Chapter 19, Problem 19.127QP

(a)

Interpretation Introduction

Interpretation:

The rate formation of ${\text{HbO}}_{\text{2}}$ has to be calculated.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Answer to Problem 19.127QP

The rate formation of ${\text{HbO}}_{\text{2}}$ is $\text{rate = 2}{\text{.5×10}}^{\text{-5}}\text{M/s}$

Explanation of Solution

The given reaction is ${\text{Hb}}_{\text{(aq)}}{\text{+O}}_{\text{2(aq)}}\stackrel{\text{k}}{\to }{\text{HbO}}_{\text{2(aq)}}$

The order of the reaction is second order

The rate law for the given reaction is ${\text{rate = k[Hb][O}}_{\text{2}}\text{]}$

Now, we have the values of rate constant and concentration of ${\text{Hb and O}}_{\text{2}}$, so substitute the values of these in quantities to get the rate of formation of ${\text{HbO}}_{\text{2}}$ as follows.

$\text{rate = }\left(\text{2}{\text{.1×10}}^{\text{6}}\text{/M}\text{.s}\right)\left(\text{8}{\text{.0×10}}^{\text{-6}}\text{/M}\right)\left(\text{1}{\text{.5×10}}^{\text{-6}}\text{/M}\right)$

$\text{rate = 2}{\text{.5×10}}^{\text{-5}}\text{M/s}$

(b)

Interpretation Introduction

Interpretation:

The rate consumption of ${\text{O}}_{\text{2}}$ has to be calculated.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Answer to Problem 19.127QP

The rate consumption of ${\text{O}}_{\text{2}}$ is $\text{rate = 2}{\text{.5×10}}^{\text{-5}}\text{M/s}$

Explanation of Solution

Above obtained rate for ${\text{HbO}}_{\text{2}}$ is $\text{rate = 2}{\text{.5×10}}^{\text{-5}}\text{M/s}$, therefore ${\text{O}}_{\text{2}}$ is also consumed at the same $\text{rate = 2}{\text{.5×10}}^{\text{-5}}\text{M/s}$, and from the reaction there will be the mole ratio between ${\text{O}}_{\text{2}}$ and ${\text{HbO}}_{\text{2}}$ is 1:1.

(c)

Interpretation Introduction

Interpretation:

The rate formation of ${\text{HbO}}_{\text{2}}$ increases to $1.4\times {10}^{-4}\text{M/s}$ during exercise to meet the demand of the increased metabolism rate, what must the oxygen concentration be to sustain this rate of ${\text{HbO}}_{\text{2}}$ formation has to be determined.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Answer to Problem 19.127QP

The rate formation of ${\text{HbO}}_{\text{2}}$ increases to $1.4\times {10}^{-4}\text{M/s}$ during exercise to meet the demand of the increased metabolism rate, what must the oxygen concentration be to sustain this rate of ${\text{HbO}}_{\text{2}}$ formation is $\left[{\text{O}}_{\text{2}}\right]\text{=8}\text{.3}\times {\text{10}}^{\text{-6 }}\text{M }$

Explanation of Solution

By increasing the rate of formation of ${\text{HbO}}_{\text{2}}$ the concentration of $\text{Hb}$ remains the same.  Let us assume that the temperature is constant and the above obtained rate constant value and the concentration of $\text{Hb}$ value is substituted in the rate law then we get the value of concentration for ${\text{O}}_{\text{2}}$ as follows.

${\text{rate = k[Hb][O}}_{\text{2}}\text{]}$

$\text{1}\text{.4 }\times {\text{10}}^{\text{-4 }}\text{M / s = }\left(\text{2}\text{.1}\times {\text{10}}^{\text{6}}\text{ / M}\text{.s}\right)\left(\text{8}\text{.0 }\times {\text{10}}^{\text{-6}}\text{ M}\right)\left({\text{O}}_{\text{2}}\right)$

$\left[{\text{O}}_{\text{2}}\right]\text{=8}\text{.3}\times {\text{10}}^{\text{-6 }}\text{M }$