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: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Short answer

The equilibrium constant for the reaction \( \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} \) at 298 K is approximately 55.6.

Step-by-step solution

Manipulate the given reactions to obtain the reaction of interest

First, we need to rearrange the given reactions (1) and (2) to obtain the reaction we are looking for. We can achieve this by reversing reaction (1) and adding it to reaction (2): Reverse reaction (1): $$ \mathrm{HbO}_{2} \longrightarrow \mathrm{Hb}+\mathrm{O}_{2} \hspace{1cm} \Delta G^{\circ}_{1R}=70 \mathrm{~kJ} $$ Add reversed reaction (1) to reaction (2): $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} \hspace{1cm} \Delta G^{\circ}=\Delta G^{\circ}_{1R}+\Delta G^{\circ}_2 $$

Calculate the standard free-energy change for the reaction of interest

Now we calculate the standard free-energy change for the reaction of interest by summing the standard free-energy changes of the manipulated reactions: $$ \Delta G^{\circ}=\Delta G^{\circ}_{1R}+\Delta G^{\circ}_2 = 70 \mathrm{~kJ} + (-80 \mathrm{~kJ}) = -10 \mathrm{~kJ} $$

Calculate the equilibrium constant at 298 K

We can calculate the equilibrium constant (K) using the following formula, relating the standard free-energy change to the equilibrium constant: $$ \Delta G^{\circ}=-RT\ln K $$ where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K). We can rearrange the formula to solve for K: $$ K=e^{-\frac{\Delta G^{\circ}}{RT}} $$ Plugging in the values: $$ K=e^{-\frac{-10\mathrm{~kJ/mol}}{(8.314 \times 10^{-3}\mathrm{~kJ/mol·K})(298\mathrm{~K})}} $$ $$ K\approx e^{4.02} \approx 55.6 $$ The equilibrium constant for the reaction at 298 K is approximately 55.6.

Key concepts

Free Energy Change

Free energy change, often denoted as \( \Delta G \), is a key concept in understanding chemical reactions and their spontaneity. It represents the maximum amount of work a thermodynamic process can perform at constant temperature and pressure. In simpler terms, it tells us whether a reaction can occur on its own.
For a reaction to be spontaneous, \( \Delta G \) should be negative. In the exercise about carbon monoxide (CO) and hemoglobin (Hb), we see two reactions with different free energies.

• \( \mathrm{Hb} + \mathrm{O}_2 \longrightarrow \mathrm{HbO}_2 \) has \( \Delta G^{\circ} = -70 \) kJ.
• \( \mathrm{Hb} + \mathrm{CO} \longrightarrow \mathrm{HbCO} \) has \( \Delta G^{\circ} = -80 \) kJ.

These values suggest that CO binds more strongly to hemoglobin because it has a more negative free energy change.
By reversing and combining these reactions, we find \( \Delta G^{\circ} = -10 \) kJ for the new reaction of interest. This small negative value indicates a spontaneous reaction, though not as strongly as each individual reaction.

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), gives insight into the balance between product and reactant concentrations at equilibrium. It is calculated from the expression \( \Delta G^{\circ} = -RT \ln K \), connecting free energy change and equilibrium behavior.
When \( \Delta G^{\circ} \) is negative, as it is in the problem, \( K \) is greater than 1, showing a higher concentration of products compared to reactants at equilibrium.
In our exercise, by determining the value of \( K \approx 55.6 \), we can conclude that at 298 K, the reaction heavily favors the formation of the products (\( \mathrm{HbCO} \) and \( \mathrm{O}_2 \)).

• A larger \( K \) value means the products are more stable.
• The previous calculations using \( \Delta G^{\circ} \) assure us that CO binding is indeed preferred over \( \mathrm{O}_2 \).

Gas Constant

The gas constant \( R \) is a vital component when linking thermodynamic properties, such as free energy, to chemical equilibrium. It's a universal constant applicable to a wide range of chemistry-related calculations.
In the formula \( \Delta G^{\circ} = -RT \ln K \), \( R \) acts as the proportionate factor that allows us to convert the energy terms into unit reactions of matter. Its value is \( 8.314 \) J/mol·K, which reflects the energy per degree per mole.
Handling these calculations:

• Temperature: The constant is used with temperature in Kelvin, ensuring consistency and accuracy in thermodynamic equations.
• Equilibrium calculations: Multiplying \( R \) by temperature gives the energy scale reference needed to determine \( K \).

Our problem used \( R \) effectively to find \( K \approx 55.6 \), showing how critical standardized constants are in computing reaction properties.