Question

. Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen

hemoglobin (aq) + O₂(aq) → hemoglobin .O₂ (aq)

aq is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol^-1s^-1. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10^-9 M and that of oxygen is 5 × 10^-5 M.

Short answer

The initial rate of oxygen binding is $4x{10}^6{\text{Ms}}^{-1}$

Step-by-step solution

Rate equation:

The equation for the reaction mentioned in the question is

$\text{Heamoglobin}\left(\text{Aq}\right)+O_2\left(\text{aq}\right)\to \text{hemoglobin}-O_2$

The reaction is first order both in the case of heamoglobin and O_{2 .}Therfore the rate expression for the reaction is

$\text{rate}=\text{k}\left[\text{Haemoglobin}\left(\text{aq}\right)\right]\left[O_2\left(\text{aq}\right)\right]........\left(1\right)$

Where k is the rate constant.

Initial rate:

The values given are

$$\begin{gathered} \text{k}=4x{10}^7{\text{Lmol}}^{-1}{s}^{-1} \\ \left[\text{Haemoglobin}\left(\text{Hb}\right)\right]=2x{10}^{-9}\text{M} \\ \left[{\text{Oxygen-O}}_{\text{2}}\left(\text{Aq}\right)\right]=5x{10}^{-5}\text{M} \\ \text{Substitutingallvaluesinequation-1} \\ \text{rate}\hspace{0.17em}=4x{10}^7{\text{Lmol}}^{-1}{s}^{-1}x\left(2x{10}^{-9}\text{M}\right)\left(5x{10}^{-5}\text{M}\right) \\ As1\text{M}=1{\text{molL}}^{-1} \\ {\text{rate=4x10}}^{-6}{\text{Ms}}^{-1} \end{gathered}$$

Therefore initial rate of oxygen binding is$4x{10}^{-6}{\text{Ms}}^{-1}$