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Chapter 15: Problem 87

Photosynthesis can be represented by $$ \begin{array}{r} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ}=2801 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of \(\mathrm{CO}_{2}\) is increased, (b) \(\mathrm{O}_{2}\) is removed from the mixture, (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (sucrose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased, (g) more sunlight falls on the plants.

Short Answer

Expert verified
In summary, increasing \( \mathrm{CO}_{2} \), removing \( \mathrm{O}_{2} \), removing \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), adding water, decreasing temperature and more sunlight will all shift the equilibrium to the right, leading to increased photosynthesis. Adding a catalyst, however, will not shift the equilibrium but will only speed up the rate at which equilibrium is achieved.

Step by step solution

01

(a) Partial pressure of \( \mathrm{CO}_{2} \) is increased

According to Le Chatelier's principle, an increase in the concentration of reactants will shift the equilibrium to the right in an effort to use up the added reactant. Therefore, increasing the partial pressure of \( \mathrm{CO}_{2} \) will lead to more \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \) in an attempt to maintain equilibrium.
02

(b) \( \mathrm{O}_{2} \) is removed from the mixture

Based on Le Chatelier's principle, removing a product will cause the equilibrium to shift to the right to replace the removed product. Thus, removing \( \mathrm{O}_{2} \) from the mixture will result in an increase in the production of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \).
03

(c) \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) (sucrose) is removed from the mixture

According to Le Chatelier's principle, removing a product will shift the equilibrium to the right to replace the removed product. Thus, removing \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) from the mixture will result in the production of more \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \).
04

(d) More water is added

Again referring to Le Chatelier's principle, adding a reactant will cause the equilibrium to shift to the right to consume some of the added reactant. Therefore, adding more water will result in more \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \).
05

(e) A catalyst is added

Catalysts affect the rate of reaction but not the position of equilibrium. Therefore, adding a catalyst does not change the equilibrium concentrations of the reactants or products. It only allows the system to reach equilibrium faster.
06

(f) Temperature is decreased

As the reaction is endothermic (as indicated by the positive ΔH), lowering the temperature effectively removes heat from the system. According to Le Chatelier's principle, the system will try to counteract this change by producing more heat. Thus, the reaction will shift to the right where it is exothermic, producing more \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \).
07

(g) More sunlight falls on the plants

The photosynthesis reaction is light-driven. More light allows more photosynthesis to take place, thereby making more reactant used up. So, the system shifts to the right, making more \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) and \( \mathrm{O}_{2} \).

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Most popular questions from this chapter

Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ If the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at \(2200^{\circ} \mathrm{C}\), what is \(K_{P} ?\)

Consider the statement: The equilibrium constant of a reacting mixture of solid \(\mathrm{NH}_{4} \mathrm{Cl}\) and gaseous \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\) is \(0.316 .\) List three important pieces of information that are missing from this statement.
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