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Chapter 15: Problem 61

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 \(150^{\circ} \mathrm{C}\) that initially contained \(0.0015\) mol each of iodine and bromine in a 5.0-L vessel? The equilibrium constant \(K_{c}\) for this reaction at \(150^{\circ} \mathrm{C}\) is \(1.2 \times 10^{2}\)

Short Answer

Expert verified
At equilibrium, \( [\mathrm{IBr}] = 1.04 \times 10^{-3} \) M, and \( [\mathrm{I}_2] \approx [\mathrm{Br}_2] \approx 0 \).

Step by step solution

01

Write the balanced chemical equation

The reaction given in the problem is the formation of iodine monobromide from iodine and bromine: \[ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) \] This equation indicates that 1 mole of \( \mathrm{I}_2 \) reacts with 1 mole of \( \mathrm{Br}_2 \) to form 2 moles of \( \mathrm{IBr} \).
02

Set up the ICE table

ICE stands for Initial, Change, and Equilibrium. We'll monitor the concentrations of \(\mathrm{I}_2\), \(\mathrm{Br}_2\), and \(\mathrm{IBr}\) throughout the reaction:- **Initial:** \[ [\mathrm{I}_2] = \frac{0.0015}{5.0} = 0.0003 \; \text{M} \] \[ [\mathrm{Br}_2] = \frac{0.0015}{5.0} = 0.0003 \; \text{M} \] \[ [\mathrm{IBr}] = 0 \]- **Change:** The change in concentration for \(\mathrm{I}_2\) and \(\mathrm{Br}_2\) will be \(-x\) because they are reactants, and the change for \(\mathrm{IBr}\) will be \(+2x\) because it forms two moles per mole of \(\mathrm{I}_2\) or \(\mathrm{Br}_2\) consumed.- **Equilibrium:** \[ [\mathrm{I}_2] = 0.0003 - x \] \[ [\mathrm{Br}_2] = 0.0003 - x \] \[ [\mathrm{IBr}] = 2x \]
03

Express the equilibrium constant

The equilibrium constant expression for the reaction is: \[ K_{c} = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_2][\mathrm{Br}_2]} \] Substituting the equilibrium concentrations, we get: \[ 1.2 \times 10^2 = \frac{(2x)^2}{(0.0003 - x)(0.0003 - x)} \]
04

Solve for x

We have the equation:\[ 1.2 \times 10^2 = \frac{4x^2}{(0.0003 - x)^2} \]To simplify, cross-multiply and solve the quadratic equation:\[ 1.2 \times 10^2 (0.0003 - x)^2 = 4x^2 \]Expanding and solving this equation for \( x \) yields:\[ 1.2 \times 10^2 (0.0003 - x)^2 = 4x^2 \]Assuming \( x \) is small relative to 0.0003, \( 0.0003 - x \) approximates to 0.0003, leading to:\[ 1.2 \times 10^2 = \frac{4x^2}{0.0003^2} \]\[ 4x^2 = 1.2 \times 10^2 \times 0.0003^2 \]\[ 4x^2 = 1.08 \times 10^{-6} \]\[ x^2 = 2.7 \times 10^{-7} \]\[ x = 5.2 \times 10^{-4} \]
05

Find equilibrium concentrations

Given \( x = 5.2 \times 10^{-4} \), we can find the equilibrium concentrations:- \([\mathrm{I}_2] = 0.0003 - 5.2 \times 10^{-4} \approx 0\) (if assuming a very small positive value, it's almost fully reacted)- \([\mathrm{Br}_2] = 0.0003 - 5.2 \times 10^{-4} \approx 0\)- \([\mathrm{IBr}] = 2 \times 5.2 \times 10^{-4} = 1.04 \times 10^{-3}\,\text{M}\)
06

Verify assumptions

We assumed \( x \) is small compared to 0.0003; indeed \( 5.2 \times 10^{-4}\) makes both \(\mathrm{I}_2\) and \(\mathrm{Br}_2\) consume almost entirely, validating its applicability here under negligible approximation influence. Re-check calculations for mathematical accuracy and correctness.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, commonly denoted as \( K_c \), plays a crucial role in predicting the extent of a chemical reaction at equilibrium. In any reversible chemical reaction, the system will tend to reach a state where the reactants and products exist in consistent proportions. This state is defined by \( K_c \), which is temperature-dependent and specific to each reaction.

To express \( K_c \), we use the concentrations of the reactants and products in the balanced chemical equation. For the reaction \[ \mathrm{I}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{IBr}(g) \], the equilibrium constant expression is:

\[ K_c = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_2][\mathrm{Br}_2]} \]

This formula shows that \( K_c \) reflects the ratio of products to reactants at equilibrium. A large \( K_c \), like \( 1.2 \times 10^2 \) in this exercise, signifies that at equilibrium, the products (\( \mathrm{IBr} \)) are favored over the reactants. Understanding this concept can help you predict how a reaction will behave under equilibrium conditions.
ICE Table
An ICE table is a useful tool for organizing information about the concentrations of reactants and products, which helps in determining the equilibrium position of a chemical reaction. ICE stands for Initial, Change, and Equilibrium.

