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: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

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: \(\Delta G^{\circ}_{\text{given reaction}} = -\Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} = 70 \mathrm{~kJ} + (-80 \mathrm{~kJ}) = -10 \mathrm{~kJ}\)

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.