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For the chemical reaction: \(3 \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_{3} \mathrm{Y}(\mathrm{g})\), the amount of \(\mathrm{X}_{3} \mathrm{Y}\) at equilibrium is affected by (a) Temperature and pressure (b) Temperature only (c) Pressure only (d) Temperature, pressure and catalyst

Short Answer

Expert verified
(a) Temperature and pressure affect the equilibrium position.

Step by step solution

01

Understand the Reaction and Equilibrium

The reaction involves gases reacting to form another gas, thus it's a gaseous equilibrium. The equilibrium position is influenced by the conditions under which the reaction occurs, such as temperature and pressure, due to Le Chatelier's Principle.
02

Analyze Effect of Temperature

Temperature affects the equilibrium of a reaction by changing the reaction rate and shifting the equilibrium. For exothermic or endothermic reactions, increasing or decreasing the temperature will shift the equilibrium to favor the reactants or products accordingly.
03

Analyze Effect of Pressure

Pressure influences gaseous equilibria, especially when there is a change in the number of moles of gas. According to Le Chatelier's Principle, an increase in pressure shifts the equilibrium towards the side with fewer gas molecules. Here, 4 moles of gases on the left decrease to 1 mole on the right, so increasing pressure favors the formation of \(\mathrm{X}_{3}\mathrm{Y}\).
04

Consider the Role of Catalysts

Catalysts increase the rate at which equilibrium is achieved but do not affect the position of the equilibrium. They help the reaction reach equilibrium faster, but they do not shift the equilibrium towards products or reactants.
05

Conclusion from Analysis

From the analysis, both temperature and pressure affect the equilibrium position as they cause shifts in equilibrium favoring either reactants or products. However, catalysts do not change the equilibrium position.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that explains how a system at equilibrium responds to external changes. If a system at equilibrium experiences a change in concentration, pressure, or temperature, the equilibrium position will shift to counteract the effect of the disturbance. Essentially, the system "pushes back" to find a new state of balance.
A simple way to think about this principle is in terms of cause and effect. If you increase the concentration of a reactant, the equilibrium will shift to produce more products, attempting to reduce the added reactant. Similarly, removing a product will cause the reaction to shift to the right to form more product.
Understanding Le Chatelier's Principle helps us predict the direction in which a reaction shifts under different conditions, making it a crucial tool for chemists.
Effect of Temperature on Equilibrium
Temperature changes can influence a reaction's equilibrium position significantly. Every chemical reaction is either exothermic (releases heat) or endothermic (absorbs heat). When the temperature changes, it affects how the reaction behaves.
In an exothermic reaction, an increase in temperature will cause the equilibrium to shift towards the reactants. This occurs because adding heat is like adding more of a product in terms of energy balance, so the system counteracts by forming more reactants.
In contrast, for an endothermic reaction, increasing the temperature favors the formation of products, as heat is seen as a "reactant" that promotes the forward reaction. Thus, adjusting temperature can effectively shift the equilibrium to either produce more products or more reactants, depending on the reaction's heat exchange.
Effect of Pressure on Equilibrium
Pressure impacts reactions involving gases, where a change in the number of moles of gas occurs. According to Le Chatelier's Principle, if pressure is increased, the equilibrium shifts towards the side with fewer moles of gas.
In the given reaction, there are 4 moles of gaseous reactants transforming into a single mole of product, \(\mathrm{X}_{3}\mathrm{Y}\). So, increasing the pressure favors the formation of \(\mathrm{X}_{3}\mathrm{Y}\), moving the equilibrium to the right. Conversely, decreasing the pressure would shift the equilibrium toward the reactants, as it favors the side with more gas moles.
Therefore, manipulating pressure is a valuable way to control the amount of product formed in reactions involving gaseous substances.
Role of Catalysts in Chemical Reactions
Catalysts play an important role in speeding up chemical reactions, but they do not affect the equilibrium position itself. Instead, they lower the activation energy required for the reaction to proceed, allowing the system to reach equilibrium more quickly.
While catalysts do not alter the concentrations of reactants or products at equilibrium, they are valuable for industrial processes where speed and efficiency are critical. By reducing the time it takes to achieve equilibrium, catalysts enhance the overall rate of production.
It's crucial to remember that although catalysts can make reactions feasible under less severe conditions, they cannot alter the basic thermodynamic properties of the reactions they catalyze.

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Most popular questions from this chapter

\(9.2 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\). At equilibrium, \(50 \%\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (g) is dissociated. What is the equilibrium constant (in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right) ?\) (molecular weight of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 92 ): (a) \(0.1\) (b) \(0.2\) (c) \(0.4\) (d) 2

Which of the following change will shift the reaction in forward direction: \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g})\) Take \(\Delta \mathrm{H}^{\circ}=+150 \mathrm{~kJ}\) (a) Increase in concentration of I (b) Increase in total pressure (c) Decrease in concentration of \(\mathrm{I}_{2}\) (d) Increase in temperature

The role of a catalyst in a reversible reaction is to: (a) Alter the equilibrium constant of the reaction (b) Increase the rate of forward reaction (c) Allow the equilibrium to be achieved quickly (d) Decrease the rate of backward reaction

Which of the following reaction will be favoured at low pressure: (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\)

For the reaction, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\left(\mathrm{K}_{\mathrm{c}}=1.8 \times 10^{-6}\right.\) at \(\left.184^{\circ} \mathrm{C}\right)\) \((\mathrm{R}=0.0831 \mathrm{~kJ} /(\mathrm{mol} \mathrm{K}))\) when \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{c}}\) are compared at \(184^{\circ} \mathrm{C}\) it is found that: (a) \(\mathrm{K}_{\mathrm{p}}\) is greater than \(\mathrm{K}_{\mathrm{c}}\) (b) \(\mathrm{K}_{\mathrm{p}}\) is less than \(\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}=K_{c}\) (d) Whether \(\mathrm{K}_{\mathrm{p}}\) is greater than, less than or equal to \(\mathrm{K}_{\mathrm{c}}\) depends upon the total gas pressure

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