Chapter 14, Problem 127AP

Interpretation Introduction

Interpretation:

The rate of formation of ${\text{HbO}}_2$

and the rate consumption of ${\text{O}}_2$

are to be calculated. Also, the concentration of ${\text{O}}_2$

required to sustain the increased rate of ${\text{HbO}}_2$

is to be determined.

Concept introduction:

The rate of chemical reaction at any instant is the decrease in concentration of the reactant(s) or increase in concentration of product(s) in unit time, at that instant, throughout the progress of the reaction.

The equation that relates rate of a reaction to the concentration of reactants is called the rate law. For the reaction as:

$a\text{A}+b\text{B}\to c\text{C}+d\text{D}$

The rate law is as follows:

$\text{rate}={\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

Rate constant for a reaction is the proportionality constant that relates the rate of reaction and the concentration of reactants in the reaction.

Answer to Problem 127AP

Solution:

$2.52\times {10}^{-5}\text{ M/s}$.

$2.52\times {10}^{-5}\text{ M/s}$.

$8.3\times {10}^{-6}\text{M}$.

Explanation of Solution

Given information: The reaction between hemoglobin and oxygen is as follows:

$\text{Hb}\left(aq\right)+{\text{O}}_2\left(aq\right)\to {\text{HbO}}_2\left(aq\right)$

The concentration of hemoglobin and oxygen is $8.0\times {10}^{-6}\text{M}$ and $1.5\times {10}^{-6}\text{M}$, respectively. Rate constant for the reaction is $2.1\times {10}^6{\text{ M}}^{-1}{\text{s}}^{-1}$.

a) The rate of formation of ${\text{HbO}}_2$

According to the reaction, the rate law is as follows:

$\text{rate}=k\left[\text{Hb}\right]\left[{\text{O}}_2\right]$

Here, $k$ is the rate constant, and $\left[\text{Hb}\right],\left[{\text{O}}_2\right]$ are the concentrations of reactants.

Substitute the values $\left[\text{Hb}\right],\left[{\text{O}}_2\right]$, and $k$ in the above equation,

$\begin{array}{c}\text{rate}=\left(2.1\times {10}^6\cancel{{\text{M}}^{-1}}{\text{s}}^{-1}\right)\left(\left[8.0\times {10}^{-6}\cancel{\text{M}}\right]\left[1.5\times {10}^{-6}\text{M}\right]\right)\\ =\left(2.1\right)\left(12.0\times {10}^{-6}\right)\\ =2.52\times {10}^{-5}\text{M/s}\end{array}$

b) The rate of consumption of ${\text{O}}_2$.

From the reaction, it can be observed that ${\text{HbO}}_2$

and ${\text{O}}_2$

bear a $1:1$ molar quantitative relation. Thus, the rate of oxygen consumed is the same as the rate of Hb consumed in the reaction.

So, the rate of oxygen consumed is $2.52\times {10}^{-5}\text{M/s}$.

c) The oxygen concentration must be to sustain the given rate of ${\text{HbO}}_2$.

The rate of formation of ${\text{HbO}}_2$

increase to $1.4\times {10}^{-4}\text{M/s}$

The rate of formation of ${\text{HbO}}_2$ increases, but the concentration of hemoglobin remains the same. Assuming that the temperature is constant for the reaction, we can use the same rate constant as in part (a). Now, to solve for oxygen concentration,

The rate law for the reaction is as follows:

$\text{rate}=k\left[\text{Hb}\right]\left[{\text{O}}_2\right]$

Substitute the values of $\left[\text{Hb}\right]$, rate, and $k$ in the above equation,

$\begin{array}{c}1.4\times {10}^{-4}\text{M/}\cancel{\text{s}}=\left(2.1\times {10}^6/\cancel{\text{M}}\text{.}\cancel{\text{s}}\right)\left(8.0\times {10}^{-6}\cancel{M}\right)\left[{\text{O}}_2\right]\\ \left[{\text{O}}_2\right]=\left(\frac{1.4\times {10}^{-4}\text{M/}\cancel{\text{s}}}{16.80\cancel{\text{s}}}\right)\\ =0.083\times {10}^{-4}\text{M}\\ =8.3\times {10}^{-6}\text{M}\end{array}$