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In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

Short Answer

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a) The amount of carbon dioxide increases, b) The amount of carbon dioxide is unaffected, c) The amount of carbon dioxide is unaffected, d) The amount of carbon dioxide decreases.

Step by step solution

01

Response to Anion Increase

(a) If the amount of bicarbonate anion (\(\mathrm{HCO}_{3}^{-}\)) is increased, according to Le Chatelier's principle, the reaction would be expected to shift in the direction that would reestablish equilibrium, thus in the direction where \(\mathrm{HCO}_{3}^{-}\) is used up, i.e. towards the formation of \(\mathrm{CO}_{2}\). Thus the amount of carbon dioxide would increase.
02

Response to Pressure Increase

(b) An increase in the pressure of \(\mathrm{CO}_{2}\) would push the reaction towards the direction that decreases the pressure. Since this reaction has equal number of moles of gas on either side of the equation, increasing the pressure would have no effect on the direction of the equilibrium. So, the amount of carbon dioxide would not change.
03

Response to Increased Enzyme

(c) For an enzyme catalyzed reaction, increasing the amount of the enzyme i.e., carbonic anhydrase, doesn't change the position of the equilibrium but increases the rate at which it is established. Thus, the increase in enzymatic content doesn't alter the amount of carbon dioxide.
04

Response to Temperature Decrease

(d) Since it's stated the reaction is endothermic, it absorbs heat. Thus, decreasing the temperature would drive the reaction to the direction that absorbs heat to increase the temperature, i.e., towards the formation of bicarbonate ions \(\mathrm{HCO}_{3}^{-}\) and thus reducing the amount of \(\mathrm{CO}_{2}\) .

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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 like a rule of thumb for predicting how a chemical system at equilibrium responds to external changes. It states that when a system at equilibrium experiences a change in concentration, temperature, volume, or pressure, the equilibrium will shift to counteract the inflicted change and re-establish a new equilibrium state.

This principle is crucial to understanding many chemical processes, including those relevant to biology, such as the carbonic anhydrase reaction in the human body. For instance, when the concentration of bicarbonate anion is increased, as per Le Chatelier's principle, the system will adjust to reduce the concentration of this anion by converting it to carbon dioxide and water. This reaction highlights how cells can regulate the concentrations of various species dynamically to maintain homeostasis.
Endothermic Reactions
Endothermic reactions are processes that absorb energy from their surroundings, typically in the form of heat. This means that such reactions require an input of energy to proceed. The carbonic anhydrase reaction involved in the interconversion of carbon dioxide is endothermic, so it absorbs heat.

This impacts the equilibrium position whenever there's a temperature change. If the temperature is decreased, the system responds by shifting the equilibrium towards the endothermic process — in this case, the formation of bicarbonate ions — to try to absorb more heat and restore balance. Understanding these reactions is important not just in biochemistry, but also in industrial processes where temperature control is critical.
Chemical Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction occur at the same rate, leading to the concentrations of the reactants and products remaining constant over time. It doesn't mean the reactants and products are in equal concentration, but rather that their ratio doesn't change.

In the carbonic anhydrase reaction, equilibrium can be affected by changes in bicarbonate anion levels, pressure, and carbonic anhydrase amounts. Le Chatelier’s principle helps predict the shifts in equilibrium with these changes; however, altering the amount of enzyme—while it speeds up the rate at which equilibrium is reached—doesn't shift the equilibrium itself.
Enzyme Catalysis
Enzymes are biological catalysts that accelerate chemical reactions in living organisms without being consumed in the process. Enzyme catalysis is vital for sustaining life because it increases the rate of reaction to a biologically useful speed. Carbonic anhydrase, for example, is an enzyme that catalyzes the rapid interconversion of carbon dioxide and bicarbonate anion.

Increasing the amount of carbonic anhydrase in a system will speed up the attainment of equilibrium but doesn't actually change the position of equilibrium. This concept is fundamental in the development of pharmaceuticals and understanding metabolic pathways, where the efficiency of enzyme-catalyzed reactions is paramount.

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Most popular questions from this chapter

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

Write an equilibrium constant, \(K_{\mathrm{p}},\) for the formation from its gaseous elements of (a) 1 mol \(\mathrm{NOCl}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{ClNO}_{2}(\mathrm{g}) ;\) (c) \(1 \mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) ;\) (d) \(1 \mathrm{mol}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Châtelier's principle; (d) effect of a catalyst on equilibrium.

If the volume of an equilibrium mixture of \(\mathrm{N}_{2}(\mathrm{g}), \mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{NH}_{3}(\mathrm{g})\) is reduced by doubling the pressure, will \(P_{\mathrm{N}_{2}}\) have increased, decreased, or remained the same when equilibrium is re established? Explain. $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$$

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