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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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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium can help us predict how changes in photosynthesis can affect the balance of reactants and products. When a chemical reaction is at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This does not mean that concentrations of reactants and products are equal, but that their rates of production are balanced. In the case of photosynthesis, where carbon dioxide and water convert into glucose and oxygen, equilibrium can be disturbed by changes in conditions. For example:
  • If we increase the concentration of carbon dioxide, the equilibrium will shift to the right to produce more glucose and oxygen. This is because the system seeks to consume the excess carbon dioxide and restore balance.
  • Removing products like oxygen also shifts the reaction to right, as the system tries to replace lost products and maintain equilibrium.
Le Chatelier's Principle helps illustrate how changes in concentration will affect the direction of equilibrium, always aiming to restore the initial state of balance.
Photosynthesis
Photosynthesis is a crucial biological process where plants, algae, and some bacteria convert light energy into chemical energy, typically in the form of glucose. This reaction is given by \[6 ext{CO}_{2} + 6 ext{H}_{2} ext{O} \rightleftharpoons ext{C}_{6} ext{H}_{12} ext{O}_{6} + 6 ext{O}_{2}\]This process not only allows plants to create their own food, which is glucose, but also ensures that oxygen is available for other living organisms. The process of photosynthesis can be influenced by several factors:
  • Sunlight: It provides the energy needed for photosynthesis. An increase in sunlight boosts the process, leading to more glucose and oxygen production.
  • Temperature and Water: Optimal conditions are crucial. Adding water can push the reaction to the right, supporting more glucose formation.
Understanding the balance and influence of various factors on photosynthesis is vital in grasping how life on Earth thrives and maintains balance.
Endothermic Reaction
An endothermic reaction absorbs heat from its surroundings; photosynthesis is a perfect example of this. The role of energy cannot be overstated, as it is crucial for breaking bonds in reactants and forming new ones in products. Given by the positive value of \(\Delta H^{\circ} = 2801 \text{kJ/mol}\), this process necessitates energy input. This means photosynthesis requires external energy, such as light, to proceed, contrary to what exothermic reactions entail, which release energy.Factors in endothermic reactions, like:
  • Temperature: When the temperature decreases, the equilibrium may shift towards the formation of more products to generate heat, as per Le Chatelier's Principle.
  • Presence of sunlight: More sunlight equates to more energy available for the reaction, driving the equilibrium forward.
Understanding endothermic reactions such as photosynthesis can help us better manipulate conditions to maintain and improve life's balance. This understanding is crucial for everything from agricultural practices to managing ecosystems.

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Most popular questions from this chapter

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Consider the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If nitrosyl bromide, NOBr, is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is \(0.25 \mathrm{~atm},\) calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

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 this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

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