Consider the following reaction: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} \quad \Delta E=-285 \mathrm{kJ} $$ The bond energy for \(A_{2}\) is one-half the amount of the AB bond energy. The bond energy of \(\mathbf{B}_{2}=432 \mathrm{kJ} / \mathrm{mol} .\) What is the bond energy of \(\mathrm{A}_{2} ?\)

Short Answer

Expert verified
The bond energy of A₂ is 49 kJ/mol.

Step by step solution

01

Write down the given information

Here are the information given in the problem: - A₂ + B₂ ⟶ 2AB, ∆E = -285 kJ - Bond energy of B₂ = 432 kJ/mol - Bond energy of A₂ = (1/2) × Bond energy of AB
02

Calculate the total bond energy of the products

Assuming that the bond energy of AB is x kJ/mol, the total bond energy of products, considering the stoichiometric coefficients, will be 2x kJ.
03

Calculate the total bond energy of the reactants

The bond energy of A₂ is (1/2) x, and the bond energy of B₂ is 432 kJ/mol. Therefore, the total bond energy of reactants is (1/2)x + 432 kJ.
04

Apply Hess's Law

Hess's law states that the energy change for the reaction (∆E) is equal to the total bond energy of products minus the total bond energy of reactants. In this case, we have: -285 kJ = (2x kJ) - [(1/2)x + 432 kJ]
05

Solve for x (bond energy of AB)

Rearrange the equation to solve for x: -285 kJ + 432 kJ = (2x kJ) - (1/2)x 147 kJ = (3/2)x x = 98 kJ/mol The bond energy of the AB molecule is 98 kJ/mol.
06

Find the bond energy of A₂

Using the relationship given, we can now find the bond energy of A₂: Bond energy of A₂ = (1/2) × Bond energy of AB Bond energy of A₂ = (1/2) × 98 kJ/mol Bond energy of A₂ = 49 kJ/mol So, the bond energy of A₂ is 49 kJ/mol.

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 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

Write Lewis structures for the following. Show all resonancestructures where applicable.a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{N}_{2} \mathrm{O}_{4} \text { exists as } \mathrm{O}_{2} \mathrm{N}-\mathrm{NO}_{2} .\right)\) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{SCN}^{-} .\) )

Name each of the following compounds. Assume the acids are dissolved in water. a. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) c. \(\mathrm{Co}_{2} \mathrm{S}_{3}\) d. ICl e. \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) f. \(\mathrm{KClO}_{3}\) g. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(\mathbf{h} . \mathrm{Sr}_{3} \mathrm{N}_{2}\) i. \(\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{3}\right)_{3}\) J. \(\mathrm{SnO}_{2}\) \(\mathbf{k} . \mathrm{Na}_{2} \mathrm{CrO}_{4}\) I. \(\mathrm{HClO}\)

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. Al and Cl b. Na and O c. Sr and F d. Ca and Se

Rationalize the following lattice energy values: $$\begin{array}{|lc|} \hline & \text { Lattice Energy } \\ \text { Compound } & \text { (kj/mol) } \\ \hline \text { CaSe } & -2862 \\ \text { Na }_{2} \text { Se } & -2130 \\ \text { CaTe } & -2721 \\ \text { Na }_{2} \text { Te } & -2095 \\ \hline \end{array}$$

Name each of the following compounds: a. CsF b. \(\mathrm{Li}_{3} \mathrm{N}\) c. \(A g_{2} S\) d. \(\mathrm{MnO}_{2}\) e. \(\mathrm{TiO}_{2}\) f. \(\mathrm{Sr}_{3} \mathrm{P}_{2}\)

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.

Sign-up for free