Problem 99 - Chemistry The Central Science In Si Units

Question

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{ccc} \mathrm{Hb}+\mathrm{O}_{2} & \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow & \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at 298 \(\mathrm{K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Step-by-Step Solution

Answer

The equilibrium constant \( K_{eq} \approx 56.42 \) at 298 K.

Step 1: Write the Equations

The expressions for the formation of HbO\(_2\) and HbCO are given: \[ \text{Hb} + \text{O}_2 \rightarrow \text{HbO}_2 \quad \Delta G^{\circ} = -70 \text{ kJ/mol} \] \[ \text{Hb} + \text{CO} \rightarrow \text{HbCO} \quad \Delta G^{\circ} = -80 \text{ kJ/mol} \] We need the equation for the given equilibrium: \[ \text{HbO}_2 + \text{CO} \rightleftharpoons \text{HbCO} + \text{O}_2 \]

Step 2: Calculate the Reaction Free Energy

To find the \( \Delta G^{\circ} \) for the reaction \( \text{HbO}_2 + \text{CO} \rightleftharpoons \text{HbCO} + \text{O}_2 \), subtract the \( \Delta G^{\circ} \) of formation of \( \text{HbO}_2 \) from \( \text{HbCO} \): \[ \Delta G^{\circ}_{\text{reaction}} = \Delta G^{\circ}_{\text{HbCO}} - \Delta G^{\circ}_{\text{HbO}_2} = (-80 \text{ kJ/mol}) - (-70 \text{ kJ/mol}) = -10 \text{ kJ/mol} \]

Step 3: Relation Between Free Energy and Equilibrium Constant

The relation between standard free energy change and the equilibrium constant \( K_{eq} \) at temperature \( T \) is given by: \[ \Delta G^{\circ} = -RT \ln K_{eq} \] Where \( R = 8.314 \text{ J/mol⋅K} \) and \( T = 298 \text{ K} \).

Step 4: Calculate the Equilibrium Constant

Rearrange the equation from the previous step to solve for \( K_{eq} \): \[ K_{eq} = e^{-\Delta G^{\circ}/RT} \] Substitute \( \Delta G^{\circ} = -10,000 \text{ J/mol} \): \[ K_{eq} = e^{-(-10,000 \text{ J/mol})/(8.314 \text{ J/mol⋅K} \times 298 \text{ K})} = e^{4.032} \approx 56.42\]

Step 5: Conclusion

The equilibrium constant \( K_{eq} \) for the reaction \( \text{HbO}_2 + \text{CO} \rightleftharpoons \text{HbCO} + \text{O}_2 \) at 298 K is approximately 56.42.

Key Concepts

Free Energy Change

Free energy change is a crucial concept in understanding chemical reactions and their spontaneity. It is represented as \( \Delta G \), which connects the thermodynamics of a process with its equilibrium properties. When \( \Delta G \) is negative, it implies that a process can proceed spontaneously. In the context of hemoglobin binding, we observe two different reactions:

• \( \text{Hb} + \text{O}_2 \rightarrow \text{HbO}_2 \) with \( \Delta G^{\circ} = -70 \text{ kJ/mol} \)
• \( \text{Hb} + \text{CO} \rightarrow \text{HbCO} \) with \( \Delta G^{\circ} = -80 \text{ kJ/mol} \)

These values indicate that the binding of carbon monoxide to hemoglobin is more spontaneous compared to oxygen binding. This greater negativity of \( \Delta G \) for CO binding highlights its stronger affinity to hemoglobin. In essence, the larger the negative \( \Delta G \), the more favorable or spontaneous the reaction is. It is this property that makes carbon monoxide a potent toxin as it can outcompete oxygen, compromising oxygen transport in the blood.

Hemoglobin Binding

But why does carbon monoxide bind more strongly to hemoglobin compared to oxygen? This question can be answered by examining the nature of hemoglobin's binding sites. Hemoglobin carries oxygen in the blood from the lungs to other parts of the body, where oxygen is released. It achieves this through a protein structure capable of binding oxygen molecules tightly and then releasing them as needed.The affinity of hemoglobin for carbon monoxide is about 200-250 times greater than for oxygen. This is because of the particular way that the carbon monoxide molecule interacts with the iron atom in hemoglobin. The molecular geometry of CO allows it to fit into the binding site more snugly than oxygen, forming a very stable complex. This stability is reflected in the lower \( \Delta G^{\circ} \) for the formation of HbCO, demonstrating a more spontaneous situation compared to HbO\(_2\).Understanding hemoglobin binding and the dangers of substances like carbon monoxide is crucial in fields such as medicine and respiratory therapy. It helps tailor approaches to treating carbon monoxide poisoning and improving oxygen delivery to tissues.

Standard Gibbs Free Energy

Standard Gibbs Free Energy (\( \Delta G^{\circ} \)) is an extension of the concept of free energy change but is specifically measured under standard conditions (e.g., 298 K, 1 atm pressure, and 1 M concentration for solutions). The standard condition makes \( \Delta G^{\circ} \) a useful tool because it allows comparisons between different reactions. In reactions involving hemoglobin, the \( \Delta G^{\circ} \) values help us compare the inherent stability and spontaneity of different bindings.The equation linking standard Gibbs Free Energy with the equilibrium constant is a cornerstone in physical chemistry:\[\Delta G^{\circ} = -RT \ln K_{eq}\]where \( R = 8.314 \text{ J/mol⋅K} \) and \( T = 298 \text{ K} \). This equation can be rearranged to calculate the equilibrium constant \( K_{eq} \), which quantifies the ratio of the concentration of products to reactants at equilibrium.In the given carbon monoxide and oxygen binding situation, \( \Delta G^{\circ} \) was calculated as \(-10 \text{ kJ/mol}\) for the reaction \( \text{HbO}_2 + \text{CO} \rightleftharpoons \text{HbCO} + \text{O}_2 \). Using the formula for equilibrium constant, this led to \( K_{eq} \approx 56.42 \). Knowing \( K_{eq} \) offers insight into how far the reaction will proceed and the dominance of products or reactants at equilibrium, emphasizing the significance of \( \Delta G^{\circ} \) in predicting reaction behavior.