Question

Consider the following biochemical reaction where C_wH_xO_yN_z is substrate, CH_αO_βN_δ is biomass, and C_jH_kO_lN_m is product.

C_wH_xO_yN_z + a O₂ + b H_gO_hN_i = c CH_αO_βN_δ + d CO₂ + e H₂O + f C_jH_kO_lN_m

Valence state: C = 4;              H = 1;              O = -2;            N = -3

(i) Develop available electron balance for the reaction above. Clearly show the steps.

(ii) Develop an equation to calculate the theoretical oxygen demand for the reaction above. Clearly show the steps.

(iii) Develop an equation to calculate the maximum possible yield.

ANSWER should be in variable form. Part (iii) is an equation for the yield left in terms of variables.

Answer 1

F2+2NO2→2NO2FF2+2NO2→2NO2F

Application of the "chemical intuition" mentioned in the above box would lead us to suspect that any process that involves breaking of the strong F–F bond would likely be slow enough to be rate limiting, and that the resulting atomic fluorines would be very fast to react with another odd-electron species:

Step 1	F2+2NO2−→k1NO2F+F+NO2F2+2NO2→k1NO2F+F+NO2	(slow, rate-determining)
Step 2	F+NO2−→k2NO2FF+NO2→k2NO2F	(very fast)

If this mechanism is correct, then the rate law of the net reaction would be that of the rate-determining step:

rate=k1[F2][NO2]