Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 29

Determine \(\Delta n_{\text {gas }}\) for each of the following reactions: (a) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{HNO}_{3}(l)+\mathrm{ClF}(g) \rightleftharpoons \mathrm{ClONO}_{2}(g)+\mathrm{HF}(g)\)

Short Answer

Expert verified
(a) 1, (b) -3, (c) 1

Step by step solution

01

- Understanding \(\text {\Delta n_{\text {gas}}}\)

\(\text {\Delta n_{\text {gas}}}\) represents the change in the number of moles of gaseous reactants and products. It is calculated using the formula: \(\text {\Delta n_{\text {gas}}} = \text{moles of gaseous products} - \text{moles of gaseous reactants}\).
02

- Calculate \(\text {\Delta n_{\text {gas}}}\) for reaction (a)

For the reaction \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s) + \mathrm{CO}_{2}(g)\): \The gaseous products: 1 mole of \(\text{CO}_2\)\The gaseous reactants: 0 moles\Thus, \(\text {\Delta n_{\text {gas}}} = 1 - 0 = 1\).
03

- Calculate \(\text {\Delta n_{\text {gas}}}\) for reaction (b)

For the reaction \(2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{O}(l)\): \The gaseous products: 0 moles\The gaseous reactants: 3 moles (2 moles of \(\mathrm{H}_2\) and 1 mole of \(\mathrm{O}_2\))\Thus, \(\text {\Delta n_{\text {gas}}} = 0 - 3 = -3\).
04

- Calculate \(\text {\Delta n_{\text {gas}}}\) for reaction (c)

For the reaction \(\mathrm{HNO}_{3}(l) + \mathrm{ClF}(g) \rightleftharpoons \mathrm{ClONO}_{2}(g) + \mathrm{HF}(g)\): \The gaseous products: 2 moles (1 mole of \(\mathrm{ClONO}_2\) and 1 mole of \(\mathrm{HF}\))\The gaseous reactants: 1 mole of \(\mathrm{ClF}\)\Thus, \(\text {\Delta n_{\text {gas}}} = 2 - 1 = 1\).

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!

Key Concepts

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

Delta n_gas
Understanding the concept of \( \Delta n_{\text {gas}} \) is crucial in chemical reactions involving gases. It represents the change in the number of moles of gas during the reaction. It's calculated using the formula: \( \Delta n_{\text {gas}} = \text{moles of gaseous products} - \text{moles of gaseous reactants} \). This value helps you understand how the amount of gas changes as reactants turn into products. For instance, if a reaction starts with 2 moles of gas and ends with 4 moles of gas, the \( \Delta n_{\text {gas}} \) would be 2 (since 4 - 2 = 2). Conversely, if a reaction starts with more gas moles than it ends with, \( \Delta n_{\text {gas}} \) would be negative.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to determine the proportions of reactants and products. When you're dealing with gases, stoichiometry can help you find out the volumes and numbers of moles involved. This is essential for calculating \( \Delta n_{\text {gas}} \), as it informs you of the exact mole quantities of gases before and after the reaction. Remember, the coefficients in a balanced chemical equation represent the ratios in which substances react or are produced.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. These processes can change the state and amount of matter involved. When gases are part of the reaction, their moles need to be carefully tracked. By observing the balanced equation of a reaction, you can determine which substances are gases and their respective amounts. This is important when calculating \( \Delta n_{\text {gas}} \). For example, in the reaction \( 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) \), there is a reduction in gas moles, indicating that \( \Delta n_{\text {gas}} = -3 \).
Moles of Gas
Understanding moles of gas is fundamental in interpreting chemical reactions. A mole is a basic unit in chemistry representing \( 6.022 \times 10^{23} \) particles of a given substance. In the context of gases, the mole concept is used to express volumes under standard conditions. When you calculate \( \Delta n_{\text {gas}} \), you focus on how many moles of gas there are before and after a reaction. For a reaction, if one mole of \( \mathrm{CO}_2 \) gas is produced from a solid \( \mathrm{MgCO}_3 \), \( \Delta n_{\text {gas}} \) would be +1, showing an increase in moles of gas.

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

Le Châtelier's principle is related ultimately to the rates of the forward and reverse steps in a reaction. Explain (a) why an increase in reactant concentration shifts the equilibrium position to the right but does not change \(K ;\) (b) why a decrease in \(V\) shifts the equilibrium position toward fewer moles of gas but does not change \(K ;\) (c) why a rise in \(T\) shifts the equilibrium position of an exothermic reaction toward reactants and changes \(K ;\) and (d) why a rise in temperature of an endothermic reaction from \(T_{1}\) to \(T_{2}\) results in \(K_{2}\) being larger than \(K_{1}\)

A toxicologist studying mustard gas, \(\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2},\) a blistering agent, prepares a mixture of \(0.675 M \mathrm{SCl}_{2}\) and \(0.973 \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{4}\) and allows it to react at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) $$ \mathrm{SCl}_{2}(g)+2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(g) $$ At equilibrium, \(\left[\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}\right]=0.350 \mathrm{M}\). Calculate \(K_{\mathrm{p}}\)

The minerals hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) and magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)\) exist in equilibrium with atmospheric oxygen: \(4 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad K_{\mathrm{p}}=2.5 \times 10^{87}\) at \(298 \mathrm{~K}\) (a) Determine \(P_{\mathrm{O}_{2}}\) at equilibrium. (b) Given that \(P_{\mathrm{O}_{2}}\) in air is 0.21 atm, in which direction will the reaction proceed to reach equilibrium? (c) Calculate \(K_{c}\) at \(298 \mathrm{~K}\).

Nitrogen dioxide decomposes according to the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ where \(K_{\mathrm{p}}=4.48 \times 10^{-13}\) at a certain temperature. If \(0.75 \mathrm{~atm}\) of \(\mathrm{NO}_{2}\) is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g) ?\)

Hydrogen fluoride, \(\mathrm{HF}\), can be made by the reaction $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ In one experiment, \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) and \(0.050 \mathrm{~mol}\) of \(\mathrm{F}_{2}(g)\) are added to a \(0.50-\mathrm{L}\) flask. Write a reaction table for this process.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

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