The second-order rate constant of oxygen uptake by hemoglobin (Hb) to form oxyhe
ID: 868896 • Letter: T
Question
The second-order rate constant of oxygen uptake by hemoglobin (Hb) to form oxyhemoglobin (HbO2) is 2.1 x 10^6 M-1s-1 at 37C. The simplfied equation for this reaction is:
Hb(aq) + O2 (aq) > HbO2 (aq)
What is the rate of formation of HbO2 if concentrations of Hb and O2 in the blood at the lungs are 8.0 x 10^-6 and 1.5 x 10^-6M?
What would be the rate constant expression if the reaction were pseudo-first order with respect to Ox? Determine the half-life under these conditions.
What is the equilibrium constant expression for this reaction?
Explanation / Answer
Hb(aq) + O2(aq) <------> HbO2(aq)
rate of reaction = rate of disappearance of Hb(aq) = rate of disappearance of O2 = rate of appearance of HbO2(aq)
or, rate of reaction = -d[Hb]/dt = -d[O2]/dt = d[HbO2]/dt
Thus, rate of reaction = rate of appearance of HbO2(aq) = d[HbO2]/dt..............(1)
Now, rate of reaction = rate constant(k)*[Hb]*[O2] = (2.1*106)*(8*10-6)*(1.5*10-6) = 2.52*10-5 M/s..........(2)
Thus, from (1) and (2)
rate of appearance of HbO2(aq) = d[HbO2]/dt = 2.52*10-5 M/s.......................(A)
If the reaction was pseudo first order w.r.t O2, then the change in concentration of O2 would be negligible
Thus, [O2] will be almost same i.e constant
Hence the rate of reaction will then be :-
rate of reaction = rate constant*[Hb]
where rate constant = (2.1*106)*[O2] = (2.1*106)*(1.5*10-6) = 3.15 s-1
Now, half life for 1st order reaction = ln2/rate constant = 0.693/3.15 = 0.22 sec....................(B)
Expression for equilibrium constant , i.e Keq = [HbO2]/{[Hb]*[O2]}..................(C)