/*! 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 4 In the reaction \(\mathrm{BaO}_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the reaction \(\mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=+\mathrm{ve} .\) In equilibrium condition, pressure of \(\mathrm{O}_{2}\) depends on (a) increase mass of \(\mathrm{BaO}_{2}\) (b) increase mass of \(\mathrm{BaO}\) (c) temperature of equilibrium (d) mass of \(\mathrm{BaO}_{2}\) and \(\mathrm{BaO}\) both

Short Answer

Expert verified
(c) The pressure of \( \text{O}_2 \) depends on the temperature of equilibrium.

Step by step solution

01

Assessing Effects of Mass Changes

First, consider how changes in the mass of the solid reactants and products affect the equilibrium. In this reaction, both \( \text{BaO}_2 \) and \( \text{BaO} \) are solids. According to Le Chatelier's Principle, changes in the mass of solid substances do not affect the equilibrium position of gaseous products. Therefore, options (a), (b), and (d) are not viable since altering the mass of either solid reactant or product will not affect the pressure of \( \text{O}_2 \).
02

Examining the Role of Temperature

Given that the enthalpy change \( \Delta H \) is positive, the reaction is endothermic. According to Le Chatelier's Principle, increasing the temperature will shift the equilibrium towards the product side to absorb the additional heat. This results in an increase in the formation of \( \text{O}_2 \), hence increasing the pressure of \( \text{O}_2 \) in the equilibrium mixture.
03

Conclusion from Analysis

Based on the analysis, the pressure of \( \text{O}_2 \) at equilibrium is affected by changes in temperature because it shifts the equilibrium position of the endothermic reaction forwards, increasing \( \text{O}_2 \) formation.

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.

Le Chatelier's Principle
Le Chatelier's Principle provides insights into how a chemical equilibrium adjusts to changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. For example, if you alter the temperature, pressure, or concentration of the reactants or products, the equilibrium adjusts to minimize that change.

In the case of the given reaction, \(\mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\), we're dealing with a situation where only the pressure of the gaseous product \( \mathrm{O}_2 \) can shift in response to external changes. If temperature increases, the equilibrium will shift to produce more \( \mathrm{O}_2 \), minimizing the change in heat.

Remember, changes in the mass of solids do not affect the equilibrium of gases, since solids are not part of the expression for the equilibrium law. This means options involving changes in the mass of solid reactants or products, like \(\text{BaO}_2\) and \(\text{BaO}\), have no influence on the equilibrium position.
Phase Changes in Reactions
Understanding phase changes in a chemical reaction is crucial for predicting how equilibrium will respond. In reactions where different states of matter are present—solids, liquids, and gases—only gases significantly impact the equilibrium state.

In our reaction, \( \mathrm{BaO}_{2} \) and \( \mathrm{BaO} \) are solids, and \( \mathrm{O}_{2} \) is a gas. Solids do not enter the equilibrium expression, so any change in their mass doesn't shift the equilibrium position. However, changes in the conditions affecting \( \mathrm{O}_{2} \), such as temperature, directly impact the equilibrium.

These phase distinctions help in predicting the behavior of reactions under different conditions and understanding why only gaseous molecules impact the equilibrium calculations.
Endothermic Reactions
An endothermic reaction is one that absorbs heat from its surroundings. It has a positive enthalpy change, denoted as \( \Delta H = +\text{ve} \). This means that the reaction requires energy input to proceed. When the temperature increases in an endothermic reaction, according to Le Chatelier’s Principle, the equilibrium will shift to absorb the extra heat supplied.

In the context of our reaction, \( \text{BaO}_2(\text{s}) \rightleftharpoons \text{BaO}(\text{s}) + \text{O}_2(\text{g}) \), as the temperature is raised, more \( \text{O}_2 \) is produced to absorb the additional heat. This increase in \( \text{O}_2 \) is reflected as an increase in its pressure within the reaction vessel at equilibrium.

Understanding why temperature influences endothermic reactions in this way is key to predicting and controlling chemical processes that involve heat absorption, such as those used in various industrial applications.

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 what manner will increase of pressure affect the following equation? \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) shift in the reverse direction (b) shift in the forward direction (c) increase in the yield of hydrogen (d) no effect

For the reaction, \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\), the equilibrium concentrations of \(\mathrm{H}_{2}, 1_{2}\) and \(\mathrm{HI}\) are 8,3 and \(28 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively. Equilibrium constant of the reaction is (a) \(32.67\) (b) \(31.67\) (c) \(34.67\) (d) \(36.67\)

If \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) \(\mathrm{K}_{\mathrm{c}}=2.5 \times 10^{30}\) \(\mathrm{NO}+\frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) \(\mathrm{K}_{c_{2}}=1.6\) find \(K_{c}\) for the reaction given below \(\frac{1}{2} \mathrm{~N}_{8}+\frac{1}{2} \mathrm{O}_{8}+\frac{1}{2} \mathrm{Br}_{\mathrm{g}} \rightleftharpoons \mathrm{NOBr}\) (a) \(1.01 \times 10^{-15}\) (b) \(2.02 \times 10^{-15}\) (c) \(1.01 \times 10^{30}\) (d) \(2.02 \times 10^{15}\)

If an inert gas is added in the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\) at constant volume, then its equilibrium (a) remains unaffected (b) favours the backward reaction (c) favours the forward reaction (d) increases the dissociation of reactants

The role of a catalyst in a reversible reaction is to (a) alter the equilibrium constant of the reaction (b) increase the rate of forward reaction (c) allow the equilibrium to be achieved quickly (d) decrease the rate of backward reaction

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.