Please help problem 1 a-d Homework 4 Problem 1 1. For production of penicillin (

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Please help problem 1 a-d

Homework 4 Problem 1 1. For production of penicillin (CidHisO N2S, molecular weight 334.4 g/mol), glucose (CeH1206) is used as substrate and phenylacetic acid (CsHs02) is added as precursor. The stoichiometry for overall synthesis is: A supply of 100 mol/hr of glucose is fed to a continuous stirred-tank reactor for the production of penicillin, along with 200 mol/hr of ammonia (NH3), 130 mol/hr of oxygen, 125 mol/hr of sulfuric acid and 100 mol/hr of phenylacetic acid. a. Determine the limiting reactant in the process. (5 points) b. Calculate the percent excess molar % of the reactant that is in largest excess (i.e. has the largest n"1). (5 points)? Assuming 85% conversion of the limiting reactant, find the reaction rate. (5 points) c. d. Determine the daily production of penicillin, in Kg/day. (5 points) Problem 2

Explanation / Answer

Part a

From the stoichiometry of the reaction

1.67 mol glucose reacts with = 2 mol NH3 = 0.5 mol O2 = 1 mol H2SO4 = 1 mol C8H8O2

1 mol glucose reacts with = 2/1.67 mol NH3 = 0.5/1.67 mol O2 = 1/1.67 mol H2SO4 = 1/1.67 mol C8H8O2

For 100 mol glucose

100 mol glucose reacts with = 2*100/1.67 mol NH3 = 0.5*100/1.67 mol O2 = 1*100/1.67 mol H2SO4 = 1*100/1.67 mol C8H8O2

100 mol glucose reacts with = 119.76 mol NH3 = 29.94 mol O2 =59.88 mol H2SO4 = 59.88 mol C8H8O2

All glucose have been reacted.

We have more NH3(200 mol) than required (119.76 mol)

We have more O2(130 mol) than required (29.94 mol)

We have more H2SO4 (125 mol) than required (59.88 mol)

We have more C8H8O2 (100 mol) than required (59.88 mol)

Limiting reactant = C6H12O6 (glucose)

Part b

% excess NH3 = (200 - 119.76)*100/119.76 = 67.00 %

% excess O2 = (130 - 29.94)*100/29.94 = 334.20 %

% excess H2SO4 = (125 - 59.88)*100/59.88 = 108.75 %

% excess C8H8O2 = (100 - 59.88)*100/59.88 = 67.00 %

O2 is in largest excess i.e. 334.20 %

Part c

Conversion of glucose = 0.85

1 mol glucose reacts with = 2/1.67 mol NH3 = 0.5/1.67 mol O2 = 1/1.67 mol H2SO4 = 1/1.67 mol C8H8O2

85 mol glucose reacts with = 2*85/1.67 mol NH3 = 0.5*85/1.67 mol O2 = 1*85/1.67 mol H2SO4 = 1*85/1.67 mol C8H8O2

85 mol glucose reacts with = 101.79 mol NH3 = 25.45 mol O2 = 50.90 mol H2SO4 = 50.90 mol C8H8O2

The reactants at outlet

Glucose = 100 - 85 = 15 mol

NH3 = 200 - 101.79 = 98.21 mol

O2 = 130 - 25.45 = 104.55 mol

H2SO4 = 125 - 50.90 = 74.1 mol

C8H8O2 = 100 - 50.90 = 49.1 mol

If the reaction is elementary

Reaction rate = rate constant x [C6H12O6]1.67 [NH3]2 [O2]0.5 [H2SO4] [C8H8O2]

= k x [15]1.67 [98.21]2 [104.55]0.5 [74.1] [49.1]

= k x 3.303 x 10^10 M/h

Part d

Penicillin produced

=( 1 mol Penicillin x 85 mol glucose/h) / (1.67 mol glucose)

= 50.898 mol/h x 334.4 g/mol

= 17020.36 g/h x 24h/day x 1kg/1000g

= 408.48 kg/day

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