2. Oxygen for metabolism is taken up by hemoglobin (Hb) to form oxyhemoglobin (H

Question

2. Oxygen for metabolism is taken up by hemoglobin (Hb) to form oxyhemoglobin (Hb0) according to the simplified equation: Hb(aq) + O2 HbO2(aq) Where the second-order rate constant is 2.1 x 10 M's' at 37 °C. For an average adult, the concentrationso (a) Calculate the rate of formation of HbO. (b) Calculate the rate of consumption of O2 (c) The rate of formation of HbO2 increases to 1.5 x 10 Mis during exercise to meet the demand of an increased metabolic rate. Assuming the Hb concentration remains the same, what oxygen concentration is necessary to sustain this rate of HbO, formation?

Explanation / Answer

let A= Oxygen and B= Hemoblogin, C= HbO2

the rate -dCA/dt is consumption of Oxygen   for second order reaction is given by

--dCA/dt= KCACB, K= rate constant. Substituting the value of [A] , [B] and rate constant K in the equation gives

-dCA/dt = 2.1*106* 8*10-6*1.5*10-6 M/s=2.52*10-5 M/s

as per the stoichiometry of the reaction -dCA/dt= dCC/dt = rate of formation of HbO2= 2.52*10-5 M/s

dCC/dt= KCACB

given dCC/dt= 1.5*10-4M/s and it is also gvien that [B]= 8*10-6 M/s

1.5*10-4= 2.1*106* [A]*8*10-6, [A]= 8.93*10-6 M

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