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Chapter 21: Problem 81

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

Short Answer

Expert verified
The equilibrium constant for the reaction HbO₂ + CO → HbCO + O₂ at 25°C is approximately 56.56.

Step by step solution

01

Determine the ΔG° for the desired reaction

By reversing the first given reaction and adding it to the second reaction, we can obtain the desired reaction as follows: Reverse the first reaction: HbO₂ → Hb + O₂, ΔG° = 70 kJ Keep the second reaction as it is: Hb + CO → HbCO, ΔG° = -80 kJ Add both reactions: ( HbO₂ → Hb + O₂ ) + ( Hb + CO → HbCO ) Which leads to: HbO₂ + CO → HbCO + O₂ Now let's determine the standard free energy change for the new reaction, ΔG°: ΔG°(desired reaction) = ΔG°(first reaction) + ΔG°(second reaction) ΔG°(desired reaction) = 70kJ + (-80 kJ) ΔG°(desired reaction) = -10 kJ
02

Calculate the equilibrium constant (K)

Now that we have the standard free energy change (ΔG°) for the desired reaction, we can use the equation ΔG° = -RT ln(K) to solve for the equilibrium constant (K): -ΔG° = RT ln(K) -(-10 kJ) = (8.314 J/mol*K) * (298 K) * ln(K) (10 kJ) * (1000 J/1 kJ) = (8.314 J/mol*K) * (298 K) * ln(K) 10000 J = 2479.572 J/mol * ln(K) Divide both sides by 2479.572 J/mol: ln(K) ≈ 4.035 Now exponentiate both sides to solve for K: K ≈ e^(4.035) K ≈ 56.56 The equilibrium constant for the reaction HbO₂ + CO → HbCO + O₂ at 25°C is approximately 56.56.

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