Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Many common nonmetallic elements exist as diatomic molecules at room temperature. When these elements are heated to \(1500 . \mathrm{K},\) the molecules break apart into atoms. A general equation for this type of reaction is \(\mathrm{E}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{E}(\mathrm{g})\) where E stands for an atom of each element. Equilibrium constants for dissociation of these molecules at \(1500 . \mathrm{K}\) are \begin{tabular}{lcll} \hline Species & \(\kappa_{c}\) & Species & \multicolumn{1}{c} {\(K_{c}\)} \\ \hline \(\mathrm{Br}_{2}\) & \(8.9 \times 10^{-2}\) & \(\mathrm{H}_{2}\) & \(3.1 \times 10^{-10}\) \\ \(\mathrm{Cl}_{2}\) & \(3.4 \times 10^{-3}\) & \(\mathrm{~N}_{2}\) & \(1 \times 10^{-27}\) \\ \(\mathrm{~F}_{2}\) & 7.4 & \(\mathrm{O}_{2}\) & \(1.6 \times 10^{-11}\) \\ \hline \end{tabular} (a) Assume that \(1.00 \mathrm{~mol}\) of each diatomic molecule is placed in a separate \(1.0-\mathrm{L}\) container, sealed, and heated to \(1500 . \mathrm{K}\). Calculate the equilibrium concentration of the atomic form of each element at \(1500 . \mathrm{K}\). (b) From these results, predict which of the diatomic elements has the lowest bond dissociation energy, and compare your results with thermochemical calculations and with Lewis structures.

Short Answer

Expert verified
Fluorine (F₂) has the lowest bond dissociation energy.

Step by step solution

01

Understand the reaction

For the general reaction \( \mathrm{E}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{E}(\mathrm{g}) \), the equilibrium constant \( K_c \) is defined in terms of concentrations as \( K_c = \frac{[E]^2}{[E_2]} \). At initial time, the concentration of \( E_2 \) is 1.00 mol/L, and \( E \) is 0.
02

Define variables for changes in concentration

Let \( x \) be the concentration of \( E \) at equilibrium. Then the equilibrium concentration of \( E_2 \) will be \( 1.00 - \frac{x}{2} \), because each \( E_2 \) molecule produces 2 \( E \) atoms.
03

Write the equilibrium expression

The equilibrium expression for the reaction is \( K_c = \frac{x^2}{1 - x/2} \). We will solve this equation to find \( x \) for each diatomic molecule.
04

Solve for each species

For each species, substitute \( K_c \) from the table into the equation and solve for \( x \):- **Bromine (Br₂):** \( K_c = 8.9 \times 10^{-2} = \frac{x^2}{1 - \frac{x}{2}} \)- **Chlorine (Cl₂):** \( K_c = 3.4 \times 10^{-3} = \frac{x^2}{1 - \frac{x}{2}} \)- **Fluorine (F₂):** \( K_c = 7.4 = \frac{x^2}{1 - \frac{x}{2}} \)- **Hydrogen (H₂):** \( K_c = 3.1 \times 10^{-10} = \frac{x^2}{1 - \frac{x}{2}} \)- **Nitrogen (N₂):** \( K_c = 1 \times 10^{-27} = \frac{x^2}{1 - \frac{x}{2}} \)- **Oxygen (O₂):** \( K_c = 1.6 \times 10^{-11} = \frac{x^2}{1 - \frac{x}{2}} \)
05

Calculate x for each element

Upon solving each equation, approximate values for \( x \), representing equilibrium concentration of \( E \), will be:- **Br: ** \( x \approx 0.27 \, \text{mol/L} \)- **Cl: ** \( x \approx 0.11 \, \text{mol/L} \)- **F: ** \( x \approx 1.70 \, \text{mol/L} \)- **H: ** \( x \approx \text{very small} \)- **N: ** \( x \approx \text{effectively zero} \)- **O: ** \( x \approx \text{very small} \)
06

Predict bond dissociation energy

From the calculated values, \( F_2 \) has the highest equilibrium concentration of \( E \) at \( 1500 \text{ K} \), suggesting it has the lowest bond dissociation energy. Generally, lower bond energies correspond to higher extent of dissociation at equilibrium at high temperatures. This finding can be compared with known thermochemical data and considering Lewis structures.

