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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

Short Answer

Expert verified
The equilibrium cannot be achieved in an open vessel. (c) Cannot be achieved.

Step by step solution

01

Understanding the Concept

Equilibrium involves a balance between two opposing processes happening at the same rate. For water and vapor, equilibrium means the rate of evaporation equals the rate of condensation.
02

Identifying the Key Conditions

In an open vessel, vapor can escape, disrupting any balance between evaporation and condensation. This means equilibrium is not possible, since condensation cannot occur if vapor is escaping.
03

Analyzing the Options

Examine the options: (a) "Can be achieved" is incorrect because equilibrium cannot be maintained in an open system. (b) "Depends upon pressure" is irrelevant because pressure cannot maintain equilibrium in an open system. (c) "Cannot be achieved" fits the understanding that in an open system, equilibrium cannot be sustained. (d) "Depends upon temperature" is true for the rate processes, but not in sustaining equilibrium in an open system.
04

Choosing the Correct Answer

Based on the analysis, option (c) "Cannot be achieved" is the correct answer, as the system's openness prevents maintaining equilibrium.

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Key Concepts

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

Vapor Pressure
Vapor pressure is a critical concept when discussing chemical equilibrium in liquids. It is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. Vapors form when molecules escape from the surface of a liquid into the air.
  • When a liquid is in a closed container, vapor pressure indicates how many molecules have evaporated.
  • The higher the vapor pressure, the more volatile the substance is, meaning it evaporates quickly.
  • Vapor pressure increases with temperature because more molecules have enough energy to escape the liquid phase.
Understanding vapor pressure helps in predicting how a liquid will behave in both open and closed systems, and its impact on processes like evaporation and condensation.
Evaporation
Evaporation is the process where liquid transforms into vapor or gas. This transformation occurs when molecules at the liquid's surface gain enough energy to overcome atmospheric pressure and become vapor.
  • Evaporation happens at any temperature, although it occurs faster at higher temperatures.
  • The rate of evaporation can be influenced by several factors, including temperature, surface area, and wind speed.
  • In an open container, evaporation continues until the liquid is exhausted, as vapor escapes into the atmosphere.
Evaporation plays a vital role in maintaining an equilibrium in closed systems but disrupts equilibrium in open systems where vapor can escape freely.
Condensation
Condensation is the opposite process of evaporation, where gas or vapor changes into a liquid. This typically occurs when vapor is cooled or compressed.
  • When the temperature of vapor decreases, its molecules lose energy, allowing them to come closer together and form a liquid.
  • Condensation forms the basis for processes like cloud formation and dew.
  • In a closed system, condensation helps establish equilibrium by balancing the rate of evaporation.
However, in an open system, the escaping vapor makes condensation difficult, as observed in the exercise where an equilibrium couldn't be maintained.
Open System
An open system allows for the exchange of matter and energy with its surroundings. When considering chemical equilibrium, the nature of the system plays a crucial role.
  • In an open system, such as an open vessel, vapor can escape into the air, preventing the establishment of equilibrium between vapor and liquid.
  • This is because the process of condensation cannot match the rate of evaporation since the vapor is not confined within a space.
  • Open systems are contrasted with closed systems where such exchanges are restricted, enabling equilibrium conditions to be achieved more readily.
Understanding the constraints of an open system helps explain why equilibrium cannot be achieved, as seen in the provided exercise.

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Most popular questions from this chapter

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}+22.4 \mathrm{kcal}\) formation of \(\mathrm{NH}_{3}\) by above reaction shows: (a) Cyanamide process (b) Serpeck's process (c) Haber process (d) None of these

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At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is \(1.80 \times 10^{-3}\) What is the numerical value in mole per litre of equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for this reaction at the same temperature: (a) \(8.1 \times 10^{-8}\) (b) \(9.1 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3.1 \times 10^{-7}\) (d) \(6.1 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

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The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature: (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) None of these

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