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) → Br₂(g) + BrF₃(l) with ΔH = -125.3 kJ2. 5 BrF(g) → 2 Br₂(g) + BrF₅(l) with ΔH = -166.1 kJ3. BrF₃(l) + F₂(g) → BrF₅(l) with ΔH = -158.0 kJ
02

- Identify the Target Reaction

Determine which reaction we need to find the enthalpy change for: BrF(g) → 1/2 Br₂(g) + 1/2 F₂(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 Br₂ and F₂.
04

- Combine and Adjust Reactions

Use the third reaction by reversing and scaling appropriately:BrF₃(l) → BrF(g) - F₂(g) with ΔH = 158.0 kJThen, use the first reaction divided by 3:1 BrF(g) → 1/3 Br₂(g) + 1/3 BrF₃(l) with ΔH = -125.3/3 kJ.Combine these two reactions to obtain:1 BrF(g) → 1/3 Br₂(g) + 1/3 BrF₃(l) → 1/3 Br₂(g) + 1/3( BrF(g) - F₂(g) ) Thus, BrF(g) → 1/3 Br₂(g) + 1/3 BrF(g) - 1/3 F₂(g) + 2/3 Br₂(g) → ...
05

- Calculate ΔH

Combining the ΔH values and reducing the equations gives: ΔH₁ + ΔH₂ = 1/3(125.3 kJ), ... giving ΔH = ... kJ.

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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 (Br₂) and bromine trifluoride (BrF₃) or pentafluoride (BrF₅). This is evident in the balanced reactions provided:
  • 3 BrF(g) → Br₂(g) + BrF₃(l)
  • 5 BrF(g) → 2 Br₂(g) + BrF₅(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 (◌H) 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 ◌H 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 (ΔH) for BrF decomposing into its elements: BrF(g) → 1/2 Br₂(g) + 1/2 F₂(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 (ΔH). It is essential to keep track of the sign and magnitude of ΔH 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) → 1/2 Br₂(g) + 1/2 F₂(g).
  • Manipulate the provided equations using algebraic methods—reverse 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.

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Most popular questions from this chapter

Two substances with the empirical formula HNO are hyponitrous acid \((\mathscr{A}=62.04 \mathrm{~g} / \mathrm{mol})\) and nitroxyl \((\mathscr{A}=31.02 \mathrm{~g} / \mathrm{mol}) .\) (a) What is the molecular formula of each species? (b) For each species, draw the Lewis structure having the lowest formal charges. (Hint: Hyponitrous acid has an \(\mathrm{N}=\mathrm{N}\) bond. \()\) (c) Predict the shape around the \(\mathrm{N}\) atoms of each species. (d) When hyponitrous acid loses two protons, it forms the hyponitrite ion. Draw cis and trans forms of this ion.

How do atomic properties account for the low densities of the Group \(1 \mathrm{~A}(1)\) elements?

Each of the following properties shows a regular trend in Group \(1 \mathrm{~A}(1) .\) Predict whether each increases or decreases up the group: (a) Melting point (b) \(\mathrm{E}-\mathrm{E}\) bond length (c) Hardness (d) Molar volume (e) Lattice energy of EBr

What is a disproportionation reaction, and which of the following fit the description? (a) \(\mathrm{I}_{2}(s)+\mathrm{KI}(a q) \longrightarrow \mathrm{KI}_{3}(a q)\) (b) \(2 \mathrm{ClO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{HClO}_{3}(a q)+\mathrm{HClO}_{2}(a q)\) (c) \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{NaClO}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2}(s) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (e) \(3 \mathrm{MnO}_{4}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$$$ 2 \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)+4 \mathrm{OH}^{-}(a q)$$ (f) \)3 \mathrm{AuCl}(s) \longrightarrow \mathrm{AuCl}_{3}(s)+2 \mathrm{Au}(s)$

An industrial chemist treats solid \(\mathrm{NaCl}\) with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and obtains gaseous \(\mathrm{HCl}\) and \(\mathrm{NaHSO}_{4}\). When she substitutes solid NaI for \(\mathrm{NaCl}\), she obtains gaseous \(\mathrm{H}_{2} \mathrm{~S},\) solid \(\mathrm{I}_{2},\) and \(\mathrm{S}_{8},\) but no \(\mathrm{HI}\). (a) What type of reaction did the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) undergo with NaI? (b) Why does NaI, but not \(\mathrm{NaCl}\), cause this type of reaction? (c) To produce HI \((g)\) by reacting NaI with an acid, how does the acid have to differ from sulfuric acid?

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