Question

Carbon monoxide and the cyanide ion are both toxic because they bind more strongly than oxygen to the iron in hemoglobin (Hb). $$\begin{array}{rl}{\mathrm{Hb}+\mathrm{O}_{2} \rightleftharpoons \mathrm{HbO}_{2}} & {K=2 \times 10^{12}} \\ {\mathrm{Hb}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}} & {K=1 \times 10^{14}}\end{array}$$ Calculate the equilibrium constant value for this reaction. $$\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}$$ Does the equilibrium favor reactants or products?

Short answer

The equilibrium constant for the target reaction is \(5 \times 10^{1}\), and the equilibrium favors the products.

Step-by-step solution

Write the individual reactions and their equilibrium constants

First, we have the two given reactions with their respective equilibrium constants:\[\mathrm{Hb} + \mathrm{O}_2 \rightleftharpoons \mathrm{HbO}_2, K_1=2 \times 10^{12}\] \[\mathrm{Hb} + \mathrm{CO} \rightleftharpoons \mathrm{HbCO}, K_2=1 \times 10^{14}\]

Write the target reaction

We are asked to find the equilibrium constant of the reaction:\[\mathrm{HbO}_2 + \mathrm{CO} \rightleftharpoons \mathrm{HbCO} + \mathrm{O}_2\]

Derive the target reaction from the given reactions

The target reaction can be obtained by reversing the first reaction and adding it to the second reaction. When a reaction is reversed, the equilibrium constant for the new reaction is the reciprocal of the original constant. So for the reversed reaction,\[\mathrm{HbO}_2 \rightleftharpoons \mathrm{Hb} + \mathrm{O}_2\]the new equilibrium constant will be \(K_1^{\prime} = \frac{1}{K_1}\). Adding this to the second reaction, we get the target reaction with the equilibrium constant \(K_{\text{target}} = \frac{K_2}{K_1}\) since we are multiplying the two reactions.

Calculate the equilibrium constant for the target reaction

Using the equation from Step 3, we calculate:\[K_{\text{target}} = \frac{K_2}{K_1} = \frac{1 \times 10^{14}}{2 \times 10^{12}} = 5 \times 10^{1}\]Therefore, the equilibrium constant for the target reaction is \(5 \times 10^{1}\).

Determine the direction of equilibrium

Since the equilibrium constant is much greater than 1 (\(5 \times 10^{1}\)), the equilibrium lies far to the right, favoring the production of products.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rates of the forward and reverse reactions in a chemical process are equal, resulting in no net change in the concentrations of reactants and products over time. It's crucial to note that chemical equilibrium is a dynamic state, where reactants are continually being converted to products and vice versa, but in such a manner that the amounts of each remain constant.

An example of chemical equilibrium is the binding of oxygen to hemoglobin, a process crucial for transporting oxygen throughout the body. In the bloodstream, hemoglobin (Hb) binds to oxygen (O₂) to form oxyhemoglobin (HbO₂). The equilibrium constant, denoted by K, is a numerical value that reflects the ratio of product concentrations to reactant concentrations at equilibrium. This value is significant as it helps predict the direction and extent of chemical reactions under given conditions.

In the provided exercise, when we refer to the calculation of the equilibrium constant value for the reaction where HbO₂ and carbon monoxide (CO) compete to bind with hemoglobin, understanding the equilibrium constants of the individual reactions is key in determining the final equilibrium state.

Le Chatelier's Principle

Le Chatelier's principle is a guiding rule to predict the effect of a change in conditions on a chemical equilibrium. When an external stress, such as a change in concentration, pressure, or temperature, is applied to a system at equilibrium, the system adjusts to minimize the stress and restore a new state of equilibrium.

For instance, if more carbon monoxide (CO) is introduced into the bloodstream, it will shift the equilibrium to form more carboxyhemoglobin (HbCO) due to its higher affinity for hemoglobin compared to oxygen. This shift can be detrimental to biological systems, as it impairs the blood's ability to carry oxygen, illustrating the importance of equilibrium concepts in understanding biological processes as well as chemical reactions.

Applying Le Chatelier's principle to chemical equilibria allows us to understand and predict how shifts in conditions can affect the outcome of reactions, which is pivotal in chemical manufacturing and in maintaining bodily functions like oxygen transport.

Hemoglobin Binding Affinity

Hemoglobin binding affinity refers to the strength with which hemoglobin binds to various molecules, such as oxygen (O₂) and carbon monoxide (CO). Hemoglobin's affinity for different ligands affects its function as an oxygen transporter in the body.

As demonstrated in the textbook exercise, hemoglobin has a different affinity for oxygen and carbon monoxide, which is quantitatively expressed by their respective equilibrium constants. A higher equilibrium constant indicates a stronger binding affinity. Therefore, hemoglobin's affinity for CO is stronger than its affinity for O₂, as seen from the larger equilibrium constant for the Hb-CO binding reaction. The balance between these affinities is a vital aspect of physiological health, as carbon monoxide can displace oxygen in hemoglobin, preventing efficient oxygen delivery to tissues.

Understanding the principles behind hemoglobin affinity is essential not only in biochemical and medical contexts but also in environmental health, where exposure to carbon monoxide and other pollutants can disrupt the delicate balance of oxygen transport in the body.