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The overall photosynthesis reaction is \(6 \mathrm{CO}_{2}(\mathrm{~g})+\) \(6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})\), and \(\Delta \mathrm{H}^{\circ}=+2802 \mathrm{~kJ}\). Suppose that the reaction is at equilibrium. State the effect that each of the following changes will have on the equilibrium composition (tends to shift toward the formation of reactants, tends to shift toward the formation of products, or has no effect). (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. (e) 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 left, (b) Shifts right, (c) Shifts right, (d) Shifts right, (e) Shifts right, (f) No effect, (g) Shifts left.

Step by step solution

01

- Understanding Le Chatelier's Principle

The problem involves applying Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle will help predict the direction in which the reaction will shift when a change is applied.
02

- Examining Change (a): Increasing \(\mathrm{O}_{2}\) Partial Pressure

Increasing the partial pressure of \(\mathrm{O}_{2}\) would increase the concentration of one of the products. According to Le Chatelier's Principle, the system will try to reduce this change by shifting the equilibrium to the left, favouring the formation of reactants.
03

- Examining Change (b): Compressing the System

Compressing the system effectively increases the pressure. Because there are more gas molecules on the reactant side (12) than on the product side (6), the equilibrium will shift to the right, towards the side with fewer gas molecules, which is towards the formation of more products.
04

- Examining Change (c): Increasing \(\mathrm{CO}_{2}\) Amount

Increasing the amount of \(\mathrm{CO}_{2}\), a reactant, will cause the equilibrium to shift toward the products to reduce the added \(\mathrm{CO}_{2}\) concentration.
05

- Examining Change (d): Increasing Temperature

Increasing the temperature adds heat to the system. Because the reaction is endothermic (as indicated by the positive \(\Delta H^\circ\)), heat can be considered a reactant. The system will counter this by shifting the equilibrium to the right, towards the products, to consume the added heat.
06

- Examining Change (e): Removing Some \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)

Removing some \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\), a product, will cause the equilibrium to shift towards the right to replace the removed product.
07

- Examining Change (f): Adding Water

Adding water, a reactant in its liquid form, will have little to no effect on the equilibrium since the reaction is in terms of gases and aqueous solutions; the concentration of liquid water does not appear in the equilibrium expression.
08

- Examining Change (g): Decreasing \(\mathrm{CO}_{2}\) Partial Pressure

Decreasing the partial pressure of \(\mathrm{CO}_{2}\) reduces the concentration of a reactant. According to Le Chatelier's Principle, the equilibrium will shift to the left to increase the partial pressure of \(\mathrm{CO}_{2}\), favouring the formation of reactants.

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that describes a state in which the rate of the forward reaction equals the rate of the reverse reaction. This balance results in no net change in the concentration of reactants and products over time, although both reactions still occur. It's important to realize that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios remain constant.

For instance, the photosynthesis reaction, which combines carbon dioxide and water to form glucose and oxygen, reaches equilibrium when the rate at which glucose and oxygen form is equal to the rate at which they break down to form carbon dioxide and water. Understanding how to manipulate this equilibrium is essential in many industrial and biological processes, where maximizing product yield is desirable.
Explaining Equilibrium Shift
An equilibrium shift occurs when a change in conditions causes the equilibrium position to move in favor of either the reactants or the products. Le Chatelier's Principle provides a framework for predicting how a change in pressure, concentration, or temperature will affect the position of equilibrium. If an external change increases the concentration of a product, for instance, the system responds by shifting the equilibrium towards the reactants to minimize this disturbance, as demonstrated in the scenario where the partial pressure of oxygen was increased in the photosynthesis equilibrium.

Effects of Pressure and Concentration

When the system is compressed, causing an increase in pressure, or when a reactant like carbon dioxide is added, the equilibrium shifts towards the side with fewer gas molecules or the side that consumes the added substance, respectively. Likewise, removing a product or reducing a reactant's concentration will shift the equilibrium in the opposite direction, aiming to replace the lost substance.
Endothermic Reactions and Equilibrium
Understanding the nature of endothermic reactions is key to interpreting their behavior under the influence of temperature changes. An endothermic reaction absorbs heat from its surroundings. Regarding photosynthesis, with a positive \(\Delta H^\circ\), the reaction requires heat to proceed. Hence, heat effectively acts as a 'reactant'.

When the temperature is increased, more heat is introduced, and the equilibrium shifts to consume this excess heat, favoring the products. This shift is an attempt to return to equilibrium by absorbing the added heat energy. Thus, an increase in temperature for an endothermic process often results in an increased yield of products, which is the case for the photosynthesis equation presented in the exercise.

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Most popular questions from this chapter

At \(500 .{ }^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\) \(\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container; then the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the amount present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) bar; (b) \(1.00\) bar.

(a) Calculate the reaction Gibbs free energy of \(\mathrm{I}_{2}(\mathrm{~g}) \rightarrow\) \(2 \mathrm{I}(\mathrm{g})\) at \(1200 . \mathrm{K}(K=6.8)\) when the partial pressures of \(\mathrm{I}_{2}\) and \(\mathrm{I}\) are \(0.13\) bar and \(0.98\) bar, respectively. (b) What is the spontaneous direction of the reaction? Explain briefly.

A \(0.100-\mathrm{mol}\) sample of \(\mathrm{H}_{2} \mathrm{~S}\) is placed in a \(10.0\) - \(\mathrm{L}\) reaction vessel and heated to \(1132^{\circ} \mathrm{C}\). At equilibrium, \(0.0285 \mathrm{~mol} \mathrm{H}_{2}\) is present. Calculate the value of \(K_{c}\) for the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) at \(1132^{\circ} \mathrm{C}\).

When \(1.00 \mathrm{~g}\) of gaseous \(\mathrm{I}_{2}\) is heated to \(1000 . \mathrm{K}\) in a \(1.00-\mathrm{L}\) sealed container, the resulting equilibrium mixture contains \(0.830 \mathrm{~g}\) of \(\mathrm{I}_{2}\). Calculate \(K_{c}\) for the dissociation equilibrium \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\).

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