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The reaction between hemoglobin, Hb, and oxygen, \(\mathrm{O}_{2},\) in red blood cells is responsible for transporting \(\mathrm{O}_{2}\) to body tissues. This process can be represented by the following equilibrium reaction: \(\mathrm{Hb}(a q)+\mathrm{O}_{2}(g) \rightleftarrows \mathrm{HbO}_{2}(a q)\) What will happen to the concentration of oxygenated hemoglobin, HbO, at high altitude, where the pressure of oxygen is 0.1 atm instead of 0.2 atm, as it is a at sea level?

Short Answer

Expert verified
The concentration of oxygenated hemoglobin (HbO2) will decrease at high altitude.

Step by step solution

01

Understand the Equilibrium Reaction

The equilibrium reaction for hemoglobin and oxygen is given by: \[\mathrm{Hb}(aq) + \mathrm{O}_{2}(g) \rightleftarrows \mathrm{HbO}_{2}(aq)\] This equation shows that hemoglobin (Hb) reacts with oxygen (O2) to form oxygenated hemoglobin (HbO2).
02

Le Chatelier's Principle

Le Chatelier's Principle states that if a stress is applied to a system at equilibrium, the system shifts in a direction to counteract that stress and restore equilibrium.
03

Identify the Change in Oxygen Pressure

At high altitude, the pressure of oxygen decreases from 0.2 atm (at sea level) to 0.1 atm.
04

Apply Le Chatelier's Principle

With the decrease in oxygen pressure, the equilibrium will shift to counteract this decrease. According to Le Chatelier's Principle, the equilibrium will shift to the left to increase the amount of O2.
05

Determine the Effect on HbO2 Concentration

When the equilibrium shifts to the left, the concentration of oxygenated hemoglobin (HbO2) will decrease because less O2 is available to react with Hb.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is fundamental in understanding how systems respond to changes. It states if a change occurs in the conditions of a system at equilibrium, the system will adjust to counteract that change and re-establish equilibrium. For our context:
  • An increase in reactants will generally push the reaction towards the products.
  • An increase in products will push the reaction towards the reactants.
  • Changes in pressure and temperature also affect the direction of the shift.
This principle helps us predict the direction in which a reaction will move when external conditions are altered.
High Altitude Effects
High altitude poses unique challenges because of lower atmospheric pressure. At high altitudes, the partial pressure of oxygen \(\mathrm{O}_2\) decreases significantly when compared to sea level. Lower oxygen pressure has several physiological effects:
  • Reduced oxygen availability for uptake by hemoglobin.
  • Less oxygen delivered to tissues, leading to potential hypoxia.
Athletes training at high altitudes take advantage of this effect to boost their red blood cell count and increase oxygen carrying capacity upon returning to lower altitudes.
Oxygen Pressure
Oxygen pressure, specifically partial pressure, is crucial in the context of oxygen-hemoglobin equilibrium. At sea level, the partial pressure of oxygen is around 0.2 atm. This ensures effective binding of oxygen to hemoglobin forming oxygenated hemoglobin \(\mathrm{HbO}_2\). At higher altitudes, oxygen pressure drops to around 0.1 atm. Consequently, less oxygen is available to bind with hemoglobin. This principle directly influences how much oxygen is transported from the lungs to the tissues.
Equilibrium Shift
When the external conditions of a system in equilibrium change, an equilibrium shift occurs. Based on Le Chatelier's Principle:
An equilibrium shift happens to offset the stress applied to the system.
Under high altitude conditions, the decreased oxygen pressure causes the equilibrium of the reaction \(\mathrm{Hb}(aq) + \mathrm{O}_2(g) \rightleftarrows \mathrm{HbO}_2(aq)\) to shift to the left. This shift means that fewer \(\mathrm{HbO}_2\) molecules are formed, reducing the concentration of oxygenated hemoglobin. Thus, the body receives less oxygenated blood under these conditions.

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Most popular questions from this chapter

What is the relationship between \(K_{\mathrm{sp}}\) and the product of the ion concentrations in terms of determining whether a solution of those ions is saturated?

What relative conditions (reactant concentrations, pressure, and temperature) would favor a high equilibrium concentration of the underlined substance in each of the following equilibrium systems? a. $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \underline{\mathrm{CO}_{2}(g)}+167 \mathrm{kJ}$$ b. $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftarrows \underline{\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)}+42 \mathrm{kJ}$$ c. $$2 \mathrm{HI}(g)+12.6 \mathrm{kJ} \rightleftarrows \mathrm{H}_{2}(g)+ \underline{\mathrm{I}_{2}(g)}$$ d. $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(g)+ \underline{2 \mathrm{Cl}_{2}(g)}+113 \mathrm{kJ}$$ e. $$\mathrm{PCl}_{5}(g)+88 \mathrm{kJ} \rightleftarrows \mathrm{PCl}_{3}(g)+ \underline{\mathrm{Cl}_{2}(g)}$$

In general, which reaction is favored (forward or reverse) if the value of \(K\) at a specified temperature is a. very small? b. very large?

At \(25^{\circ} \mathrm{C}\) , the value of \(K\) is \(1.7 \times 10^{-13}\) for the following reaction. \(2 \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)\) It is determined that \(\left[\mathrm{N}_{2} \mathrm{O}\right]=0.0035 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.0027 \mathrm{mol} / \mathrm{L}\) . Using this information, what is the concentration of \(\mathrm{NO}(g)\) at equilibrium?

Predicting Outcomes When gasoline burns in an automobile engine, nitric oxide is formed from oxygen and nitrogen. Nitric oxide is a major air pollutant. High temperatures, such as those found in a combustion engine, are needed for the following reaction: \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) Kfor the reaction is 0.01 at \(2000^{\circ} \mathrm{C} .\) If 4.0 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) 0.1 \(\mathrm{mol}\) of \(\mathrm{O}_{2},\) and 0.08 \(\mathrm{mol}\) of NO are placed in a 1.0 -L vessel at \(2000^{\circ} \mathrm{C}\) , predict which reaction will be favored.

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