Problem 97 - Chemistry: The Central Science

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

Step-by-Step Solution

Answer

The change in standard free energy for the given reaction is ΔG = -10 kJ. By using the Gibbs free energy formula and substituting the values, we calculate the equilibrium constant, K, at 298 K to be approximately K ≈ 56.21.

Step 1: Calculate the change in standard free energy for the given reaction

Based on the two provided reactions, we can formulate the desired reaction as: HbO2 + CO ⇌ HbCO + O2 To determine the change in standard free energy, ΔG, we need to find the difference between the standard free energy of the products and the reactants in the equilibrium reaction. Thus, the equation becomes: ΔG = ΔGᵇ - ΔGᵃ Using the provided data: ΔGᵇ = -80 kJ ΔGᵃ = -70 kJ Therefore, the change in standard free energy for the given reaction is: ΔG = (-80 kJ) - (-70 kJ) = -10 kJ

Step 2: Calculate the equilibrium constant

To calculate the equilibrium constant, K, we will use the Gibbs free energy formula at constant temperature and pressure: ΔG = -RT ln(K) Where: ΔG = Change in standard free energy R = Gas constant (8.314 J/mol*K) T = Temperature (K) K = Equilibrium constant Rearrange the equation above to find K: K = exp(-ΔG / RT) Before calculating the equilibrium constant, convert ΔG to J/mol to match the unit of R: ΔG = (-10 kJ/mol) * (1000 J/1 kJ) = -10000 J/mol Now, substitute the values into the equation: K = exp(-(-10000 J/mol) / (8.314 J/mol*K * 298 K)) K ≈ exp(4.032) The equilibrium constant, K, at 298 K is approximately: K ≈ 56.21

Key Concepts

Carbon Monoxide Toxicity

Carbon monoxide (CO) is a dangerous gas due to its ability to bind very strongly to the iron in hemoglobin within our red blood cells. Hemoglobin is a protein that normally carries oxygen ( O_2 ) throughout the body. However, CO binds to this protein significantly more strongly than oxygen does.

When CO binds to hemoglobin, it forms a complex called carboxyhemoglobin (HbCO). This bond is extremely stable compared to the oxygen-hemoglobin bond that forms oxyhemoglobin (HbO_2). As a result, even tiny amounts of CO can outcompete oxygen, preventing oxygen from being transported effectively.

Some possible effects of this competition for hemoglobin include:

• Impaired oxygen delivery to vital organs and tissues.
• Symptoms such as headache, dizziness, and in severe cases, death.
• The critical disruption of cellular respiration and energy production.

This is why exposure to carbon monoxide can be extremely harmful, as it suffocates affected tissues slowly by blocking oxygen transport.

Standard Free Energy

In chemistry, standard free energy change ( ΔG^ heta ) is a concept that expresses the energy change associated with a chemical reaction. It helps predict the direction in which a reaction will naturally proceed.

To understand how standard free energy works, consider two reactions involving hemoglobin: one with oxygen and the other with carbon monoxide. For the reaction forming oxyhemoglobin ( HbO_2 ), we have a standard free energy change of -70 kJ/mol. Comparatively, forming carboxyhemoglobin ( HbCO ) exhibits a standard free energy change of -80 kJ/mol.

The more negative the value of ΔG^ heta , the more energetically favorable the reaction. This means:

• The HbCO formation is more favorable than HbO_2 formation at standard conditions due to its more negative free energy.
• Reactions with negative ΔG^ heta are spontaneous, meaning they proceed naturally without requiring energy input.

Understanding standard free energy is crucial in predicting the balance of different competing reactions in biological systems, like those involving oxygen and carbon monoxide binding with hemoglobin.

Equilibrium Constant Calculation

The equilibrium constant ( K ) of a reaction expresses the ratio of product concentrations to reactant concentrations at equilibrium. The calculation of K depends on knowing the standard free energy change, ΔG , for the reaction, and it's computed using the Gibbs Free Energy equation: ΔG = -RT ln(K) where:

• ΔG is the change in standard free energy (J/mol).
• R is the universal gas constant ( 8.314 ext{ J/mol} ext{ }K ).
• T is the temperature in Kelvin.

To find K for the reaction involving hemoglobin, CO, and O_2 , we first calculate ΔG which is -10 kJ/mol or -10000 J/mol to be consistent with the units of R . Using these values in the equation:
K = exp(-( -10000 J/mol) / (8.314 J/mol*K * 298 K)) Calculating this, we get:
K ≈ exp(4.032) ≈ 56.21 This means at equilibrium, the reaction strongly favors the production of HbCO and O_2 because of the relatively high value of the equilibrium constant. This insight underscores the substantial binding affinity of CO over O_2 to hemoglobin.