from stiochiomentry 1 mol of HCHO + 1/2 mol of O₂ = 1 mol of HCOOH

Product composition contains 0.6 mol of HCOOH,

hence from stoichiometry, Total no. moles of HCHO reacted (in reaction ii) to form HCOOH are: 0.6 mol ....... (a)

& Total no. of moles of O₂ reacted (in reaction ii) to form HCOOH are: 0.3 mol ................................................(b)

Total no. moles of HCHO formed before the second reaction are:moles of HCHO in product + moles of HCHO consumed to form HCOOH = 3.1 +0.6 = 3.7 moles.

From reaction (i)

CH₃OH + 1/2 O₂ = HCHO + H₂O

From stoichimetry: 1 mol CH₃OH + 1/2 mol O₂ = 1 mol HCHO + 1 mol H₂O

Total no. of moles of of HCHO formed are : 3.7 mol

Hence from stoichiometry, total no. moles of CH₃OH reacted to form HCHO are: 3.7 mol .......................................(c)

& total no. moles of O₂ reacted to form HCHO are: 1.85 mol ...................................................................................(d)

Now, total no. O₂ consumed in reaction (i) & (ii) are: 1.85 + 0.3 = 2.15 ...................................................................(e)

a) % conversion of methanol to formaldehyde is given by:

% Conversion = (moles of methanol reacted to form HCHO / Total no. moles of methanol in the feed)

=(3.7 / (3.7+8.6)) * 100 = (3.7/12.3)*100 = 30.08

% Conversion of methanol to HCHO is 30.08

b) % Conversion of methanol to HCOOH:

Two reaction:   CH₃OH + 1/2 O₂ = HCHO + H₂O

  HCHO + 1/2 O₂ = HCOOH

Overall reaction CH₃OH + O₂ = HCOOH + H₂O

Total no. moles of HCOOH formed: 0.6 mol

Hence Total no, moles of methanol reacted to form HCOOH is: 0.6 mol (From stoichiometry)

% Conversion of methanol to HCOOH= (moles of CH₃OH reacted to form HCOOH/ total no. moles of CH₃OH in feed)

= (0.6/12.3) * 100 = 4.88

% Conversion of methanol to HCOOH is 4.88 %

C) Molar ratio of Air to Methanol

Total no. of moles of methanol in feed= 12.3 mol

Total no. of mol of N₂ in product = 68 mol

So, Total no. of moles of air in feed = (68/.79) = 86.08 mol (Assume: Air compostion as 21% O₂ & 79 % N₂)

So, Air to methanol ratio= 86.08 / 12.3 = 6.998 = 7

d) Yeild of Formeldehyde

Yeild = (Moles of HCHO Formed) /(Total moles of Methanol in feed - Total moles of Methanol in product)

= 3.1 / (12.3-8.6) = 0.8378

Selectivity of Formaldehyde

Selectivity = moles of desired product formed (HCHO) / moles of undesired product formed (HCOOH)

= 3.1 / 0.6 = 5.17