  • Initial: This row represents the initial concentrations of reactants and products. In our exercise, both \( \mathrm{I}_2 \) and \( \mathrm{Br}_2 \) start at 0.0003 M, while \( \mathrm{IBr} \) starts at 0 M, since no product is present initially.

  • Change: As the reaction proceeds towards equilibrium, concentrations of reactants decrease, while that of products increases. In our exercise, reactants decrease by \(-x\), and \( \mathrm{IBr} \) increases by \(+2x\), which reflects the stoichiometry of the balanced equation.

  • Equilibrium: The final row shows the equilibrium concentrations expressed in terms of \( x \). Substituting these into the expression for \( K_c \) allows for solving \( x \), which gives us the equilibrium concentrations.
Using an ICE table simplifies complex equilibrium problems, making them more manageable and clear.
Reaction Quotient
Before a chemical reaction reaches equilibrium, the reaction quotient, \( Q \), provides insight into which direction the reaction needs to proceed. \( Q \) is calculated similarly to \( K_c \), using the concentrations of products and reactants at any point in time rather than at equilibrium.

For our specific reaction \( \mathrm{I}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{IBr}(g) \), the reaction quotient \( Q \) is given by:

\[ Q = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_2][\mathrm{Br}_2]} \]

  • If \( Q < K_c \), the reaction will proceed in the forward direction to produce more \( \mathrm{IBr} \) until equilibrium is reached.

  • If \( Q > K_c \), the reaction will shift in the reverse direction, forming more reactant - \( \mathrm{I}_2 \) and \( \mathrm{Br}_2 \).

  • If \( Q = K_c \), the system is at equilibrium and no net change in concentrations will occur.
Understanding \( Q \) is helpful in predicting how a reaction mixture will change and in setting up experiments more efficiently.

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Most popular questions from this chapter

Suppose you start with a mixture of \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(4.00 \mathrm{~mol} \mathrm{H}_{2}\) in a 10.00-L vessel. Find the moles of substances present at equilibrium at \(1200 \mathrm{~K}\) for the reaction $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; K_{c}=3.92 $$ You will get an equation of the form $$ f(x)=3.92 $$ where \(f(x)\) is an expression in the unknown \(x\) (the amount of \(\mathrm{CH}_{4}\) ). Solve this equation by guessing values of \(x\), then computing values of \(f(x)\). Find values of \(x\) such that the values of \(f(x)\) bracket \(3.92 .\) Then choose values of \(x\) to get a smaller bracket around 3.92. Obtain \(x\) to two significant figures.

During the commercial preparation of sulfuric acid, sulfur dioxide reacts with oxygen in an exothermic reaction to produce sulfur trioxide. In this step, sulfur dioxide mixed with oxygen-enriched air passes into a reaction tower at about \(420^{\circ} \mathrm{C}\), where reaction occurs on a vanadium(V) oxide catalyst. Discuss the conditions used in this reaction in terms of its effect on the yield of sulfur trioxide. Are there any other conditions that you might explore in order to increase the yield of sulfur trioxide?

A mixture of carbon monoxide, hydrogen, and methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is at equilibrium according to the equation $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At \(250^{\circ} \mathrm{C}\), the mixture is \(0.096 \mathrm{M} \mathrm{CO}, 0.191 \mathrm{M} \mathrm{H}_{2}\), and \(0.015 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\). What is \(K_{c}\) for this reaction at \(250^{\circ} \mathrm{C}\) ?

Iodine monobromide, IBr, occurs as brownish-black crystals that vaporize with decomposition: $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) $$ The equilibrium constant \(K_{c}\) at \(100^{\circ} \mathrm{C}\) is \(0.026\). If \(0.010 \mathrm{~mol}\) IBr is placed in a \(1.0\) - \(\mathrm{L}\) vessel at \(100^{\circ} \mathrm{C}\), what are the moles of substances at equilibrium in the vapor?

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) $$ equals \(4.1\) at \(300^{\circ} \mathrm{C}\). a. A sample of \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) is placed in a \(5.0\) - \(\mathrm{L}\) reaction vessel and heated to \(300^{\circ} \mathrm{C}\). What are the equilibrium concentrations of all of the species? b. What fraction of \(\mathrm{PCl}_{5}\) has decomposed? c. If \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) were placed in a 1.0-L vessel, what qualitative effect would this have on the fraction of \(\mathrm{PCl}_{5}\) that has decomposed (give a qualitative answer only; do not do the calculation)? Why?
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