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Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\)

Short Answer

Expert verified
For the given reactions upon increasing the partial pressure of H₂: (a) ΔG decreases. (b) ΔG increases. (c) ΔG decreases.

Step by step solution

01

(a) Analyzing the increase in H₂ partial pressure in the first reaction

For reaction (a), N₂(g) + 3 H₂(g) → 2 NH₃(g), it is important to check how the reaction will shift when the partial pressure of H₂, one of the reactants, is increased. According to Le Chatelier's principle, when we increase the pressure (or concentration) of a reactant, the system will try to counteract the change by shifting towards the products. In this case, the reaction will shift to the right to produce more NH₃. Because the reaction shifts towards the products, this signifies ΔG for the reaction will decrease in value.
02

(b) Analyzing the increase in H₂ partial pressure in the second reaction

For reaction (b), 2 HBr(g) → H₂(g) + Br₂(g), when the partial pressure of H₂, one of the products, is increased, we'll evaluate how the reaction shifts according to Le Chatelier's principle. With the increase in the partial pressure of H₂, the system will attempt to counteract the change by shifting towards the reactants to consume the excess H₂. In this case, the reaction will shift to the left to produce more HBr. As the reaction shifts towards the reactants, this implies that ΔG for the reaction will increase in value.
03

(c) Analyzing the increase in H₂ partial pressure in the third reaction

For reaction (c), 2 H₂(g) + C₂H₂(g) → C₂H₆(g), we have to determine how a change in the partial pressure of H₂, one of the reactants, will influence the reaction's shift. As per Le Chatelier's principle, increasing the pressure (or concentration) of a reactant will drive the system to counteract the change by shifting towards the products. In this case, the reaction will shift to the right, forming more C₂H₆. The shift of the reaction toward the products suggests that ΔG for the reaction will decrease in value.
04

Summary

For the given reactions upon increasing the partial pressure of H₂: (a) ΔG decreases. (b) ΔG increases. (c) ΔG decreases.

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Key Concepts

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

Le Chatelier's Principle and Changes in Pressure
Le Chatelier's Principle helps us understand how a system at equilibrium responds to changes in pressure or concentration. This principle states that when an external stress is applied to a system at equilibrium, the system adjusts itself to relieve that stress and restore a new equilibrium. This adjustment can lead to the shifting of equilibrium either to the left or to the right, depending on the nature of the stress.

In the context of gases, an increase in partial pressure of a reactant or product will influence the direction in which the balance will shift.
  • If the pressure of a reactant is increased, the system tends to shift towards the side where fewer gas molecules are formed (usually the products) to minimize the stress.
  • Conversely, if the pressure of a product is increased, the system will shift to form more reactants.
For instance, in reaction (a) of the exercise where more NH₃ is formed when H₂ pressure is increased, the system shifts to produce more NH₃ to relieve the change, causing a decrease in Gibbs Free Energy, ΔG.
Chemical Equilibrium and the Role of Gibbs Free Energy
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, resulting in constant concentrations of the reactants and products. At equilibrium, the system’s Gibbs Free Energy, ΔG, is at its minimum value for a given set of conditions.

Gibbs Free Energy change, ΔG, measures the capacity of a system to perform work and determines the spontaneity of a reaction.
  • A negative ΔG indicates a spontaneous reaction that favors the formation of products.
  • A positive ΔG implies that the reaction is non-spontaneous, favoring the formation of reactants instead.
As noted in the solution's analysis, adjusting the partial pressure of either reactants or products will influence ΔG. As the system shifts in response to pressure changes, ΔG may either increase or decrease depending on the reaction direction: decreasing ΔG when a reaction shifts towards products, and increasing ΔG when reversing towards reactants.
Partial Pressure Effects on Reaction Direction
In any gaseous equilibrium, the partial pressure effects play a crucial role in steering the reaction. The partial pressure is the pressure that each gas in a mixture would exert if it occupied the whole volume alone at the same temperature. Altering the partial pressure of a gas shifts the equilibrium according to Le Chatelier's Principle.

When we increase the partial pressure of one of the gases involved:
  • For a reactant, the equilibrium shifts towards products as seen in reactions (a) and (c) from the exercise, resulting in more products and decreasing ΔG.
  • For a product, such as in reaction (b), it shifts the equilibrium towards reactants, increasing ΔG as more reactants are favored.
These shifts occur because the system tends to minimize the effect of pressure changes by shifting the equilibrium composition. The changes in partial pressures are thus reflected in variations of ΔG, guiding us towards the preferred direction of the reaction.

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Most popular questions from this chapter

The standard entropies at \(298 \mathrm{~K}\) for certain of the group \(4 \mathrm{~A}\) elements are as follows: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) \(\mathrm{Si}(s)=18.81 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) and \(\operatorname{Sn}(s)=\) \(51.818 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) All but \(\mathrm{Sn}\) have the diamond structure. How do you account for the trend in the \(S^{\circ}\) values?

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