The photosynthesis reaction is \(6 \mathrm{CO}_{2}(\mathrm{~g})+\) \(6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{\mathrm{g}} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})\), and \(\Delta H^{\circ}=+2802 \mathrm{~kJ}\). Suppose that the reaction is at cquilibrium. State the effect (tend to shift toward the formation of reactants, tend to shift toward the formation of products, or have no effect) that each of the following changes will have on the equilibrium composition: (a) the partial pressure of \(\mathrm{O}_{2}\) is increased; (b) the system is compressed; (c) the amount of \(\mathrm{CO}_{2}\) is increased; (d) the temperature is increased; (c) some of the \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is removed; \((f)\) water is added; \((g)\) the partial pressure of \(\mathrm{CO}_{2}\) is decreased.

Short Answer

Expert verified
(a) Shifts toward reactants. (b) Shifts toward products. (c) Shifts toward products. (d) Shifts toward products. (e) Shifts toward products. (f) No significant effect. (g) Shifts toward reactants.

Step by step solution

01

- Analyze Reaction and Its Equilibrium

Understand that the photosynthesis reaction as given can be at equilibrium when the rates of the forward reaction (photosynthesis) and the reverse reaction (cellular respiration) are equal. The reaction is endothermic since \(\Delta H^\circ=+2802 \:kJ\). This means that the reaction requires heat to proceed.
02

- Le Chatelier's Principle

According to Le Chatelier's Principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This is done to re-establish equilibrium.
03

- The Effect of Increasing \(\mathrm{O}_2\) Partial Pressure

Increasing the partial pressure of \(\mathrm{O}_2\) will cause the equilibrium to shift towards the reactants to reduce the pressure of \(\mathrm{O}_2\) in accordance with Le Chatelier’s Principle.
04

- The Effect of Compressing the System

Compressing the system will increase the pressure overall. Since the number of gas moles is greater on the reactant side, the equilibrium will shift towards the products to reduce the pressure.
05

- The Effect of Increasing \(\mathrm{CO}_2\)

Increasing the amount of \(\mathrm{CO}_2\) will cause the equilibrium to shift towards the products as the system seeks to reduce the concentration of \(\mathrm{CO}_2\).
06

- The Effect of Increasing Temperature

Increasing the temperature will provide more heat to the system. Since the reaction is endothermic, the equilibrium will shift towards the products to absorb the extra heat.
07

- The Effect of Removing \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_{6}\)

Removing some of the \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_{6}\) will cause the equilibrium to shift towards the products to replace the \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_{6}\) that was removed.
08

- The Effect of Adding Water

Adding water does not change the equilibrium significantly because it is a pure liquid, and its concentration in terms of activities remains effectively constant.
09

- The Effect of Decreasing \(\mathrm{CO}_2\) Partial Pressure

Decreasing the partial pressure of \(\mathrm{CO}_2\) will cause the equilibrium to shift towards the reactants in order to increase the \(\mathrm{CO}_2\) concentration.

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Key Concepts

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

Le Chatelier's Principle
In the dance of chemical reactions, Le Chatelier's Principle serves as the choreographer, ensuring that all participants adapt gracefully to changes in their environment. To put it simply, whenever a system in equilibrium faces a change in pressure, temperature, or concentration, this principle predicts that the system will respond to counteract the disturbance and restore balance.

Imagine a seesaw equally balanced with kids on both ends. This equilibrium is disrupted when a new child joins one side. Le Chatelier's Principle is like the invisible hand that adjusts the balance by adding or removing weight from either side of the seesaw. In our photosynthesis reaction, when external conditions like gas pressure or temperature change, the reaction swings in the direction that minimizes those changes, ensuring that equilibrium is maintained. Keeping this principle in mind will be pivotal as we explore further on how various adjustments affect the delicate balance of the photosynthesis reaction.
Chemical Equilibrium
Chemical equilibrium is a state of stability in the chemical world, much like a perfectly calm sea, where forward and reverse reactions occur at exactly the same rate. This doesn't mean the reactions have stopped; they're just perfectly synced, like dancers moving in harmony.

The photosynthesis reaction reaches this state when the production of glucose and oxygen equals the rate at which they are consumed. It's crucial to understand that equilibrium does not imply equal concentrations of reactants and products. It's more about the rate at which they convert into each other reaching a balance. This sets the stage for exploring how changes affect this precise harmony without changing the overall amount of substances involved.
Endothermic Reactions
These are the reactions that thrive on warmth, akin to sun-loving plants. Endothermic reactions require heat to proceed, so they soak up energy from their surroundings much like a sponge absorbs water.

In the context of our plant-powered photosynthesis, it's an endothermic affair, with a positive heat change, \( \Delta H^\circ=+2802 \:kJ \). Essentially, this reaction relies on the sun's heat to convert carbon dioxide and water into glucose and oxygen. Since it's absorbing heat, an increase in temperature can cause the equilibrium to lean towards producing more glucose and oxygen, essentially saying 'thank you' for the extra heat by moving forward.
Equilibrium Shifts
The concept of equilibrium shifts is like adjusting the sails of a boat to catch the wind better. It's all about reaction adjustments when the equilibrium balance is nudged by external factors. These shifts can occur in two main ways: towards the reactants or the products.

In our study of photosynthesis, when you increase the presence of one of the reactants like carbon dioxide, the reaction compensates by producing more products. Conversely, increasing oxygen (a product) causes the reaction to favor the reactants to use up the excess oxygen. It's a balancing act, where changes in pressure, temperature, and concentration lead the reaction to find a new point of equilibrium, keeping the system stable and balanced.

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Most popular questions from this chapter

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

\( \mathrm{Ar} 500^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g}) .\) Calculate the value of \(K_{\varepsilon}\) at \(500^{\circ} \mathrm{C}\) for the following reactions. (a) \(\frac{1}{6} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{NH}_{3}(g)=\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g)\) (c) \(4 \mathrm{~N}_{2}(\mathrm{~g})+12 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 8 \mathrm{NH}_{3}(\mathrm{~g})\)

The equilibrium constant for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g})\) has the value \(K=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\). Find the value of \(K\) for each of the following reactions at the same temperature. (a) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g})\) (b) \(\mathrm{SO}_{3}(\mathrm{~g})=\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) (c) \(3 \mathrm{SO}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{SO}_{3}(\mathrm{~g})\)

\( \mathrm{~A} 3.00-\mathrm{L}\) reaction vessel is filled with \(0.150 \mathrm{~mol}\) \(\mathrm{CO}, 0.0900 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.180 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\). Equilibrium is reached in the presence of a zinc oxidechromium(III) oxide catalyst; and at \(300^{\circ} \mathrm{C}, K_{c}=\) \(1.1 \times 10^{-2}\) for the reaction, \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\). (a) As the reaction approaches equilibrium, will the molar concentration of \(\mathrm{CH}_{3} \mathrm{OH}\) increase, decrease, or remain unchanged? (b) What is the equilibrium composition of the mixture?

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