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Hemoglobin (Hb) can form a complex with both \(\mathrm{O}_{2}\) and \(\mathrm{CO}\). For the reaction $$ \mathrm{HbO}_{2}(\mathrm{aq})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{HbCO}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g}) $$ at body temperature, \(K\) is about \(200 .\) If the ratio \(|\mathrm{HbCO}| /\left|\mathrm{HbO}_{2}\right|\) comes close to \(1,\) death is probable. What partial pressure of CO in the air is likely to be fatal? Assume the partial pressure of \(\mathbf{O}_{2}\) is \(0.20 \mathrm{atm}\)

Short Answer

Expert verified
Fatal CO pressure is 0.001 atm.

Step by step solution

01

Identify Constants

Given, the equilibrium constant \( K \) for the reaction is 200, and the partial pressure of \( \mathrm{O}_2 \) is 0.20 atm.
02

Set Up the Reaction Quotient

The equilibrium constant \( K \) for the reaction is defined as \[K = \frac{[\mathrm{HbCO}] \cdot P_{\mathrm{O}_2}}{[\mathrm{HbO}_2] \cdot P_{\mathrm{CO}}}\]where \([\mathrm{HbCO}]\) and \([\mathrm{HbO}_2]\) denote the concentrations of \( \mathrm{HbCO} \) and \( \mathrm{HbO}_2 \) respectively, and \( P_{\mathrm{O}_2} \) and \( P_{\mathrm{CO}} \) are the partial pressures of \( \mathrm{O}_2 \) and \( \mathrm{CO} \).
03

Analyze Ratios

The problem asks for the partial pressure of CO when \( \left|\mathrm{HbCO}\right| / \left|\mathrm{HbO}_2\right| \approx 1 \). This implies that \([\mathrm{HbCO}] \approx [\mathrm{HbO}_2]\).
04

Substitute and Solve for CO Pressure

Substitute \( [\mathrm{HbCO}] = [\mathrm{HbO}_2] \) into the equilibrium expression:\[K = \frac{[\mathrm{HbCO}] \cdot 0.20}{[\mathrm{HbO}_2] \cdot P_{\mathrm{CO}}} = \frac{[\mathrm{HbO}_2] \cdot 0.20}{[\mathrm{HbO}_2] \cdot P_{\mathrm{CO}}}\]Cancel \([\mathrm{HbO}_2]\) and solve for \( P_{\mathrm{CO}} \):\[200 = \frac{0.20}{P_{\mathrm{CO}}} \Rightarrow P_{\mathrm{CO}} = \frac{0.20}{200} = 0.001 \, \text{atm}\]
05

Conclusion

The partial pressure of CO in the air that is likely to be fatal, assuming \( \left|\mathrm{HbCO}\right| / \left|\mathrm{HbO}_2\right| \approx 1 \), is 0.001 atm.

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Key Concepts

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

Equilibrium constant
In chemistry, the equilibrium constant, denoted as \( K \), is a crucial concept when dealing with reversible reactions. It reflects the ratio at equilibrium between the reactants and products of a reaction. This constant shows how far a reaction proceeds before reaching a state where the rates of the forward and reverse reactions are equal.
In the context of hemoglobin reactions, the equilibrium constant can determine how readily oxygen (\( \mathrm{O}_2 \)) or carbon monoxide (\( \mathrm{CO} \)) binds to hemoglobin. The existence of a large equilibrium constant, such as 200 in the case of \( \mathrm{HbO}_2 + \mathrm{CO} \rightleftharpoons \mathrm{HbCO} + \mathrm{O}_2 \), indicates that the formation of carboxyhemoglobin (\( \mathrm{HbCO} \)) is favored over the oxygen-bound form of hemoglobin (\( \mathrm{HbO}_2 \)) at equilibrium. This strong affinity for CO is a primary reason why carbon monoxide is such a potent competitor for oxygen in the bloodstream.
The use of the equilibrium expression, \[ K = \frac{[\mathrm{HbCO}] \cdot P_{\mathrm{O}_2}}{[\mathrm{HbO}_2] \cdot P_{\mathrm{CO}}} \], allows us to explore the conditions under which hemoglobin preferentially binds to CO instead of \( \mathrm{O}_2 \), helping in identifying hazardous exposure levels of CO in the air.
Partial pressure
Partial pressure is the pressure that a gas in a mixture of gases contributes to the total pressure. It is an important concept when dealing with gases in solutions, like oxygen and carbon monoxide in blood. The partial pressure of a gas indicates its concentration, influencing how gases interact with liquids such as blood.
In the context of breathing, humans inhale a mixture of gases where nitrogen, oxygen, and trace gases coexist. Since oxygen is crucial for cellular functions, its partial pressure in the blood is critical to ensuring that tissues receive adequate oxygen. It is commonly around 0.20 atm under standard conditions in the air.

