Question

Carbon monoxide is toxic because it bonds much more strongly to the iron in haemoglobin than does ${\mathit{O}}_{\mathbf{2}}$. Consider the following reactions and approximate standard free energy changes:

$\begin{array}{l}\mathbf{H}\mathbf{g}\mathbf{b}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }\mathbf{H}\mathbf{g}\mathbf{b}{\mathbf{O}}_{\mathbf{2}}\mathbf{\Delta }{\mathbf{G}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{70}\mathbf{k}\mathbf{J}\\ \mathbf{H}\mathbf{g}\mathbf{b}\mathbf{+}\mathbf{C}\mathbf{O}\mathbf{\to }\mathbf{H}\mathbf{g}\mathbf{b}\mathbf{C}\mathbf{O}\mathbf{\Delta }{\mathbf{G}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{80}\mathbf{k}\mathbf{J}\end{array}$

Using these data, estimate the equilibrium constant value at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$for the following reaction:

$\mathit{H}\mathit{g}\mathit{b}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathit{C}\mathit{O}\mathbf{\to }\mathit{H}\mathit{g}\mathit{b}\mathit{C}\mathit{O}\mathbf{+}{\mathit{O}}_{\mathbf{2}}$

Short answer

The equilibrium constant value at ${25}^{\circ }\text{C}$is.$56.59$

Step-by-step solution

Definition of equilibrium constant

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature is defined as equilibrium constant.

Calculation of equilibrium constant value

The given reaction is:

$\begin{array}{l}\text{Hgb}+{\text{O}}_{\text{2}}\to {\text{HgbO}}_{\text{2}}{\text{\hspace{0.17em}ΔG}}^{\text{o}}=-70\text{kJ}\\ \text{Hgb}+\text{CO}\to {\text{HgbCO\hspace{0.17em}\hspace{0.17em}ΔG}}^{\text{o}}=-80\text{kJ}\end{array}$

The required reaction is${\text{HgbO}}_{\text{2}}+\text{CO}\to \text{HgbCO\hspace{0.17em}}+{\text{O}}_{\text{2}}$

The required reaction is obtained by reversing reaction 1 and add it to reaction 2

$\begin{array}{l}\text{Hgb}+{\text{O}}_{\text{2}}\to {\text{HgbO}}_{\text{2}}{\text{\hspace{0.17em}ΔG}}^{\text{o}}=-(-70\text{kJ)}\\ \underset{\_}{\text{Hgb}+\text{CO}\to {\text{HgbCO\hspace{0.17em}\hspace{0.17em}ΔG}}^{\text{o}}=-80\text{kJ}}\\ {\text{HgbO}}_{\text{2}}+\text{CO}\to \text{HgbCO\hspace{0.17em}}+{\text{O}}_{\text{2}};{\text{\hspace{0.17em}ΔG}}^{\text{o}}=-10\text{\hspace{0.17em}kJ}\end{array}$

The equilibrium constant for the reaction is calculated as${\text{ΔG}}^{\text{o}}=-\text{RTlnK}$

Converting temperature from${}^{\circ }\text{C\hspace{0.17em}to\hspace{0.17em}K}$

$25+273=298\text{\hspace{0.17em}K}$

Substituting the values,

$\begin{array}{c}-10\times {10}^3\text{\hspace{0.17em}J/mol}=-\left(8.3145\text{\hspace{0.17em}J/mol-K}\right)\left(298\text{\hspace{0.17em}K}\right)\ln \left(\text{K}\right)\\ \ln \left(\text{K}\right)=\frac{10\times {10}^3}{\left(8.3145\right)\left(298\right)}\\ \text{K}=56.59\end{array}$