Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 30

In the reaction \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftarrows 2 \mathrm{NO}_{2}\), the degree of dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) increases with (a) increase in pressure (b) decrease in temperature (c) increase in volume (d) presence of catalyst

Short Answer

Expert verified
a) Increase in pressure b) Decrease in temperature c) Increase in volume d) Presence of catalyst Answer: (b) Decrease in temperature and (c) Increase in volume.

Step by step solution

01

Brief overview of Le Chatelier's principle and degree of dissociation

Le Chatelier's principle states that if a stress (such as a change in pressure, temperature, or concentration) is applied to a system at equilibrium, the system will adjust to counteract the stress and restore equilibrium. The degree of dissociation refers to the extent to which a substance has dissociated into its constituent parts.
02

Effect of increasing pressure (Option A)

When pressure is increased, the system tends to shift the equilibrium to the side with fewer moles of gas to counteract the change in pressure. In the given reaction, there are fewer moles of gas on the left side (1 mole N2O4) compared to the right side (2 moles NO2). Thus, increasing the pressure will shift the equilibrium to the left side, decreasing the degree of dissociation of N2O4.
03

Effect of decreasing temperature (Option B)

The given reaction is an endothermic reaction, as it requires the intake of energy to dissociate N2O4 into two molecules of NO2. When the temperature is decreased, the system will try to counteract this change by shifting the equilibrium toward the side that absorbs the heat, which is the right side in this case. Thus, decreasing the temperature will increase the degree of dissociation of N2O4.
04

Effect of increasing volume (Option C)

When the volume is increased, the pressure decreases. As discussed in Option A, lowering the pressure will favor the side with a higher number of gas moles, which is the right side in the given reaction. Thus, increasing the volume will lead to an increase in the degree of dissociation of N2O4.
05

Effect of presence of catalyst (Option D)

The presence of a catalyst increases the rate of both the forward and the reverse reactions. However, it does not change the position of the equilibrium or the degree of dissociation. Therefore, the presence of a catalyst has no effect on the degree of dissociation of N2O4.
06

Final Answer

Based on our analysis, the degree of dissociation of N2O4 increases with: (b) decrease in temperature (c) increase in volume

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

In the reaction: \(\mathrm{NO}_{2}+\mathrm{CO} \rightleftarrows \mathrm{NO}+\mathrm{CO}_{2}\), the equilibrium state may be recognized by the constancy of colour.

The minimum energy which two molecules should possess so that their collisions result in a chemical reaction is called_________ energy.

For each of the questions, four choices have been provided. Select the correct alternative. In the formation of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) from \(\mathrm{NO}_{2}\) the rates of production of (a) \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) are equal (b) \(\mathrm{NO}\) is double the rate of consumption of \(\mathrm{NO}_{2}\) (c) \(\mathrm{NO}\) is twice the rate of production of \(\mathrm{O}_{2}\) (d) \(\mathrm{O}_{2}\) is twice the rate of production of \(\mathrm{NO}\)

For each of the questions, four choices have been provided. Select the correct alternative. For a reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\), the rate law is written as \(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}[\mathrm{~B}] .\) Doubling the concentration of "A" without changing concentration of "B" increases the rate of reaction by (a) 2 times (b) 4 times (c) 8 times (d) 16 times

For each of the questions, four choices have been provided. Select the correct alternative. If an activated complex is formed in chemical reactions according to the collision theory, which of the following is true with respect to its stability? (a) It is highly stable because it has high energy. (b) It is less stable because it has lower energy. (c) It is less stable because it has high energy. (d) None of the above.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

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