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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: $$\begin{array}{cl}\mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ}\end{array}$$ 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 (K) for the reaction \(\mathrm{HbO}_{2} + \mathrm{CO} \leftrightarrow \mathrm{HbCO} + \mathrm{O}_{2}\) under the given conditions is approximately \(6.4 \times 10^{12}\).

Step by step solution

01

Analyze the Given Reactions and the Target Reaction

The given reactions are: \(\mathrm{Hb} + \mathrm{O}_{2} \rightarrow \mathrm{HbO}_{2}\) \(\Delta G° = -70 kJ\) (Reaction 1) \(\mathrm{Hb} + \mathrm{CO} \rightarrow \mathrm{HbCO}\) \(\Delta G° = -80 kJ\) (Reaction 2) The target reaction is: \(\mathrm{HbO}_{2} + \mathrm{CO} \leftrightarrow \mathrm{HbCO} + \mathrm{O}_{2}\) (Reaction 3) We can see that Reaction 3 is the reverse of Reaction 1 + Reaction 2.
02

Calculate the Standard Free Energy Change for the Target Reaction

Since Reaction 3 is simply the sum of Reaction 1 and 2 (with Reaction 1 being run in reverse), we can calculate the standard free energy for Reaction 3 by subtracting the \(\Delta G°\) for Reaction 1 from the \(\Delta G°\) for Reaction 2. Applying the principle of thermodynamics that, if a reaction is reversed, the sign of \(\Delta G°\) changes, the free-energy change for the backward reaction of Reaction 1 would be +70 kJ. Thus, \(\Delta G°\) for Reaction 3 is given by: \(\Delta G_{3}° = \Delta G_{2}° - \Delta G_{1}°\) Substituting the given values, \(\Delta G_{3}° = -80 kJ - 70 kJ = -150 kJ\)
03

Convert the Temperature to Kelvin

To use the ΔG° = -RT ln K equation, we need the temperature in Kelvin rather than Celsius. \(T = 25°C + 273.15 = 298.15 K\)
04

Calculate the Equilibrium Constant

Substitute the values of ΔG°, R, and T into the equation ΔG° = -RT ln K: We also convert kJ to J, because the universal gas constant R's unit in this case is J (R = 8.314 J/mol K) \(-150000 J = -(8.314 J/mol K)(298.15 K)lnK\) Solve for ln K: \(lnK = \frac{-150000 J}{(8.314 J/mol K)(298.15 K)}\) Then, take the exponential of both sides to solve for K: \(K = e^{lnK}\) Calculate the final value for K on a calculator. This final result is the equilibrium constant for the target reaction involving hemoglobin, O2, and CO under the given conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Energy Change
Free energy change, represented as \( \Delta G \), is a concept in chemistry that measures the amount of energy available to do work in a chemical reaction. It reflects the difference between the energy in the reactants and the energy in the products. A negative free energy change means a reaction is spontaneous and can proceed without any additional energy input.
In this exercise, the free energy change for the formation of hemoglobin-bound oxygen (\(\Delta G° = -70 \mathrm{kJ}\)) and hemoglobin-bound carbon monoxide (\(\Delta G° = -80 \mathrm{kJ}\)) were given. These values help understand how favorably hemoglobin binds to different gases.
Hemoglobin Binding
Hemoglobin is a protein in red blood cells responsible for transporting oxygen throughout the body. Its binding capacity is very specific and crucial for physiological processes. This exercise focuses on comparing its binding to oxygen and carbon monoxide.
Carbon monoxide has a greater affinity for hemoglobin than oxygen, making it toxic since it can outcompete oxygen for binding sites on hemoglobin, leading to reduced oxygen delivery to tissues. By examining the free energy changes, we can deduce which gas hemoglobin prefers under standard conditions.
Gibbs Free Energy
Gibbs free energy, often referred to merely as free energy, is a thermodynamic quantity that determines the spontaneity of a process. In terms of biochemical reactions, it indicates how favorable a reaction is.
  • A negative Gibbs free energy implies the process is thermodynamically favorable.
  • Changes in Gibbs free energy are used to compute equilibrium constants for reactions, which tell us about the ratio of products to reactants at equilibrium.
For hemoglobin binding reactions, this concept helps predict the likelihood of hemoglobin binding to either oxygen or carbon monoxide.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products, often involving energy changes. In biological contexts, these reactions are responsible for sustaining life processes.
In this scenario, reactions involve hemoglobin binding with \(\mathrm{O}_{2}\) and \(\mathrm{CO}\). The transformation described is of key interest when observing how different molecules interact with hemoglobin. These insights help us understand potential hazards such as carbon monoxide poisoning and develop solutions.
Thermodynamics
Thermodynamics is a branch of physics that deals with heat and temperature and their relation to energy and work. It provides the foundation for understanding energy changes during chemical reactions.
With the principle \( \Delta G° = -RT \ln K \), this exercise demonstrates how changes in Gibbs free energy are linked to equilibrium constants.
  • \( R \) is the universal gas constant.
  • \( T \) is the temperature in Kelvin.
Thus, thermodynamics helps predict whether a chemical reaction is likely to occur under certain conditions.
Reaction Equilibria
Reaction equilibria refer to the state where the forward and reverse reactions occur at the same rate, so the concentration of reactants and products remains constant over time.
The equilibrium constant \((K)\) provides a measure of this balance. A larger \( K \) indicates a reaction favors product formation. In our analysis, calculating \( K \) helps understand which hemoglobin complex is more stable under physiological conditions, essential for medical and environmental health considerations.

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