Question

Carbon monoxide poisoning results when carbon monoxide replaces oxygen bound to hemoglobin. The oxygenated form of hemoglobin, \(\mathrm{Hb} \cdot \mathrm{O}_{2}\) carries \(\mathrm{O}_{2}\) to the lungs. $$ \mathrm{Hb} \cdot \mathrm{O}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Hb} \cdot \mathrm{CO}(a q)+\mathrm{O}_{2}(g) $$ At \(98.6^{\circ} \mathrm{F}\left(37^{\circ} \mathrm{C}\right), \Delta G^{\circ}\) for the reaction is about \(-14 \mathrm{~kJ}\). What is the ratio of \(\left[\mathrm{Hb} \cdot \mathrm{O}_{2}\right]\) to \([\mathrm{Hb} \cdot \mathrm{CO}]\) when the pressure of \(\mathrm{CO}\) is the same as that of \(\mathrm{O}_{2}\) ?

Short answer

Answer: The ratio of Hb·O2 to Hb·CO is approximately 55.4 when the partial pressures of CO and O2 are equal.

Step-by-step solution

Calculate the equilibrium constant K using Gibbs free energy and the equation ΔG = -RT ln(K)

Given that ΔG° = -14 kJ at 37 °C, we can determine the equilibrium constant K using the equation ΔG = -RT ln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (37°C = 310.15 K). First, convert ΔG to J/mol: ΔG° = -14,000 J/mol. Then, plug the values into the equation: -14,000 J/mol = -(8.314 J/mol·K)(310.15 K) ln(K) Now, solve for K: $$ K = e^{\frac{14000}{(8.314)(310.15)}} \approx 3068 $$

Set up an ICE table

Now that we have the equilibrium constant K, set up an ICE (Initial-Change-Equilibrium) table for the reaction to find the equilibrium concentrations of Hb·O2 and Hb·CO: | | Hb·O2 | + | CO | <-> | Hb·CO | + | O2 | |----|-------|-------|----|-------|-------|-------|----| | I | x | | p | | 0 | | p | | C | -y | | -y | | y | | y | | E | x - y | | p-y | | y | | p+y | Here, the initial concentration of Hb·O2 is represented as x and that of CO and O2 are both represented as p (the same partial pressure as oxygen).

Write the expression for K and substitute the equilibrium concentrations

Next, write the expression for K in terms of the equilibrium concentrations: $$ K = \frac{[\mathrm{Hb} \cdot \mathrm{CO}] [\mathrm{O}_{2}]}{[\mathrm{Hb} \cdot \mathrm{O}_{2}] [\mathrm{CO}]} $$ Now substitute the values from the ICE table into the K expression: $$ 3068 = \frac{(y)(p + y)}{(x - y)(p - y)} $$

Solve for the ratio of Hb·O2 to Hb·CO

We need to find the ratio of Hb·O2 to Hb·CO at equilibrium, i.e. \(\frac{(x - y)}{y}\). However, the equation from Step 3 still has two unknown variables, x and y. We can rewrite the ratio expression as \(\frac{1}{\frac{y}{x - y}}\). Since we are given the partial pressures of oxygen and CO are the same, we can assume x is roughly equal to p. Hence, we can rewrite the K expression again using this assumption: $$ 3068 = \frac{(y)(p + y)}{(p - y)(p - y)} $$ Now, make y the subject of the equation: $$ y = \frac{p}{\sqrt{3068} + 1} $$ Next, substitute this expression for y back into the ratio expression: $$ \text{ratio} = \frac{(x - y)}{y} = \frac{p - \frac{p}{\sqrt{3068} + 1}}{\frac{p}{\sqrt{3068} + 1}} = \sqrt{3068} = \boxed{55.4} $$ So the ratio of Hb·O2 to Hb·CO is approximately 55.4 when the partial pressures of CO and O2 are equal.

Key concepts

Gibbs Free Energy and Chemical Equilibrium

Gibbs free energy, denoted as \( \Delta G \), is a thermodynamic quantity used to predict the direction of chemical reactions and the conditions under which they are spontaneous. A spontaneous process is one that occurs without the need for energy input from the surroundings; it has the potential to proceed on its own under the given conditions.

Gibbs free energy is related to the equilibrium constant, \( K \), through the relationship \( \Delta G = -RT \ln(K) \), where \( R \) is the universal gas constant and \( T \) the temperature in Kelvin. When \( \Delta G \) is negative, the process is spontaneous in the forward direction, leading to products, and \( K > 1 \), indicating products are favored at equilibrium. Conversely, when \( \Delta G \)> 0, the reverse process is spontaneous, and \( K < 1 \), suggesting reactants are favored.

In the context of our textbook exercise, the negative \( \Delta G \) indicates that the binding of carbon monoxide (CO) to hemoglobin is spontaneous under standard conditions, and the equilibrium constant \( K \) is much greater than 1, signifying a strong preference for the formation of carboxyhemoglobin (Hb·CO) over oxyhemoglobin (Hb·O2).

Equilibrium Constant and Reaction Quotient

The equilibrium constant, \( K \), is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. It is a measure of the extent of the reaction and is determined only by the temperature.

For the reaction given in our exercise, the equilibrium constant is defined as \[ K = \frac{[\mathrm{Hb} \cdot \mathrm{CO}] [\mathrm{O}_{2}]}{[\mathrm{Hb} \cdot \mathrm{O}_{2}] [\mathrm{CO}]} \]. The large value of \( K \) approximated to 3068 signifies that, at equilibrium, the concentration of carboxyhemoglobin is much higher than oxyhemoglobin, which aligns with the clinical aspect of carbon monoxide poisoning, where CO preferentially binds to hemoglobin over oxygen.

It is important to understand that the equilibrium constant is connected to the concentrations or pressures of reactants and products at equilibrium; it does not provide information on the speed at which equilibrium is reached. Students sometimes confuse this with the reaction rate, which is a separate concept.

Using ICE Tables to Solve Equilibrium Problems

The ICE table is a systematic method used to organize information and solve for unknowns in equilibrium problems. ICE stands for Initial, Change, and Equilibrium concentrations or pressures. This method requires setting up a table that represents the initial amounts of reactants and products, the changes that occur as the reaction progresses to equilibrium, and the final equilibrium concentrations or pressures.

In our exercise, the initial concentration of oxyhemoglobin (Hb·O2) is represented by \( x \) and the initial pressures of CO and O2 are both \( p \) due to the assumption that they are equal. As the system reaches equilibrium, these concentrations change by an amount \( y \). At equilibrium, the concentrations are \( x - y \) for Hb·O2 and \( y \) for Hb·CO. It is crucial to note that the 'C' in the ICE table corresponds to stoichiometric ratios; however, in this simple exchange reaction, the ratio is 1:1 for all components.

The final step involves plugging the equilibrium concentrations into the equilibrium expression and solving for the unknown. In our scenario, the ratio of \( [\mathrm{Hb} \cdot \mathrm{O}_{2}] \) to \( [\mathrm{Hb} \cdot \mathrm{CO}] \) when CO and O2 pressures are equal is a critical value representing the preference of hemoglobin for CO or O2. This application of the ICE table highlights its importance in quantitative analysis of equilibrium scenarios.