Question

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin \((\mathrm{Hb})\) 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: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$

Short answer

The equilibrium constant at 25°C for the reaction HbO₂(aq) + CO(g) ⟷ HbCO(aq) + O₂(g) is approximately: K ≈ 56.23

Step-by-step solution

Determine the final reaction by combining the given reactions

First, we will obtain the reaction of interest by combining the given reactions. This can be done by: 1. Reversing the first reaction 2. Adding the reversed first reaction to the second reaction The reversed first reaction is: HbO₂(aq) ⟶ Hb + O₂(g), ΔG° = +70 kJ Now, let's add the reversed first reaction and the second reaction together: HbO₂(aq) ⟶ Hb + O₂(g) Hb + CO(g) ⟶ HbCO(aq) The final reaction is: HbO₂(aq) + CO(g) ⟷ HbCO(aq) + O₂(g)

Determine the standard free energy change of the combined reaction

Since we added the reversed first reaction and the second reaction, we must add their standard free energy changes as well. So, the standard free energy change for the desired reaction, ΔG°, would be: ΔG° = (+70 kJ) + (-80 kJ) = -10 kJ/mol

Convert standard free energy change to equilibrium constant

We will use the relation between standard free energy change and equilibrium constant K, which is given by the formula: ΔG° = -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. We are given the temperature in Celsius, so we need to convert it to Kelvin. At 25°C, the temperature in Kelvin is: T = 25 + 273.15 = 298.15 K Now, let's solve for the equilibrium constant, K: -10,000 J/mol = -(8.314 J/mol∙K) × (298.15 K) × ln(K) Rearrange to isolate ln(K): ln(K) = -10,000 J/mol / (-8.314 J/mol∙K × 298.15 K) ≈ 4.030 Now, take the exponent on both sides to solve for K: K = e^(4.030) ≈ 56.23

Write the final answer for the equilibrium constant

The equilibrium constant at 25°C for the given reaction HbO₂(aq) + CO(g) ⟷ HbCO(aq) + O₂(g) is approximately: K ≈ 56.23

Key concepts

Standard Free Energy Change

The standard free energy change, denoted by \( \Delta G^{\circ} \), represents the shift in free energy during a reaction under standard conditions (often 1 atm pressure and 298 K). This value helps in predicting whether a reaction will occur spontaneously.
- A negative \( \Delta G^{\circ} \) suggests that the reaction is spontaneous and can proceed without any external energy input.
- Conversely, a positive \( \Delta G^{\circ} \) indicates that the reaction is non-spontaneous under standard conditions.
To determine the equilibrium constant \( K \) for a reaction, we utilize the relationship:\[ \Delta G^{\circ} = -RT \ln(K) \]where

• \( R \) is the gas constant, approximately 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin.

In the exercise, the \( \Delta G^{\circ} \) for the formation of HbCO from HbO₂ and CO is specified as \(-10 \text{ kJ/mol} \). By rearranging the equation and solving for \( K \), we obtain an equilibrium constant that provides insights into how far the reaction will proceed under given conditions. This is crucial for understanding chemical behavior in biological systems like oxygen transport in blood.

Carbon Monoxide Toxicity

Carbon monoxide (CO) is a colorless, odorless, and highly toxic gas. Its toxicity is primarily due to its ability to bind with high affinity to hemoglobin (a protein in blood responsible for oxygen transport). This is dangerous because:
- CO tends to bind to the iron in hemoglobin about 200 to 250 times more strongly than oxygen (\( \text{O}_2 \)).
- Because of this strong binding, CO competes with and displaces oxygen, leading to reduced oxygen transport in the bloodstream.
Symptoms of CO poisoning range from mild headaches and nausea to severe consequences like unconsciousness or death, depending on exposure levels. The competitive binding of CO on hemoglobin diminishes the body's ability to supply necessary \( \text{O}_2 \) to tissues and organs, which can cause critical damage if not treated promptly. Therefore, understanding the equilibrium and kinetic parameters of CO and hemoglobin interaction is vital for healthcare and safety science.

Hemoglobin Binding

Hemoglobin (Hb) is a complex protein found in red blood cells. Its primary role is to transport oxygen from the lungs to tissues, and carbon dioxide back to the lungs for exhalation. The process of hemoglobin binding involves:
- **Oxygen Binding:** Each hemoglobin molecule can bind up to four oxygen molecules, forming oxyhemoglobin (HbO₂). This reversible binding is crucial for efficient oxygen transport.
- **Carbon Monoxide Binding:** When carbon monoxide is present, it can attach to hemoglobin's iron sites, forming carboxyhemoglobin (HbCO).
The affinity of hemoglobin for oxygen is generally regulated by factors such as pH level and carbon dioxide concentration. However, CO can disrupt this regulation due to its higher affinity for hemoglobin. This process not only limits oxygen delivery but also alters the structural conformation of hemoglobin, affecting its usual function.
Managing conditions and understanding the mechanisms of hemoglobin's binding properties are vital to unraveling the physiological impacts of various gases, like CO, on human health.