Answer to Question #255977 in General Chemistry for angel

Question

Write the stoichiometric equation lor aerobic heterotrophic microbial growth on a carbohydrate using ammonia as the nitrogen source, under conditions such that the true growth yield (YH) is 0.71 mg of biomass COD formed per mg of carbohydrate COD removed.

Answer 1

R = R., i; • R, - i; ■ R, t; = Y„ = 0.71 f = 1.00 0.71 = 0.29 Therefore

The electron donor is carbohydrate and the acceptor is oxygen. Thus, from Table

R, = 1/4 CH..O + 1/4 H O = 1/4 CO, + H + e R, = 12 H O = 1/4 O, + H f- e

Since ammonia is the nitrogen source, R, is:

R, = 1/20 C,H,0;N + 9/20 H:0 = 1/5 CO, + 1/20 HCO,

Applying Eq. 3.14 gives:

-0.29 R„ = 0.0725 O, + 0.29 H' + 0.29 e = 0.145 H,0

-0.71 R, = 0.142 CO, + 0.0355 HCO, + 0.0355 NH; + 0.71 H' + 0.71 e

= 0.0355 C,H,0:N + 0.3195 H:0

R = 0.25 CH,0 + 0.0725 O, + 0.0355 NH; + 0.0355 HCO,

= 0.0355 CsH;0,N + 0.108 CO, + 0.2145 H,0

This can be normalized to one mole of carbohydrate by dividing through by 0.25, giving Eq. 3.6, which was the starting point of