/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 If a reaction is reversible, whe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a reaction is reversible, when can it be said to have reached equilibrium?

Short Answer

Expert verified
A reversible reaction is at equilibrium when the rates of the forward and reverse reactions are equal and the concentrations of reactants and products remain constant.

Step by step solution

01

Understanding Reversible Reactions

A reversible reaction is one where the reactants form products, which can themselves react to form the original reactants. In this type of reaction, the forward and backward reactions occur simultaneously.
02

Defining Chemical Equilibrium

Chemical equilibrium occurs when the rate at which the reactants are converting into products is equal to the rate at which the products are converting back into reactants. At this point, the concentrations of reactants and products remain constant over time.
03

Recognizing Equilibrium State

Equilibrium can be said to have been reached in a reversible reaction when observable properties of the system such as concentration, color, pressure, and pH become constant. It is important to understand that the reactions are still occurring but at the same rate in both directions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Reversible Reactions
Reversible reactions are fascinating chemical processes where the products formed can revert back into the original reactants. Think of it like a dance where two partners separate and come back together in rhythm. In the chemical world, for instance, when ice melts to form water, given the right conditions, that water can freeze back into ice. This occurs because the transformation isn't a one-way street; it can go back and forth depending on various factors like temperature and pressure.

In the context of a chemical experiment, you might observe a reversible reaction as a mixture that seems to stop changing after a while. However, it’s vital to note that the reaction actually continues; it’s just that the transformations in both directions balance each other out. This balance provides a perfect segue into the next concept of chemical equilibrium, where things get even more interesting.
Equilibrium State
Imagine a busy intersection where the same number of cars enter and leave every hour. This flow creates a sort of 'traffic equilibrium'. Similarly, in chemistry, an equilibrium state is reached in a reversible reaction when the forward reaction (reactants turning into products) and the backward reaction (products reverting to reactants) proceed at an identical pace. It's like the chemical version of a standstill in our intersection analogy.

At equilibrium, while the microscopic action never stops, macroscopic properties such as color and pressure stabilize. This does not imply that the reactants and products are present in equal amounts but rather that their concentrations no longer change over time. It’s important for students to recognize that equilibrium represents a dynamic balance between competing processes, not a complete halt of chemical activity.
Reaction Rates
The rate of a reaction is measured by how quickly the concentrations of reactants or products change over time. It's like tracking how fast a runner completes a lap; the timing tells us much about their speed. In chemistry, these rates are influenced by several factors, including temperature, concentration, and the presence of a catalyst.

When studying equilibrium, both the forward and reverse reaction rates are pivotal. At the beginning of a reversible reaction, the forward rate is high because there are plenty of reactants. As products accumulate, the forward rate slows and the reverse rate increases. Equilibrium is reached when these two rates become equal. Visualizing a scale that balances itself as the added weight on one side is canceled out by an equal weight on the other side can help in understanding this concept. Understanding reaction rates helps students to grasp why reaching equilibrium does not mean that reactions stop, but rather that the dance between reactants and products reaches a harmonious tempo.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the equilibrium \(4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g)\) (a) What is the expression for the equilibrium constant \(\left(K_{c}\right)\) of the reaction? (b) How must the concentration of \(\mathrm{NH}_{3}\) change to reach equilibrium if the reaction quotient is less than the equilibrium constant? (c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of \(\mathrm{NO}_{2} ?\) (d) If the change in the pressure of \(\mathrm{NO}_{2}\) is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of \(\mathrm{O}_{2}\) change?

Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst. (a) Write the expression for the equilibrium constant \(\left(K_{c}\right)\) for the reversible reaction \(2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \quad \Delta H=-90.2 \mathrm{kJ}\) (b) What will happen to the concentrations of \(\mathrm{H}_{2}, \mathrm{CO},\) and \(\mathrm{CH}_{3} \mathrm{OH}\) at equilibrium if more \(\mathrm{H}_{2}\) is added? (c) What will happen to the concentrations of \(\mathrm{H}_{2}, \mathrm{CO},\) and \(\mathrm{CH}_{3} \mathrm{OH}\) at equilibrium if CO is removed? (d) What will happen to the concentrations of \(\mathrm{H}_{2}, \mathrm{CO},\) and \(\mathrm{CH}_{3} \mathrm{OH}\) at equilibrium if \(\mathrm{CH}_{3} \mathrm{OH}\) is added? (e) What will happen to the concentrations of \(\mathrm{H}_{2}, \mathrm{CO},\) and \(\mathrm{CH}_{3} \mathrm{OH}\) at equilibrium if the temperature of the system is increased?

A reaction is represented by this equation: \(2 \mathrm{W}(a q) \rightleftharpoons \mathrm{X}(a q)+2 \mathrm{Y}(a q) \quad K_{c}=5 \times 10^{-4}\) (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations of \(\leq 1 M,\) identify two sets of concentrations that describe a mixture of \(\mathrm{W}, \mathrm{X},\) and \(\mathrm{Y}\) at equilibrium.

Acetic acid is a weak acid that reacts with water according to this equation: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)\) Will any of the following increase the percent of acetic acid that reacts and produces \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) ion? (a) Addition of HCl (b) Addition of NaOH (c) Addition of \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\)

Write the mathematical expression for the reaction quotient, \(Q_{c}\) for each of the following reactions: (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{g})=\mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(g)\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)=2 \mathrm{NO}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) (d) \(\operatorname{BaSO}_{3}(s)=\operatorname{BaO}(s)+\operatorname{SO}_{2}(g)\) (e) \(\mathrm{P}_{4}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(s)\) (f) \(\operatorname{Br}_{2}(g)=2 \operatorname{Br}(g)\) (g) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I)\) (h) \(\operatorname{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)=\mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.