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Photosynthesis can be represented by: $$\begin{aligned} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=2801 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ 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) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased.

Short Answer

Expert verified
(a) Right; (b) Right; (c) Right; (d) Right; (e) No change; (f) Left.

Step by step solution

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01

Analyze the Reaction

The chemical equation for photosynthesis is \(6 \text{CO}_2 (g) + 6 \text{H}_2\text{O} (l) \rightleftharpoons \text{C}_6\text{H}_{12}\text{O}_6 (s) + 6\text{O}_2 (g)\). This is an endothermic reaction with \(\Delta H^\circ = 2801 \text{kJ/mol}\), meaning it absorbs heat.
02

Effect of Increased \(\text{CO}_2\) Pressure

Increasing the partial pressure of \(\text{CO}_2\) increases the concentration of reactants, causing the equilibrium to shift to the right to produce more \(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\), according to Le Chatelier's principle.
03

Effect of Removing \(\text{O}_2\)

Removing \(\text{O}_2\) decreases the products' concentration. The equilibrium shifts to the right to produce more \(\text{O}_2\) and \(\text{C}_6\text{H}_{12}\text{O}_6\) to counteract this change.
04

Effect of Removing \(\text{C}_6\text{H}_{12}\text{O}_6\)

Removing \(\text{C}_6\text{H}_{12}\text{O}_6\) shifts the equilibrium to the right, as the system attempts to replenish the removed product and maintain equilibrium.
05

Effect of Adding More Water

Increasing the concentration of water will shift the equilibrium to the right, favoring the formation of more products \(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\) according to Le Chatelier's principle.
06

Effect of Adding a Catalyst

A catalyst speeds up both the forward and backward reactions equally but does not shift the equilibrium position. Therefore, equilibrium remains unaffected.
07

Effect of Decreasing Temperature

Since the reaction is endothermic, decreasing the temperature will shift the equilibrium to the left, favoring the reactants and limiting the production of \(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\).

Key Concepts

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

Photosynthesis
Photosynthesis is a process by which plants, algae, and some bacteria convert light energy, usually from the sun, into chemical energy in the form of glucose. During this process, carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)) are used as reactants to produce glucose (\(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\)) and oxygen (\(\mathrm{O}_2\)).
This process can be represented by the chemical equation:
  • 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)
Photosynthesis is crucial because it is the primary source of organic matter for virtually all organisms, and it also contributes to oxygen production in the environment.
Le Chatelier's Principle
Le Chatelier's Principle helps in predicting the response of a chemical equilibrium to changes in concentration, pressure, or temperature. When a system at equilibrium experiences a change or stress, it will adjust in a way that counteracts the change.
Let's apply this principle to photosynthesis:
  • If the concentration of \(\mathrm{CO}_2\) increases, the equilibrium will shift to the right to produce more glucose and \(\mathrm{O}_2\).
  • Remove \(\mathrm{O}_2\) and the system shifts right to produce more \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) and restore balance.
  • Reducing glucose prompts the reaction to move right, creating more \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\).
  • Addition of water results in the equilibrium also favoring product formation.
    In each scenario, the system naturally adjusts to maintain equilibrium.
Endothermic Reaction
Photosynthesis is an endothermic reaction, meaning it absorbs energy from its surroundings. This energy, in the form of sunlight, is essential to drive the production of glucose and oxygen. The reaction's enthalpy change (\(\Delta H^{ oday}\)) is positive, signifying this absorption of heat.
An endothermic reaction poses special considerations:
  • When the temperature decreases, the equilibrium tends to shift toward the reactants to capture more heat, thereby reducing the formation of \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\).
  • On the contrary, raising the temperature generally drives the reaction to the products' side, enhancing the yield.
Understanding this concept is crucial for predicting the outcomes in various environmental conditions.
Catalyst Effect
Catalysts are substances that speed up chemical reactions without being consumed in the process. In photosynthesis, adding a catalyst would enhance the rate of both the forward and reverse reactions.
Despite their role, catalysts do not alter the position of equilibrium itself; they only affect how fast equilibrium is achieved. Here's how they work:
  • Their primary function is to lower the activation energy required for reactions, allowing the process to proceed quicker.
  • In biological systems, enzymes often act as catalysts, facilitating the conversion of reactants to products efficiently.
Even though catalysts expedite reactions, they cannot shift equilibrium toward the reactants or products. They simply help the system reach its equilibrium more rapidly.
Temperature Effect
Temperature plays a significant role in chemical equilibria, impacting reaction pathways and outcomes profoundly. In the context of photosynthesis, which is endothermic, temperature changes have notable effects:
  • When temperature decreases, equilibrium shifts toward the reactants (\(\mathrm{CO}_2, \mathrm{H}_2\mathrm{O}\)), meaning less glucose and oxygen are formed.
  • Conversely, increasing the temperature shifts the equilibrium toward products, encouraging more glucose production.
Understanding the temperature effect is crucial in discussions about plant growth, agricultural productivity, and adapting to climate changes.
It showcases the need to consider environmental and external conditions, especially how plants might respond to seasonal and climate variations.

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Most popular questions from this chapter

Write the expressions for the equilibrium constants \(K_{P}\) of the following thermal decomposition reactions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftarrows 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\).

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l)\) \(\Longrightarrow \mathrm{Hg}(g)\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor (in \(\mathrm{mg} / \mathrm{m}^{3}\) ). Does this concentration exceed the safety limit of \(0.05 \mathrm{mg} / \mathrm{m}^{3}\) ? (Ignore the volume of furniture and other objects in the laboratory.)

In the uncatalyzed reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) $$ the pressure of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377\) atm and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm}\) at \(100^{\circ} \mathrm{C}\). What would happen to these pressures if a catalyst were added to the mixture?

Industrial production of ammonia from hydrogen and nitrogen gases is done using the Haber process. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g) \quad \Delta H^{\circ}=-92.6 \mathrm{~kJ} / \mathrm{mol}\) Based on your knowledge of the principles of equilibrium, what would the optimal temperature and pressure conditions be for production of ammonia on a large scale? Are the same conditions also optimal from the standpoint of kinetics? Explain.

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 M,\left[\mathrm{~N}_{2}\right]=0.11 M,\) and \(\left[\mathrm{H}_{2}\right]=1.91 M\) Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftarrows \mathrm{NH}_{3}(g)\)

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