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The following reaction represents the binding of oxygen by the protein hemoglobin (Hb): $$\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{Hb}: \mathrm{O}_{2}(\mathrm{aq}) \quad \Delta H<0$$ Explain how each of the following affects the amount of \(\mathrm{Hb}: \mathrm{O}_{2}:\) (a) increasing the temperature; (b) decreasing the pressure of \(\mathrm{O}_{2} ;\) (c) increasing the amount of 6 hemoglobin.

Short Answer

Expert verified
Increasing the temperature or decreasing the pressure of \(\mathrm{O}_{2}\), would decrease the amount of \(\mathrm{Hb}: \mathrm{O}_{2}\), while increasing the amount of hemoglobin would increase the amount of \(\mathrm{Hb}: \mathrm{O}_{2}\). These conclusions are based on Le Chatelier's principle, which states that a system at equilibrium responds to disturbances in a way that minimizes the effect of the disturbance.

Step by step solution

01

Effect of Temperature

First, consider the effect of temperature. The given reaction is exothermic, which means it releases heat. According to Le Chatelier's principle, if an exothermic reaction is heated, the system responds by favouring the reverse reaction that absorbs the excess heat. Therefore, increasing the temperature would decrease the amount of \(\mathrm{Hb}: \mathrm{O}_{2}\).
02

Effect of Pressure

Next, consider the effect of pressure. According to Le Chatelier's principle, when a system at equilibrium is subjected to a change in pressure, the system shifts in the direction that counteracts the pressure change. Here, decreasing the pressure of oxygen would lead to a decrease in \(\mathrm{Hb}: \mathrm{O}_{2}\), as a lower pressure will lead the reaction to shift towards the side with more moles of gas, i.e., the left-hand side with unbound oxygen.
03

Effect of Hemoglobin Concentration

Finally, consider the effect of the amount of hemoglobin. When the amount of a reactant is increased, according to Le Chatelier's principle, the equilibrium shifts to the side to counter the increase. Hence increasing the amount of hemoglobin would increase the amount of \(\mathrm{Hb}: \mathrm{O}_{2}\), as the reaction would shift to the right to accommodate for the increased hemoglobin concentration.

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Key Concepts

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

Hemoglobin Oxygen Binding
Hemoglobin (Hb) is a protein in red blood cells responsible for transporting oxygen throughout the body. Oxygen binds to hemoglobin in a reversible reaction, forming oxyhemoglobin (\text{HbO}\(_2\)). Understanding this binding process is crucial for grasping how changes in environmental conditions could impact oxygen transport.

When you breathe in, oxygen molecules (\text{O}\(_2\)) enter your bloodstream and bind to hemoglobin. This binding is an equilibrium process: \text{Hb} + \text{O}\(_2\) \rightleftharpoons \text{HbO}\(_2\). Various factors can shift this equilibrium, affecting how well hemoglobin can bind to oxygen and thus carry it to tissues that need it.
Exothermic Reactions
Exothermic reactions are chemical processes in which energy is released into the surroundings, usually in the form of heat. This release of energy is often accompanied by an increase in temperature of the reaction mixture.

The reaction between hemoglobin and oxygen is exothermic. When oxyhemoglobin forms, it releases energy, which is a fundamental concept for predicting how changes in temperature can affect this reaction. According to the principle of energetics, the amount of oxyhemoglobin will be influenced by temperatures as they can affect the rate and direction of this heat-releasing reaction.
Equilibrium Shift
An equilibrium shift occurs when a system that is in a state of balance is disturbed by a change in conditions such as concentration, temperature, or pressure. According to Le Chatelier's principle, the system will adjust to partially counteract the imposed change, thereby establishing a new equilibrium position.

For the hemoglobin-oxygen reaction, shifts in equilibrium can alter the amount of \text{HbO}\(_2\). Understanding the direction of the shift is pivotal for predicting the reaction's response to different stressors, which, in turn, can impact how hemoglobin carries oxygen.
Effect of Temperature on Equilibrium
Temperature changes have a profound impact on equilibrium systems. For exothermic reactions, such as the binding of oxygen to hemoglobin, increasing the temperature typically favors the reverse reaction in accordance with Le Chatelier's principle. This means that as temperature rises, the equilibrium shifts to the left, reducing the formation of oxyhemoglobin (\text{HbO}\(_2\)).

This is due to the system absorbing excess heat by favoring the endothermic process—the dissociation of oxygen from hemoglobin—mitigating the rise in temperature.
Effect of Pressure on Equilibrium
Pressure changes can also shift chemical equilibria, especially for reactions involving gases. For our hemoglobin example, decreasing the pressure of oxygen leads the equilibrium to favor the side with more gas molecules. This is because a reduction in pressure causes the system to counteract the change by shifting the balance toward the production of more gas molecules—in this case, dissociating oxygen from hemoglobin.

Thus, a lower oxygen pressure results in fewer \text{HbO}\(_2\) molecules, as the equilibrium moves left toward unbound hemoglobin and oxygen.
Hemoglobin Concentration Effect
The concentration of hemoglobin in the blood also plays a role in the oxygen-binding equilibrium. Increasing the amount of hemoglobin prompts the equilibrium to shift right, towards the formation of more oxyhemoglobin (\text{HbO}\(_2\)) molecules, according to Le Chatelier's principle.

This occurs as the system seeks to balance the increased concentration of one of the reactants, highlighting the dynamic nature of equilibrium and the continuous adaptation of chemical systems to maintain balance.

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Most popular questions from this chapter

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.

An equilibrium mixture at 1000 K contains an equilibrium mixter \(0.276\ \mathrm{mol}\ \mathrm{H}_{2}, 0.276 \mathrm{mol}\ \mathrm{CO}_{2}, 0.224\ \mathrm{mol}\ \mathrm{CO},\) and \(0.224\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{O}\) $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Show that for this reaction, \(K_{\mathrm{c}}\) is independent of the reaction volume, \(V\) (b) Determine the value of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\)

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

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