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!

Key Concepts

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

Diatomic Molecules
Diatomic molecules are a specific type of molecule composed of only two atoms, either of the same or different chemical elements. Many common nonmetallic elements exist in this diatomic form under standard conditions, such as oxygen (\(O_2)\), nitrogen (\(N_2)\), and hydrogen (\(H_2)\).
These molecules are significant in chemistry because they represent the simplest molecular forms of elements that are essential for life and industrial processes.
At high temperatures, like \(1500 \text{ K}\), these diatomic molecules can dissociate into their atomic forms. For example, the general reaction for a diatomic molecule \(E_2\) disassociating into two atoms of \(E\) is represented as:- \(E_2(g) \rightleftharpoons 2E(g)\)
This type of reaction emphasizes the dynamic nature of chemical equilibria when diatomic molecules break apart and re-form under different conditions.
Understanding diatomic molecules and their behavior at different temperatures helps us grasp broader concepts in chemical equilibrium and molecular chemistry.
Equilibrium Constant
The equilibrium constant (\(K_c)\) is a fundamental concept in chemical equilibrium that quantifies the ratio of product concentrations to reactant concentrations at equilibrium.
For the dissociation of diatomic molecules \(E_2\) into atoms \(E\), \(K_c\) is expressed as:- \(K_c = \frac{[E]^2}{[E_2]}\)
Here, the concentration of products (\(E\) atoms) is squared because two atoms are produced from each diatomic molecule.
The larger the \(K_c\), the more the reaction favors the dissociation of diatomic molecules into atoms at equilibrium. In contrast, a smaller \(K_c\) implies minimal dissociation and a stronger preference for the diatomic form.
By comparing the \(K_c\) values of different diatomic molecules, one can infer how readily they dissociate at high temperatures. This, in turn, provides insights into their bond strength and stability.
Bond Dissociation Energy
Bond dissociation energy (BDE) is the energy required to break a chemical bond between two atoms in a molecule. It gives an indication of the bond's strength.
In the context of diatomic molecules, BDE directly correlates with the extent to which a molecule disassociates at high temperatures.
  • A low BDE means less energy is required to break the bond, resulting in a higher equilibrium concentration of atomic species.
  • A high BDE suggests that more energy is needed, therefore fewer atoms are present at equilibrium.
Understanding BDE is crucial for predicting chemical behavior during reactions. For instance, \(F_2\) with the highest equilibrium concentration of atoms at \(1500 \text{ K}\) indirectly indicates the lowest BDE among the tested diatomic molecules.
Comparing BDE helps chemists ascertain why certain molecules readily break apart under specific conditions and aids in designing reactions to improve yield and efficiency.

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 decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) is endothermic. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ (a) Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. (b) If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a sealed flask in which this equilibrium exists, how is the equilibrium affected? (c) What if some additional \(\mathrm{NH}_{3}\) is placed in a sealed flask containing an equilibrium mixture? (d) What will happen to the partial pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{~S}\) is removed from the flask?

Write a chemical equation for an equilibrium system that would lead to each expression \((\mathrm{a}-\mathrm{c})\) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2} \mathrm{~S}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{2}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{\left(P_{\mathrm{IF}}\right)}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{Cl}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\)

Define the terms chemical equilibrium and dynamic equilibrium.

Write equilibrium constant expressions for these reactions. For gases, use either pressures or concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{~g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{~g})\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)

Suppose \(0.086 \mathrm{~mol} \mathrm{Br}_{2}\) is placed in a 1.26-L flask. The flask is sealed and heated to \(1756 \mathrm{~K}\), a temperature at which the \(\mathrm{Br}_{2}\) dissociates to atoms $$ \mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Br}(\mathrm{g}) $$ If \(\mathrm{Br}_{2}\) is \(3.7 \%\) dissociated at this temperature, calculate \(K_{c}\)

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