Question

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Short answer

The equilibrium constant value for the reaction \(\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons\mathrm{HgbCO}+\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) is approximately 3.53.

Step-by-step solution

Determine the standard free energy change for the given reaction

We have the standard free energy changes for two reactions: 1) \(\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow\mathrm{HgbO}_{2}\): \(\Delta G^{\circ}_{1} = -70 \mathrm{~kJ}\) 2) \(\mathrm{Hgb}+\mathrm{CO} \longrightarrow\mathrm{HgbCO}\): \(\Delta G^{\circ}_{2} = -80 \mathrm{~kJ}\) We need to find the standard free energy change for the reaction: \(\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons\mathrm{HgbCO}+\mathrm{O}_{2}\) Notice that the second reaction is the reverse of Reaction 1 and adding the first reaction. To obtain this reaction, we can reverse Reaction 1 and add it to Reaction 2. Let's calculate the standard free energy change for the reversed Reaction 1: \(-\Delta G^{\circ}_{1} = 70 \mathrm{~kJ}\) Now, let's add the standard free energy changes for the reversed Reaction 1 and Reaction 2: \(\Delta G^{\circ}_{\text{given reaction}} = -\Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} = 70 \mathrm{~kJ} + (-80 \mathrm{~kJ}) = -10 \mathrm{~kJ}\)

Calculate the equilibrium constant value

To calculate the equilibrium constant value (\(K\)), we can use the relationship between the standard free energy change and the equilibrium constant: \(\Delta G^{\circ} = -RT\ln K\) Where \(R\) is the gas constant (8.314 J/(mol·K)), \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant. First, let's convert the temperature to Kelvin: \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Now, we can rearrange the equation and solve for \(K\): \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) Plugging in the values for \(\Delta G^{\circ}\), \(R\), and \(T\): \(K = e^{-\frac{-10 \times 10^{3} \mathrm{J/mol}}{(8.314 \mathrm{J/(mol\cdot K)})(298.15 \mathrm{K})}}\) \(K \approx 3.53\) So, the equilibrium constant value for the given reaction at \(25^{\circ} \mathrm{C}\) is approximately 3.53.

Key concepts

Standard Free Energy Change

The standard free energy change, denoted as \( \Delta G^{\circ} \), is a critical parameter in chemistry that helps predict the direction of a chemical reaction. It combines enthalpy and entropy changes at a given temperature to understand how spontaneous a process is.
If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous under standard conditions (favorable). Conversely, a positive \( \Delta G^{\circ} \) suggests that the reaction isn’t spontaneous.
For the reaction involving hemoglobin (Hgb), CO, and \( \text{O}_2 \), the process of finding the \( \Delta G^{\circ} \) involves reversing known reactions. Here, the \( \Delta G^{\circ} \) value calculated at \(-10 \text{ kJ} \) indicates that the reaction can proceed slightly favorably as written, potentially due to the favorable conditions generated by reverse bonding interactions.

Equilibrium Constant

The equilibrium constant, represented as \( K \), quantifies the ratio of products to reactants when a reaction reaches equilibrium. It essentially determines how far a reaction proceeds before becoming stable in terms of concentrations.
In thermodynamics, \( K \) is directly related to the standard free energy change through the equation \( \Delta G^{\circ} = -RT\ln K \).
Here, using hemoglobin’s reactions with CO and \( \text{O}_2 \), we derived an equilibrium constant of approximately 3.53. This indicates that at 25°C, the formation of \( \text{HgbCO} \) and \( \text{O}_2 \) is slightly favored, due to its greater binding affinity, particularly highlighting CO’s strong attachment to hemoglobin compared to oxygen.

Hemoglobin

Hemoglobin, abbreviated as Hgb, is a vital protein in red blood cells responsible for transporting oxygen from the lungs to tissues in the body and returning carbon dioxide back to the lungs.
This molecule’s functionality is dictated by its ability to bind, and subsequently release, oxygen and other small molecules, such as carbon monoxide (CO).
In our scenario, hemoglobin’s affinity for CO greatly exceeds that for \( \text{O}_2 \), leading to its toxicological effects. This preferential bonding to CO makes it difficult for oxygen to bind to hemoglobin in sufficient amounts, potentially causing life-threatening conditions due to inhibited oxygen transport.

Thermodynamics

Thermodynamics is the branch of physical science dealing with the relations between heat and various energy forms in chemical processes. It provides a macroscopic framework for understanding energy transfer, efficiency, and direction of spontaneous processes.
Using thermodynamic principles, we can determine how energy changes affect chemical reactions, exemplified in the hemoglobin reactions.
By using thermodynamics laws, such as those dealing with free energy, we gain insights into reaction spontaneity and equilibrium, enabling us to predict reaction behaviors under varying conditions. In studying hemoglobin’s interactions, thermodynamics helps explain why CO outcompetes \( \text{O}_2 \) for binding, as seen in the computed \( \Delta G^{\circ} \) and equilibrium constant values, further illustrating the driving forces behind these biological critical reactions.