/*! 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 122 Bromine monofluoride (BrF) dispr... [FREE SOLUTION] | 91影视

91影视

Bromine monofluoride (BrF) disproportionates to bromine gas and bromine trifluoride or pentafluoride. Use the following to find \(\Delta H_{\mathrm{rxn}}^{0}\) for the decomposition of BrF to its elements: $$\begin{aligned}3 \mathrm{BrF}(g) & \longrightarrow \mathrm{Br}_{2}(g)+\mathrm{BrF}_{3}(l) & & \Delta H_{\mathrm{rn}}=-125.3 \mathrm{~kJ} \\\5 \mathrm{BrF}(g) & \longrightarrow 2 \mathrm{Br}_{2}(g)+\mathrm{BrF}_{5}(l) & & \Delta H_{\mathrm{rn}}=-166.1 \mathrm{~kJ} \\\\\mathrm{BrF}_{3}(I)+\mathrm{F}_{2}(g) & \longrightarrow \mathrm{BrF}_{5}(l) & & \Delta H_{\mathrm{rn}}=-158.0 \mathrm{~kJ}\end{aligned}$$

Short Answer

Expert verified
Delta H is calculated as exact values and calculated ones.

Step by step solution

01

- Understand the Reactions Given

First, note the three reactions provided and their respective enthalpy changes: 1. 3 BrF(g) 鈫 Br鈧(g) + BrF鈧(l) with 螖H = -125.3 kJ2. 5 BrF(g) 鈫 2 Br鈧(g) + BrF鈧(l) with 螖H = -166.1 kJ3. BrF鈧(l) + F鈧(g) 鈫 BrF鈧(l) with 螖H = -158.0 kJ
02

- Identify the Target Reaction

Determine which reaction we need to find the enthalpy change for: BrF(g) 鈫 1/2 Br鈧(g) + 1/2 F鈧(g). This is the decomposition of BrF into its elements.
03

- Manipulate the Given Reactions

Adjust the given reactions to find the enthalpy of the decomposition. In order to get the decomposition reaction, the goal is to find a combination of reactions that lead to BrF decomposing into Br鈧 and F鈧.
04

- Combine and Adjust Reactions

Use the third reaction by reversing and scaling appropriately:BrF鈧(l) 鈫 BrF(g) - F鈧(g) with 螖H = 158.0 kJThen, use the first reaction divided by 3:1 BrF(g) 鈫 1/3 Br鈧(g) + 1/3 BrF鈧(l) with 螖H = -125.3/3 kJ.Combine these two reactions to obtain:1 BrF(g) 鈫 1/3 Br鈧(g) + 1/3 BrF鈧(l) 鈫 1/3 Br鈧(g) + 1/3( BrF(g) - F鈧(g) ) Thus, BrF(g) 鈫 1/3 Br鈧(g) + 1/3 BrF(g) - 1/3 F鈧(g) + 2/3 Br鈧(g) 鈫 ...
05

- Calculate 螖H

Combining the 螖H values and reducing the equations gives: 螖H鈧 + 螖H鈧 = 1/3(125.3 kJ), ... giving 螖H = ... kJ.

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 91影视!

Key Concepts

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

Disproportionation Reaction
A disproportionation reaction is a type of redox reaction where a single substance is simultaneously oxidized and reduced, forming two different products. In this exercise, Bromine monofluoride (BrF) undergoes disproportionation to form bromine gas (Br鈧) and bromine trifluoride (BrF鈧) or pentafluoride (BrF鈧). This is evident in the balanced reactions provided:
  • 3 BrF(g) 鈫 Br鈧(g) + BrF鈧(l)
  • 5 BrF(g) 鈫 2 Br鈧(g) + BrF鈧(l)
Understanding the concept of disproportionation is crucial because it helps in balancing redox reactions and calculating the associated energy changes. By noting how BrF is both oxidized and reduced, we can further comprehend the enthalpy calculations required in these reactions.
Enthalpy Calculations
Enthalpy (鈼孒) is a measure of the total heat content in a system. It is crucial for predicting the feasibility and direction of chemical reactions, especially those involving heat exchange. In our specific problem, we are interested in finding 鈼孒 of the given reactions for the decomposition of BrF. This involves manipulating the equations to derive the target reaction. Start by considering the given reactions and recognizing that we need to find enthalpy change (螖H) for BrF decomposing into its elements: BrF(g) 鈫 1/2 Br鈧(g) + 1/2 F鈧(g).The process involves reversing and scaling reactions to match the target decomposition equation. For example, dividing and reversing the given reactions, adding their enthalpy changes properly, and finally arriving at the required enthalpy change (螖H). It is essential to keep track of the sign and magnitude of 螖H when reactions are reversed or scaled.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. This process is governed by conservation laws like mass and energy balance. In the context of our exercise, we deal with multiple reactions involving BrF decomposing into various products. Understanding how to balance these reactions and combining them to derive new reactions is key to solving problems involving chemical synthesis.
  • Identify the target reaction: BrF(g) 鈫 1/2 Br鈧(g) + 1/2 F鈧(g).
  • Manipulate the provided equations using algebraic methods鈥攔everse and scale reactions if necessary.
  • Combine these equations to eliminate intermediate products and get the desired outcome.By performing these steps, students can solve enthalpy-related problems efficiently and understand the underlying principles governing chemical reactions. Mastery of these concepts enables deeper insights into reaction mechanisms and energy changes during chemical transformations.

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

The oxygen and nitrogen families have some obvious similarities and differences. (a) State two general physical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (b) State two general chemical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (c) State two chemical similarities between \(\mathrm{P}\) and \(\mathrm{S}\). (d) State two physical similarities between \(\mathrm{N}\) and \(\mathrm{O}\). (e) State two chemical differences between \(\mathrm{N}\) and \(\mathrm{O}\).

Bismuth(V) compounds are such powerful oxidizing agents that they have not been prepared in pure form. How is this fact consistent with the location of Bi in the periodic table?

The bond angles in the nitrite ion, nitrogen dioxide, and the nitronium ion \(\left(\mathrm{NO}_{2}^{+}\right)\) are \(115^{\circ}, 134^{\circ},\) and \(180^{\circ},\) respectively. Explain these values using Lewis structures and VSEPR theory.

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right),\) the "laughing gas" used as an anesthetic by dentists, is made by thermal decomposition of solid \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). Write a balanced equation for this reaction. What are the oxidation states of \(\mathrm{N}\) in \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) and in \(\mathrm{N}_{2} \mathrm{O} ?\)

Give the name and symbol or formula of a Group \(6 \mathrm{~A}(16)\) element or compound that fits each description or use: (a) Unstable allotrope of oxygen (b) Oxide having sulfur with the same O.N. as in sulfuric acid (c) Air pollutant produced by burning sulfur-containing coal (d) Powerful dehydrating agent (e) Compound used in solution in the photographic process

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.