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

To do this, we will first rewrite the given reactions as their corresponding equilibrium reactions by reversing the direction of the second reaction: \(Hb + O_2 \rightleftharpoons HbO_2 \) with \(\Delta G_1^{\circ} = -70 \mathrm{~kJ} \) \(HbCO \rightleftharpoons Hb + CO \) with \(\Delta G_2^{\circ} = 80 \mathrm{~kJ} \) Now, we will add these equilibrium reactions to get the desired equilibrium reaction: \[ (Hb + O_2) + (HbCO) \rightleftharpoons (HbO_2) + (Hb + CO) \] Which simplifies to: \[ HbO_2 + CO \rightleftharpoons HbCO + O_2 \] To find the overall standard free-energy change, we will add the standard free-energy changes of both reactions: \[ \Delta G_{eq}^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ} = (-70 \mathrm{~kJ}) + (80 \mathrm{~kJ}) = 10\mathrm{~kJ} \] So, the overall standard free-energy change for the given equilibrium reaction is 10 kJ.

To find the equilibrium constant (K), we will use the equation that relates the equilibrium constant to the standard free-energy change at a specific temperature, which is: \[ \Delta G^{\circ} = -RT \ln K \] We're given the temperature (T) as 298 K. We'll also need the gas constant R, which is 8.314 J/(mol*K). We can rearrange the equation to solve for K: \[ K = \exp(-\frac{\Delta G^{\circ}}{RT}) \] Plugging in the values, we get: \[ K = \exp(-\frac{10,000 \mathrm{~J}}{(8.314 \mathrm{~J/(mol*K)})(298 \mathrm{~K})}) \] \[ K = \exp(-\frac{10,000}{2,478.012}) = \exp(-4.036) \] Now using a calculator to find the exponential value, we obtain: \[ K = 0.0179 \] So, the equilibrium constant for the given reaction at 298 K is approximately 0.0179.