Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O_2. Consider the following reactions and approximate standard free energy changes: $$\begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{array}$$ 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 are given the standard free energy changes for reactions (1) and (2) as follows: (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 want to find the standard free energy change for the following reaction: \[ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}. \] To get the desired reaction, we have to reverse reaction (1) and add it to reaction (2): \(-1 \times \mathrm{(1)}\) : \[\mathrm{HgbO}_{2} \longrightarrow \mathrm{Hgb}+\mathrm{O}_{2}, \Delta G^{\circ}_1'=70 \mathrm{kJ}\]. \(+1 \times \mathrm{(2)}\) : \[\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO}, \Delta G^{\circ}_2=-80 \mathrm{kJ}\]. Adding the two reactions: \[\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}, \Delta G^{\circ}=\Delta G^{\circ}_1'+\Delta G^{\circ}_2=70 \mathrm{kJ}- 80 \mathrm{kJ} = -10 \mathrm{kJ}\]. The standard free energy change for the given reaction is -10 kJ.

The relation between standard free energy change and the equilibrium constant is given by: \[\Delta G^{\circ}=-RT \ln K\] Here, \(R = 8.3145 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\) is the gas constant, \(T = 25^{\circ}\mathrm{C} = 298\mathrm{K}\) is the temperature, and \(K\) is the equilibrium constant. We need to find the value of \(K\). First, convert the standard free energy change from kJ to J: \[\Delta G^{\circ}=-10\ \mathrm{kJ} = -10,000\ \mathrm{J}\]. Now, rearrange the equation and solve for \(K\): \[K = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{\frac{10,000\ \mathrm{J}}{(8.3145\ \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}})(298\ \mathrm{K})}} \approx 1.10 \times 10^3\]. The estimated equilibrium constant value at \($25^{\circ}\mathrm{C}\) for the given reaction is \(1.10 \times 10^3\).