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The reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\) is an example of: (a) Backward reaction (b) Forward reaction (c) Irreversible reaction (d) Reversible reaction

Short Answer

Expert verified
The reaction is a reversible reaction (d).

Step by step solution

01

Understanding the Concept of a Reversible Reaction

In chemistry, a reversible reaction is one where the products can react together to regenerate the reactants. This means the reaction can proceed in both forward and backward directions, and is represented using the double arrow symbol (\(\rightleftharpoons\)).
02

Analyzing the Given Reaction

The reaction \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(s) + \text{Cl}_2(g)\) includes a double arrow (\(\rightleftharpoons\)), which indicates that it is a reversible reaction. This symbol shows that the reaction can proceed both in the forward direction (from left to right, forming \(\text{PCl}_3\) and \(\text{Cl}_2\)) and in the backward direction (from right to left, forming \(\text{PCl}_5\)).
03

Identifying the Correct Option

Given the use of the double arrow \(\rightleftharpoons\), the reaction is based on characteristics of reversible reactions. Among the options provided, the correct choice considering its bidirectional nature is (d) Reversible reaction.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fascinating concept in the study of reversible reactions. It occurs when the rates of the forward and reverse reactions are equal, meaning that the concentrations of the reactants and products remain constant over time. Even though reactions are constantly happening in both directions, there is no net change in the system's composition.
This state of balance is achieved only in a closed system where no reactants or products can escape.
  • Equilibrium doesn't mean reactants and products are equal in concentration, just that their ratios no longer change.
  • At equilibrium, the reaction is still occurring, but the amounts of reactants and products are static.
  • The concept of equilibrium helps predict how changes in conditions (like temperature, pressure, or concentration) affect the position of the equilibrium, according to Le Chatelier's principle.
Recognizing equilibrium is crucial for understanding how reactions behave in practical scenarios, such as industrial chemical production and biological systems.
Reaction Direction
Understanding the direction of a reaction is key in determining how a chemical equation behaves under different conditions. In reversible reactions, there's potential for changes in direction, driven by changes in external conditions like temperature and pressure.
The direction a reversible reaction favors is dictated by several factors:
  • The concentration of reactants and products can shift the reaction direction. Adding more reactants, for instance, can push the reaction forward.
  • Temperature changes can favor endothermic or exothermic reactions, influencing the direction in which equilibrium shifts.
  • Pressure impacts reactions involving gases; increasing pressure typically favors the side of the reaction with fewer gas molecules.
These factors collectively illustrate how a reversible reaction's direction isn't fixed, but rather dynamic, responding to its environment according to principles like Le Chatelier's principle.
PCl5 Decomposition
The decomposition of phosphorus pentachloride (\( \text{PCl}_5 \)) is a classic case of a reversible reaction. In this process, \( \text{PCl}_5 \) decomposes to form phosphorus trichloride (\( \text{PCl}_3 \)) and chlorine gas (\( \text{Cl}_2 \)): \( \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (s) + \text{Cl}_2 (g) \).
Here are key points about this decomposition:
  • This equilibrium reaction is sensitive to temperature changes, often requiring careful control to maintain balance between the forward and backward processes.
  • In the forward reaction, \( \text{PCl}_5 \) breaks down, but in the reverse, \( \text{PCl}_3 \) and \( \text{Cl}_2 \) combine to reform \( \text{PCl}_5 \).
  • Because the reverse reaction allows for regeneration of \( \text{PCl}_5 \), this process exemplifies chemical equilibrium and reaction dynamics well.
Understanding \( \text{PCl}_5 \) decomposition helps illustrate the practical applications of reversible reactions in predicting how changes in conditions affect reaction balance and direction.

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Most popular questions from this chapter

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

HI was heated in sealed tube at \(400^{\circ} \mathrm{C}\) till the equilibrium was reached. HI was found to be \(22 \%\) decomposed. The equilibrium constant for dissociation is: (a) \(1.99\) (b) \(0.0199\) (c) \(0.0796\) (d) \(0.282\)

The chemical equilibrium of a reversible reaction is not influenced by: (a) Temperature (b) Pressure (c) Catalyst (d) Concentration

The equilibrium between water and its vapour, in an open vessel: (a) Can be achieved (b) Depends upon pressure (c) Cannot be achieved (d) Depends upon temperature

In the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(0.5 \mathrm{~mol} \mathrm{~L}^{-1^{2}}\) and concentration of \(\mathrm{H}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) while concentration of \(\mathrm{S}_{2}\) is \(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\) in one litre vessel. The value of equilibrium constant of the reaction is: (a) \(0.016\) (b) \(0.013\) (c) \(0.020\) (d) \(0.030\)

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