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?

Answer 1

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 K_eq = [HbO₂]/{[Hb]*[O₂]}..................(C)