Formaldehyde is manufactured by the catalytic oxidation of methanol, using exces
ID: 1049828 • Letter: F
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 2Explanation / Answer
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