Question

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin ( \((H b)\) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{cl} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Short answer

The equilibrium constant for the reaction \(\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}\) at 298 K is approximately 1.7 x 10^3.

Step-by-step solution

Understand Gibbs free energy equation and its relationship to equilibrium constant

To estimate the equilibrium constant, we will use the Gibbs free energy equation, which relates the standard free-energy change, the equilibrium constant, and the temperature of a reaction: \[\Delta G^{\circ} = -RT \ln K\] where \(\Delta G^{\circ}\) represents the standard free-energy change, \(R\) is the gas constant (8.314 J/(molK)), \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant.

Combine the given reactions to get the desired reaction

In order to find the equilibrium constant for the desired reaction, we will manipulate the given reactions by reversing the first and adding it to the second. Given Reactions: 1) \(\mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} \hspace{1cm} \Delta G^{\circ}_1 = -70 \mathrm{~kJ}\) 2) \(\mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} \hspace{1cm} \Delta G^{\circ}_2 = -80 \mathrm{~kJ}\) Reverse Reaction 1 1) \(\mathrm{HbO}_{2} \longrightarrow \mathrm{Hb}+\mathrm{O}_{2} \hspace{1cm} \Delta G^{\circ}_1' = +70 \mathrm{~kJ}\) Now we can add the reversed Reaction 1 and Reaction 2 to form the desired reaction: \[\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}\] To find the standard free-energy change for the desired reaction, add the \(\Delta G^{\circ}\) values of the reversed Reaction 1 and Reaction 2: \[\Delta G^{\circ}_3 = \Delta G^{\circ}_1' + \Delta G^{\circ}_2 = +70 \mathrm{~kJ/mol} -80 \mathrm{~kJ/mol} = -10 \mathrm{~kJ/mol}\]

Calculate the equilibrium constant using Gibbs free energy equation

Now we have the standard free-energy change for the desired reaction at the given temperature (298 K), and we can use this to find the equilibrium constant using the Gibbs free energy equation: \[\Delta G^{\circ} = -RT \ln K\] First, make sure to convert the standard free-energy change to J/mol: \[\Delta G^{\circ} = -10^4 \mathrm{J/mol}\] Now, substitute the known values into the equation and solve for the equilibrium constant \(K\): \[-10^4 = -(8.314)(298) \ln K\] Now, solve for \(K\): \[K = \exp\left(\frac{10^4}{(8.314)(298)}\right)\] \(K \approx 1.7\times 10^3\) The equilibrium constant for the reaction \(\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}\) at 298 K is approximately 1.7 x 10^3.

Key concepts

Gibbs Free Energy

Gibbs free energy is a concept that helps us understand the spontaneity of a reaction. It combines the system's enthalpy and entropy into a single value that indicates whether or not a reaction will occur without external energy. The formula for calculating Gibbs free energy is: \[ \Delta G = \Delta H - T\Delta S \]Where:

• \(\Delta G\) is the change in Gibbs free energy
• \(\Delta H\) is the change in enthalpy
• \(T\) is the temperature in Kelvin
• \(\Delta S\) is the change in entropy

When \(\Delta G\) is negative, the reaction tends to proceed spontaneously. If it is positive, the reaction is non-spontaneous and requires energy input. Gibbs free energy is significant because it considers both energy changes and disorder, providing a full picture of a reaction's feasibility.

Standard Free-Energy Change

The standard free-energy change, denoted as \(\Delta G^{\circ}\), is the Gibbs free energy change measured under standard conditions (1 atm pressure, 1 M concentration, and a temperature of 298 K). In chemical reactions like those involving hemoglobin and gases like CO and \(O_2\), \(\Delta G^{\circ}\) helps predict which reaction will occur more readily. In the provided problem:

• \(\Delta G^{\circ}\) for \(Hb + O_2 \rightarrow HbO_2\) is \(-70\, kJ/mol\)
• \(\Delta G^{\circ}\) for \(Hb + CO \rightarrow HbCO\) is \(-80\, kJ/mol\)

This implies that hemoglobin binds more strongly with CO than with \(O_2\), making CO a potent toxin as it inhibits oxygen transport in blood by competing with oxygen for binding sites.

Equilibrium Reactions

Equilibrium reactions occur when reactants and products are present in concentrations that have no further tendency to change with time. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The concept of equilibrium is closely tied to the equilibrium constant \(K\), given by the equation:\[ \Delta G^{\circ} = -RT \ln K \]Where:

• \(R\) is the gas constant (8.314 J/molK)
• \(T\) is temperature in Kelvin
• \(K\) is the equilibrium constant

This formula allows us to calculate \(K\) using \(\Delta G^{\circ}\). Thus, it determines how far a reaction will proceed before reaching equilibrium. In the context of hemoglobin and CO, a large \(K\) value indicates an equilibrium that heavily favors the formation of the \(HbCO\) complex, affecting oxygen delivery throughout the body.

Carbon Monoxide Toxicity

Carbon monoxide (CO) is a harmful gas primarily because of its ability to bind strongly with hemoglobin (Hb), the molecule responsible for oxygen transport in the blood. When CO binds to hemoglobin, it forms carboxyhemoglobin (HbCO), which occupies the oxygen binding sites on the hemoglobin molecules. This high affinity is due to the more negative standard free-energy change for forming \(HbCO\) compared to \(HbO_2\). Since \(HbCO\) forms preferentially, very low concentrations of CO can significantly reduce the amount of oxygen transported in the blood. Symptoms of carbon monoxide toxicity can include:

• Headaches
• Dizziness
• Confusion
• Even death at higher concentrations

Understanding chemical equilibrium and the free energy differences helps us grasp why CO is such a powerful and dangerous poison, despite its relatively simple molecular structure.