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Suppose we set up a system in which water is poured into a vessel having a hole in the bottom. If the rate of water inflow is adjusted so that it matches the rate at which water drains through the hole, the amount of water in the vessel remains constant over time. Is this an equilibrium system? Explain.

Short Answer

Expert verified
Yes, it is an equilibrium system because the rate of water inflow equals the rate of outflow, resulting in a constant amount of water in the vessel over time, which aligns with the definition of equilibrium.

Step by step solution

01

Understanding the System

Assess the characteristics of the system. Observe that water is being poured into the vessel at a certain rate and it is also leaving the vessel at the same rate. Since the inflow and outflow rates are balanced, the amount of water in the vessel remains unchanged over time.
02

Define Equilibrium

Understand the definition of an equilibrium system. An equilibrium is a state where there are no net changes in the system as the forward and reverse processes are occurring at the same rate. Even though there may be movement or activity, no overall change is observed.
03

Applying the Equilibrium Concept to the System

Apply the definition of equilibrium to the water vessel system. Since the rate of water flowing in matches the rate of water flowing out, there are no net changes in the volume of water in the vessel over time, thus conforming to the definition of an equilibrium system.

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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 fundamental concept in chemistry that occurs when the rates of the forward and reverse chemical reactions are equal, resulting in no net change in the concentrations of the reactants and products over time. It's a balance, much like a see-saw that is perfectly level, where the movement of materials into and out of the reactions occurs at the same pace.

Imagine a situation where you have a reaction where substance A turns into substance B, and B can also revert back to a certain extent. When the rate at which A transforms into B equals the rate at which B transforms back into A, you've reached a state of chemical equilibrium. This doesn't mean the reactions have stopped—far from it. They are still occurring but are counterbalancing each other, leading to a dynamic, yet stable situation.

For students, analyzing chemical equilibrium involves understanding that it's about the rates of reactions and not the amounts of substances. A common misconception is that equilibrium means the reactants and products are present in equal amounts, which is not necessarily true; rather, their rates of formation are what's equated.
Dynamic Equilibrium
Dynamic equilibrium is a type of equilibrium that is particularly important in the context of chemistry and other sciences involving physical processes. It is 'dynamic' because it involves continuing activity or change, while the system as a whole remains in an unchanged state. This is similar to walking up a down escalator with the speed that exactly matches the escalator's descent—the escalator keeps moving, and so do you, but your position in space remains the same.

In the context of the water vessel system mentioned in the exercise, dynamic equilibrium is achieved because the process of water entering and leaving the vessel is continuous. The key to understanding dynamic equilibrium is to realize that even though individual molecules are moving in and out, there is no overall effect on the system's status — the level of water in this case. This is analogous to many reversible reactions in chemistry where reactants and products are converting back and forth at equal rates, maintaining a steady-state condition.
System Analysis in Chemistry
System analysis in chemistry involves breaking down and examining the processes within a chemical reaction or any physical system to understand its behavior and predict its responses to changes. It is akin to an engineer examining each part of an engine to see how it contributes to the engine's overall functionality.

One aspect of system analysis is understanding the interaction between different components and how they impact overall stability. In the vessel water system, for example, system analysis would involve considering factors such as the rate of inflow and outflow, the size of the hole, and the capacity of the vessel. Through systematic analysis, chemists determine how variations in these factors would affect the system and how to achieve or disrupt equilibrium, providing critical insights into the control and manipulation of chemical reactions.
Rate of Reaction
Rate of reaction is a measure of how fast a chemical reaction occurs. It is often expressed as the change in concentration of a reactant or product per unit time. Faster reactions transform reactants to products quickly, while slower ones take more time to reach the same transformation. The rate can be influenced by various factors, including temperature, pressure, the presence of a catalyst, concentration, and the physical state of reactants.

To illustrate, a higher concentration of reactants usually increases the reaction rate because more reactant molecules are available to collide and form products. Understanding the reaction rate is crucial for controlling processes in industry, medicine, and many other fields. In the textbook water vessel example, if the inflow rate increases suddenly and exceeds the outflow rate, the system will no longer be at equilibrium — this is analogous to a sudden increase in reactant concentration speeding up a reaction and shifting the equilibrium in a chemical system.

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Most popular questions from this chapter

Write the equilibrium law for each of the following reactions in terms of molar concentrations: (a) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) (b) \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{HCl}(g)\) How does \(K_{\mathrm{c}}\) for reaction (a) compare with \(K_{\mathrm{c}}\) for reaction (b)?

Sketch a graph showing how the concentrations of the reactant \(A\) and product \(P\) of a typical chemical reaction \((A \rightleftharpoons P)\) vary with time during the course of the reaction. Assume that no products are present at the start of the reaction. Indicate on the graph where the system has reached equilibrium.

Under what conditions does the reaction quotient equal \(K_{c}\)??

Write the equilibrium law corresponding to \(K_{c}\) for each of the following heterogeneous reactions. (a) \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (b) \(2 \mathrm{NaHSO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) (c) \(2 \mathrm{C}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HF}(g) \rightleftharpoons \mathrm{C}_{2}(g)+\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\)

At high temperature, \(2.00 \mathrm{~mol}\) of \(\mathrm{HBr}\) was placed in a 4.00 L container where it decomposed in the reaction: $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{Br}_{2}\) was measured to be \(0.0955 \mathrm{M}\). What is \(K_{\mathrm{c}}\) for this reaction at this temperature?

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