Problem 99 - Chemistry: The Central Science in SI Units, Global Edition

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

Step-by-Step Solution

Answer

The overall standard free-energy change for the equilibrium reaction HbO2 + CO ⇌ HbCO + O2 is 10 kJ. To calculate the equilibrium constant, we use the equation ΔG° = -RT ln K. At 298 K, the equilibrium constant is approximately 0.0179.

Step 1: Find the overall standard free-energy change for the equilibrium reaction

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.

Step 2: Calculate the equilibrium constant

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.

Key Concepts

Standard Free-Energy Change

The standard free-energy change (\( \Delta G^{\circ} \)) is a critical factor in understanding chemical reactions. It indicates whether a reaction proceeds spontaneously under standard conditions. A negative value suggests spontaneity, signaling that the products have lower energy than the reactants. Conversely, a positive value means the reaction is non-spontaneous, requiring an energy input to proceed.

In the provided exercise, we have two reactions involving hemoglobin:

• The binding of \( O_2 \): \( \Delta G^{\circ} = -70 \mathrm{~kJ} \)
• The binding of CO: \( \Delta G^{\circ} = -80 \mathrm{~kJ} \)

Despite both reactions being spontaneous, CO binds more strongly to hemoglobin than \( O_2 \). This is evident as the \( \Delta G^{\circ} \) for CO is more negative, indicating a more energetically favorable reaction and thus a stronger binding.Breaking down these energy changes helps us analyze which molecules prefer to bind under equilibrium conditions and provides a basis for further calculations, such as determining the equilibrium constant tied to these reactions.

Carbon Monoxide Toxicity

Carbon monoxide (CO) is a colorless, odorless gas, dangerous primarily due to its strong affinity for hemoglobin in the blood. This high affinity is reflected in the standard free-energy change value, demonstrating a more favorable binding to hemoglobin compared to oxygen.

Once CO binds to hemoglobin, it forms carboxyhemoglobin (HbCO), severely diminishing hemoglobin's ability to carry oxygen continuously. This causes significant health issues, leading to tissue hypoxia since organs and tissues get insufficient oxygen. Symptoms of CO poisoning can vary from mild headaches and dizziness to severe outcomes such as death.

Being aware of this toxicity is critical as even low concentrations can be harmful. Immediate access to fresh air and administration of 100% oxygen are essential in suspected CO poisoning cases. It’s vital to install CO detectors in homes and regularly maintain heating systems to prevent CO exposure.

Hemoglobin Binding

Hemoglobin is a protein in red blood cells responsible for transporting oxygen from the lungs to the body's tissues and returning carbon dioxide from the tissues to the lungs. This transport is due to the ability to bind both \( O_2 \) and CO. However, their binding strength varies significantly.

While oxygen is the primary molecule hemoglobin should bind for healthy function, carbon monoxide can occupy these binding sites far more tenaciously, as evidenced by the more negative free-energy change for CO binding. Once occupied by CO, hemoglobin transports less \( O_2 \), which is crucial for cellular respiration.

Understanding hemoglobin's properties and binding affinities allows us to comprehend not just physiological processes, but also the underlying science of medical conditions like CO poisoning. It’s also the foundation for developing treatments, such as using hyperbaric oxygen therapy to displace CO from hemoglobin in affected patients.

Chemical Equilibrium Calculation

In chemistry, equilibrium describes a state where the concentrations of reactants and products remain constant over time, meaning the forward and reverse reactions occur at the same rate. Calculation of an equilibrium constant (\( K \)) provides insights into the reaction's position, whether it favors reactants or products at equilibrium.

Using the equation \( \Delta G^{\circ} = -RT \ln K \) helps find the equilibrium constant based on the standard free-energy change (\( \Delta G^{\circ} \)), the temperature (\( T \)), and the gas constant (\( R \)). In our exercise, inserting these values gives us the \( K \), explaining the favorability of the reaction at that specific temperature.

• The calculated equilibrium constant \( K \approx 0.0179 \) suggests that the reaction strongly favors reactants at equilibrium, meaning the conversion of \( HbO_2 \) and CO into \( HbCO \) and \( O_2 \) is not favored under standard conditions.

For students, mastering these calculations is fundamental. It relates thermodynamic concepts to practical situations, like the dangers of CO exposure in biological systems.