Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 75

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

Short Answer

Expert verified
Answer: The flaw in the statement is the incorrect assumption that a high equilibrium constant indicates a very fast reaction. The equilibrium constant reflects the position of equilibrium and has no direct relation to the reaction rate or speed.

Step by step solution

01

Understanding the Equilibrium Constant

The equilibrium constant (K) is a measure of the relative amounts of reactants and products present at equilibrium. It indicates the position of equilibrium but does not provide any information about the rate (speed) at which the reaction takes place.
02

Distinguishing between Equilibrium Constant and Reaction Rate

It's important to differentiate between equilibrium constant (K) and reaction rate. A high equilibrium constant means that the reaction reaches an equilibrium state where the concentration of products is higher compared to the reactants. On the other hand, reaction rate relates to how fast the reaction reaches the equilibrium.
03

Assessing the Given Statement

The given statement claims that because the equilibrium constant for the reaction is 792 at 400 K, it must be a very fast reaction. But as discussed earlier, the equilibrium constant merely indicates the position of equilibrium and does not provide any information about the rate or speed of the reaction. Therefore, the statement is incorrect in making the assumption that a high equilibrium constant translates into a fast reaction rate.
04

Conclusion

The flaw in the statement is the assumption that a high equilibrium constant (792 in this case) indicates a very fast reaction. The equilibrium constant reflects the position of equilibrium, and it has no direct relation to the reaction rate or speed.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

WEB At \(1123 \mathrm{~K}\), methane and hydrogen sulfide gases react to form carbon disulfide and hydrogen gases. At equilibrium the concentrations of methane, hydrogen sulfide, carbon disulfide, and hydrogen gas are \(0.00142 M, 6.14 \times 10^{-4} M, 0.00266 M\), and \(0.00943 M\), respectively. (a) Write a balanced equation for the formation of one mole of carbon disulfide gas. (b) Calculate \(K\) for the reaction at \(1123 \mathrm{~K}\).

For the decomposition of \(\mathrm{CaCO}_{3}\) at \(900^{\circ} \mathrm{C}, K=1.04\). $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g)$$ Find the smallest mass of \(\mathrm{CaCO}_{3}\) needed to reach equilibrium in a 5.00-L vessel at \(900^{\circ} \mathrm{C}\).

WEB Consider the following reaction at \(122^{\circ} \mathrm{C}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ (a) Write an equilibrium constant expression for the reaction and call the constant \(K_{1}\). (b) Write an equilibrium constant expression for the decomposition of one mole of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) and call the constant \(K_{2}\). (c) Relate \(K_{1}\) and \(K_{2}\).

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{B} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 30 s? After 90 s?

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation
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.

Sign-up for free