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

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

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) \(1 \mathrm{mol}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

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