/*! 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 88 A vessel originally contained \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A vessel originally contained \(0.0200 \mathrm{~mol}\) iodine monobromide (IBr), \(0.050 \mathrm{~mol} \mathrm{I}_{2}\), and \(0.050 \mathrm{~mol} \mathrm{Br}_{2}\). The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ is \(1.2 \times 10^{2}\) at \(150^{\circ} \mathrm{C}\). What is the direction (forward or reverse) needed to attain equilibrium at \(150^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The reaction will proceed forward to reach equilibrium.

Step by step solution

01

Write the Reaction Quotient

The reaction quotient, \( Q_c \), is calculated using the initial concentrations of the reactants and products. The expression is given by: \[ Q_c = \frac{[\text{IBr}]^2}{[\text{I}_2][\text{Br}_2]} \].
02

Calculate Initial Concentrations

The initial concentrations are given as \( [\text{IBr}] = 0.0200 \text{ mol} \), \( [\text{I}_2] = 0.050 \text{ mol} \), and \( [\text{Br}_2] = 0.050 \text{ mol} \). Assuming the volume is consistent, these mole amounts will be directly used in the calculation of \( Q_c \).
03

Compute the Reaction Quotient (Qc)

Substitute the initial concentrations into the \( Q_c \) expression: \[ Q_c = \frac{(0.0200)^2}{(0.050)(0.050)} \]. This calculates to \( Q_c = \frac{0.0004}{0.0025} = 0.16 \).
04

Compare Qc and Kc

The value of \( Q_c \) is 0.16 and the equilibrium constant \( K_c \) is given as 120. If \( Q_c < K_c \), the reaction will proceed in the forward direction to achieve equilibrium.

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Key Concepts

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

Reaction Quotient (Qc)
Understanding the reaction quotient, denoted as \( Q_c \), is crucial in predicting the direction in which a chemical reaction will proceed. The reaction quotient is similar to the equilibrium constant \( K_c \), but it uses the initial concentrations rather than the concentrations at equilibrium. For the given reaction:\[ \mathrm{I}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) \] \( Q_c \) is expressed as:
  • \( Q_c = \frac{[\text{IBr}]^2}{[\text{I}_2][\text{Br}_2]} \)
This formula comes from the law of mass action, which allows us to calculate \( Q_c \) using the initial amounts of each substance. By understanding \( Q_c \), students can determine how the initial system compares to the equilibrium state and thus predict the reaction's behavior. The calculated \( Q_c = 0.16 \) indicates how the initial products and reactants are distributed before equilibrium is achieved.
Equilibrium Constant (Kc)
The equilibrium constant, denoted as \( K_c \), serves as a cornerstone in studying chemical equilibria. It reflects the ratio of product to reactant concentrations at static equilibrium. It is a fixed value at a particular temperature, depicting the reaction's innate capacity to reach a state where the rates of forward and reverse reactions are equal.For our specific reaction, the given \( K_c \) is \( 1.2 \times 10^{2} \). This large value suggests a position of equilibrium which strongly favors the formation of products, meaning at equilibrium, there are more products than reactants. Understanding \( K_c \) helps in:
  • Predicting extent of a reaction: Large \( K_c \) implies product favorability.
  • Understanding the system’s position at equilibrium.
Through \( K_c \), students see how the reaction naturally stabilizes in terms of the concentration of substances.
Direction of Reaction
Determining the direction of a reaction to reach equilibrium involves comparing \( Q_c \) with \( K_c \). This comparison indicates whether the system will shift to favor products or reactants.- If \( Q_c < K_c \), the reaction moves in the forward direction to produce more products, achieving equilibrium.- If \( Q_c > K_c \), the reaction shifts in the reverse direction to produce more reactants.- If \( Q_c = K_c \), the system is already at equilibrium.In this exercise, with \( Q_c = 0.16 \) and \( K_c = 120 \), we find that \( Q_c < K_c \). Thus, the reaction will proceed in the forward direction, creating more \( \text{IBr} \) to reach equilibrium.By understanding these principles, students gain insight into reaction dynamics and equilibrium adjustments in response to initial conditions.

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Most popular questions from this chapter

Iodine, \(\mathrm{I}_{2},\) and bromine, \(\mathrm{Br}_{2}\), react to produce iodine monobromide, IBr. $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ A starting mixture of \(0.5000 \mathrm{~mol} \mathrm{I}_{2}\) and \(0.5000 \mathrm{~mol} \mathrm{Br}_{2}\) reacts at \(150^{\circ} \mathrm{C}\) to produce \(0.4221 \mathrm{~mol} \mathrm{IBr}\) at equilibrium. What would be the equilibrium composition (in moles) of a mixture that starts with \(1.000 \mathrm{~mol} \mathrm{I}_{2}\) and \(2.000 \mathrm{~mol} \mathrm{Br}_{2}\) ?

When 0.0322 mol of \(\mathrm{NO}\) and \(1.52 \mathrm{~g}\) of bromine are placed in a 1.00 - \(\mathrm{L}\) reaction vessel and sealed, the mixture reacts and the following equilibrium is established: $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ At \(25^{\circ} \mathrm{C}\) the equilibrium pressure of nitrosyl bromide is \(0.438 \mathrm{~atm} .\) What is \(K_{p} ?\)

The following reaction has an equilibrium constant \(K_{c}\) equal to 3.59 at \(900^{\circ} \mathrm{C}\) $$ \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) $$ For each of the following compositions, decide whether the reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go. a \(\left[\mathrm{CH}_{4}\right]=1.26 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.32 M,\left[\mathrm{CS}_{2}\right]=1.43 \mathrm{M}\) \(\left[\mathrm{H}_{2}\right]=1.12 \mathrm{M}\) b) \(\left[\mathrm{CH}_{4}\right]=1.25 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.52 M,\left[\mathrm{CS}_{2}\right]=1.15 M\) \(\left[\mathrm{H}_{2}\right]=1.73 \mathrm{M}\) c \(\left[\mathrm{CH}_{4}\right]=1.30 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.41 M,\left[\mathrm{CS}_{2}\right]=1.10 \mathrm{M}\) \(\left[\mathrm{H}_{2}\right]=1.20 \mathrm{M}\) d) \(\left[\mathrm{CH}_{4}\right]=1.56 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.43 \mathrm{M},\left[\mathrm{CS}_{2}\right]=1.23 \mathrm{M}\) \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M}\)

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{[\mathrm{NO}]^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

Iodine and bromine react to give iodine monobromide, IBr $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ What is the equilibrium composition of a mixture at \(145^{\circ} \mathrm{C}\) that initially contained 0.0019 mol each of iodine and bromine in a 5.0-L vessel? The equilibrium constant \(K_{c}\) for this reaction at \(145^{\circ} \mathrm{C}\) is 108 .

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