When examining carbon monoxide presence, it is vital to analyze its partial pressure. The partial pressure of CO directly impacts how much it can bind to hemoglobin. The fatal partial pressure of CO, as calculated, turns out to be a minute \( 0.001 \text{ atm} \). This signifies how even a small presence of CO in the atmosphere can significantly disturb the transport of oxygen in the blood by forming \( \mathrm{HbCO} \), reducing tissue oxygen availability.
Carbon monoxide poisoning
Carbon monoxide poisoning occurs when carbon monoxide gas interferes with the body's ability to distribute oxygen. It is a serious condition that can lead to death if not addressed promptly. This happens because carbon monoxide binds to hemoglobin much more readily than oxygen, forming carboxyhemoglobin (\( \mathrm{HbCO} \)).
When \( \mathrm{HbCO} \) levels rise, they hinder the blood's capacity to carry oxygen (\( \mathrm{O}_2 \)) to tissues and organs. This can be especially dangerous as CO is colorless, odorless, and tasteless, often going unnoticed until symptoms manifest. These symptoms may include headache, dizziness, and in severe cases, loss of consciousness.
  • Even at extremely low concentrations, such as 0.001 atm, CO can lead to life-threatening conditions since \( \mathrm{HbCO} \) formation is favored.
  • This high affinity forms part of why carbon monoxide is such a dangerous compound, as its presence in the air can quickly affect populations over a short span, often in enclosed spaces like homes with poor ventilation.
Preventing CO poisoning requires good ventilation, installing CO detectors, and ensuring that combustion appliances are well-kept to minimize accidental CO buildup. Awareness of CO's effects on hemoglobin chemistry is also crucial for addressing and preventing carbon monoxide poisoning effectively.

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Most popular questions from this chapter

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \mathrm{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K_{c}\) is 0.190 at \(73^{\circ} \mathrm{C}\). Suppose you place \(0.500 \mathrm{mol}\) of \(\mathrm{COBr}_{2}\) in a 2.00 -L. flask and heat it to \(73^{\circ} \mathrm{C}\) (see Study Question 17 ). After equilibrium has been achieved, you add an additional 2.00 mol of CO. (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of \(\mathrm{COBr}_{2}\) \(\mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) (c) How has the addition of CO affected the percentage of \(\mathrm{COBr}_{2}\) that decomposed?

The reaction $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ occurs at high temperatures. At \(700^{\circ} \mathrm{C},\) a 200.0 - \(\mathrm{L}\) tank contains 1.0 mol of \(\mathrm{CO}, 0.20\) mol of \(\mathrm{CO}_{2}\), and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K_{c}\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K_{c}\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the 200.0 -L tank are 1.0 mol of \(\mathrm{CO}, 0.20\) mol of \(\mathrm{CO}_{2}\), and \(0.80 \mathrm{mol}\) of \(\mathrm{C}\) (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K_{c} ?\) Explain.

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ is an endothermic process. Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a flask in which this equilibrium exists, how is the equilibrium affected? What if some additional \(\mathrm{NH}_{3}\) is placed in the flask? What will happen to the pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{S}\) is removed from the flask?

Write equilibrium constant expressions for the following reactions. For gases, use either pressures or concentrations. (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) (b) \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) (d) \(\mathrm{NiO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Ni}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\)

Characterize each of the following as product- or reactant-favored at equilibrium. $$\begin{aligned} &\text { (a) } \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=1.2 \times 10^{45} \end{aligned}$$ $$\begin{aligned} &\text { (b) } \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=9.1 \times 10^{-41} \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=6.5 \times 10^{11} \end{aligned}$$

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