Chapter 20, Problem 20.69QP

As stated in the chapter, carbon monoxide has a much higher affinity for hemoglobin than oxygen does. (a) Write the equilibrium constant expression (K_c) for the following process:

$\text{CO}(g)+{\text{HbO}}_2(aq)\rightleftharpoons {\text{O}}_2(g)+\text{HbCO}(ag)$

where HbO₂ and HbCO are oxygenated hemoglobin and carboxyhemoglobin, respectively.

(b) The composition of a breath of air inhaled by a person smoking a cigarette is 1.9 × 10⁻⁶ mol/L CO and 8.6 × 10⁻³ mol/L O₂. Calculate the ratio of [HbCO] to [HbO₂], given that K_c is 212 at 37°C.

Interpretation Introduction

Interpretation:

The equilibrium constant expression ${\text{K}}_{\text{c}}$ for the given reaction has to be written.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Answer to Problem 20.69QP

The equilibrium constant expression ${\text{K}}_{\text{c}}$ for given reaction is ${\text{K}}_{\text{c}}\text{=}\frac{\text{[O ][HbCO]}}{{\text{[CO][HbO}}_{\text{2}}\text{]}}$

Explanation of Solution

Given,

The balanced reaction between carbon monoxide and hemoglobin is given as

${\text{CO}}_{\text{(g)}}\text{+}{\text{HbO}}_{\text{2(aq)}}\underset{}{\overset{}{\rightleftharpoons }}{\text{O}}_{\text{2(g)}}\text{+}{\text{HbCO}}_{\text{(aq)}}$

The equilibrium constant expression is the ratio of product of concentration of oxygen and $\text{HbCO}$ to product of concentration of carbon monoxide and hemoglobin.  It is given as

${\text{K}}_{\text{c}}\text{=}\frac{\text{[O ][HbCO]}}{{\text{[CO][HbO}}_{\text{2}}\text{]}}$

Interpretation Introduction

Interpretation:

From the equilibrium constant ${\text{K}}_{\text{c}}$, the ratio of $\text{[HbCO] to [HbO2]}$  has to be calculated.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Answer to Problem 20.69QP

The ratio of $\text{[HbCO] to [HbO2]}$  is $\text{0}\text{.047}$

Explanation of Solution

Given,

$\begin{array}{l}{\text{[O}}_{\text{2}}\text{]}\text{=}\text{8}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ \text{[CO]}\text{=}\text{1}\text{.9}\text{×}{\text{10}}^{\text{-6}}\text{M}\end{array}$

The balanced reaction between carbon monoxide and hemoglobin is given as

${\text{CO}}_{\text{(g)}}\text{+}{\text{HbO}}_{\text{2(aq)}}\underset{}{\overset{}{\rightleftharpoons }}{\text{O}}_{\text{2(g)}}\text{+}{\text{HbCO}}_{\text{(aq)}}{\text{K}}_{\text{c}}\text{=}\text{212}$

The equilibrium constant expression is given as,

${\text{K}}_{\text{c}}\text{=}\frac{\text{[O ][HbCO]}}{{\text{[CO][HbO}}_{\text{2}}\text{]}}$

The ratio of $\text{[HbCO] to [HbO2]}$ is calculated as,

$\begin{array}{l}\text{212}\text{=}\frac{{\text{[O}}_{\text{2}}\text{][HbCO]}}{{\text{[CO][HbO}}_{\text{2}}\text{]}}\text{=}\frac{\text{[8}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{][HbCO]}}{\text{[1}\text{.9}\text{×}{\text{10}}^{\text{-6}}{\text{][HbO}}_{\text{2}}\text{]}}\\ \\ \frac{\text{[HbCO]}}{{\text{[HbO}}_{\text{2}}\text{]}}\text{=}\frac{\text{(212)(1}\text{.9}\text{×}{\text{10}}^{\text{-6}}\text{)}}{\text{(8}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{)}}\text{=}\text{0}\text{.047}\end{array}$