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

Suppose 0.086 mol of Bra is placed in a 1.26-L. flask and heated to \(1756 \mathrm{K}\), a temperature at which the halogen 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 \(K_{c}\)

Short Answer

Expert verified
\( K_c \approx 0.000386 \text{ M} \)

Step by step solution

01

Determine initial concentrations

The initial concentration of \( \text{Br}_2 \) is calculated using the formula: \( [\text{Br}_2]_0 = \frac{0.086 \text{ mol}}{1.26 \text{ L}} \). This gives the initial concentration as \( \ [\text{Br}_2]_0 = 0.068 \text{ M} \). Since \( \mathrm{Br}_{2} \) is dissociating at the start, the initial concentration for \( \mathrm{Br} \) is zero.
02

Calculate change in concentration

The problem states that \( \mathrm{Br}_{2} \) is 3.7% dissociated. Calculate the concentration that dissociates: \( 0.068 \times 0.037 = 0.002516 \text{ M} \).
03

Determine equilibrium concentrations

Since 3.7% of \( \text{Br}_2 \) dissociates, the equilibrium concentration of \( \text{Br}_2 \) is: \[ [\text{Br}_2]_{eq} = 0.068 - 0.002516 = 0.065484 \text{ M} \] and for \( \text{Br} \), it is twice the dissociated amount (due to stoichiometry): \[ [\text{Br}]_{eq} = 2 \times 0.002516 = 0.005032 \text{ M}. \]
04

Write expression for equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) is given by: \[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \] In this chemical reaction, the stoichiometry of the bromine molecules gives us this relation as there are two \( \text{Br} \) atoms formed for every \( \text{Br}_2 \) molecule dissociated.
05

Substitute concentrations into \( K_c \) expression

Substitute the equilibrium concentrations into the \( K_c \) expression: \[ K_c = \frac{(0.005032)^2}{0.065484}. \] Calculate this value to find \( K_c \).
06

Calculate \( K_c \)

Compute the value: \[ K_c = \frac{0.000025320}{0.065484} \approx 0.000386 \text{ M}. \] This is the equilibrium constant for the dissociation of \( \text{Br}_2 \) at \( 1756 \text{ K} \).

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.

Chemical Equilibrium
Chemical equilibrium refers to a state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction. A system at equilibrium displays no net change in the concentration of reactants or products over time. This does not mean concentrations are equal, but they remain constant. Understanding equilibrium is essential as it allows us to predict how a reaction behaves when certain conditions are altered, such as temperature or pressure. When we speak about chemical equilibrium in the context of this problem, it involves the moment where the bromine gas (\(Br_2\)) dissociates to form bromine atoms (\(Br\)), and the reaction no longer experiences a net change in concentrations despite both processes—dissociation and recombination—continuing to occur at equal rates. By studying equilibrium, we can derive the equilibrium constant (\(K_c\)), which quantifies the ratio of product concentrations to reactant concentrations.
Dissociation Reactions
Dissociation reactions occur when a compound breaks down into its constituent particles, such as atoms or simpler molecules. For the reversible dissociation of bromine gas (\(Br_2 ightarrow 2Br\)), the extent of dissociation can be affected by factors like temperature. In the given problem, it is noteworthy that only 3.7% of the bromine molecules initially present dissociate. The conditions (e.g., high temperature of 1756 K) favor the breakdown of the diatomic bromine into monoatomic bromine molecules. It is crucial to understand that the percentage of dissociation helps in determining how much of the original compound remains compared to how much it has dissociated into its elements or simpler forms.
Concentration Calculations
Concentration calculations are essential for assessing how much of a reaction's participants are present at any time. Initially, you must calculate the concentration of reactants using the formula \([X]_0 = \frac{\text{moles of } X}{\text{volume of solution}}\). In the exercise, the initial concentration of \(Br_2\) is calculated by dividing its amount in moles by the volume of the flask, resulting in 0.068 M. When a certain percentage of this is dissociated (3.7% as given), we determine the exact change in concentration needed to reach equilibrium. The equilibrium concentrations are calculated by subtracting the dissociated amount from the initial for \(Br_2\) and adding twice that amount (due to the stoichiometric coefficient) for \(Br\). These calculations are key to finding the equilibrium concentrations and subsequently allow us to determine the equilibrium constant.
Stoichiometry
Stoichiometry involves the quantitative relationships of reactants and products in chemical reactions. It is a vital concept in determining how reactants convert to products according to their balanced equation. In dissociation reactions, such as the one involving bromine gas (\(Br_2 ightarrow 2Br\)), stoichiometry simplifies the task of predicting the equilibrium concentrations of the involved substances. The balanced equation tells us that one molecule of \(Br_2\) produces two atoms of \(Br\). Therefore, when considering the dissociation percentages and calculating equilibrium concentrations, we multiply the dissociated fraction by the stoichiometric ratios to determine the concentration of \(Br\). Stoichiometry ensures that our calculations remain consistent with the reaction's inherent requirements and laws of conservation, making it indispensable for accurate chemical analysis.

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

A limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=3.87 .\) If pure \(\mathrm{CaCO}_{3}\) is placed in a 5.00 -L flask and heated to \(1000^{\circ} \mathrm{C},\) what quantity of \(\mathrm{CaCO}_{3}\) must decompose to achieve the equilibrium pressure of \(\mathrm{CO}_{2} ?\)

The equilibrium constant for the butane \(\rightleftarrows\) isobutane isomerization reaction is 2.5 at \(25^{\circ} \mathrm{C}\). If 1.75 mol of butane and 1.25 mol of isobutane are mixed, is the system at equilibrium? If not, when it proceeds to equilibrium, which reagent increases in concentration? Calculate the concentrations of the two compounds when the system reaches equilibrium.

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \mathrm{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K_{c}\) is 0.190 at \(73^{\circ} \mathrm{C}\). Suppose you place \(0.500 \mathrm{mol}\) of \(\mathrm{COBr}_{2}\) in a 2.00 -L. flask and heat it to \(73^{\circ} \mathrm{C}\) (see Study Question 17 ). After equilibrium has been achieved, you add an additional 2.00 mol of CO. (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of \(\mathrm{COBr}_{2}\) \(\mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) (c) How has the addition of CO affected the percentage of \(\mathrm{COBr}_{2}\) that decomposed?

Equal numbers of moles of \(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask and heated to \(700^{\circ} \mathrm{C}\). The initial concentration of each gas is \(0.0088 \mathrm{mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) is consumed when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K_{c}\) for this reaction.

Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) The magnitude of the equilibrium constant is always independent of temperature. (b) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (c) The equilibrium constant for a reaction has the same value as \(K\) for the reverse reaction. (d) Only the concentration of \(\mathrm{CO}_{2}\) appears in the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) (e) For the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\) \(\mathrm{CO}_{2}(\mathrm{g}),\) the value of \(K\) is numerically the same, whether the amount of \(\mathrm{CO}_{2}\) is expressed as moles/liter or as gas pressure.

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