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Chapter 16: Problem 15

The equilibrium constant for the dissociation of iodine molecules to iodine atoms $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) $$ is \(3.76 \times 10^{-3}\) at \(1000 \mathrm{K}\). Suppose 0.105 mol of \(\mathrm{I}_{2}\) is placed in a \(12.3-\mathrm{L}\). flask at \(1000 \mathrm{K}\). What are the concentrations of \(\mathrm{I}_{2}\) and \(\mathrm{I}\) when the system comes to equilibrium?

Short Answer

Expert verified
\( [\mathrm{I}_2] = 0.00624 \text{ M}, [\mathrm{I}] = 0.0046 \text{ M} \) at equilibrium.

Step by step solution

01

Write the Equilibrium Expression

The equilibrium expression for the dissociation of iodine molecules into iodine atoms is given by \( K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \), where \( [\mathrm{I}] \) and \( [\mathrm{I}_2] \) are the equilibrium concentrations of iodine atoms and molecules, respectively, and \( K_c = 3.76 \times 10^{-3} \).
02

Calculate Initial Concentration of \( \mathrm{I}_2 \)

Calculate the initial concentration of \( \mathrm{I}_2 \) by using \( \text{[I}_2]_0 = \frac{n}{V} = \frac{0.105 \text{ mol}}{12.3 \text{ L}} = 0.00854 \text{ M} \).
03

Define Change in Concentrations

Set up a change in concentration for the reaction assuming \( x \) is the amount dissociated: \( [\mathrm{I}_2] = 0.00854 - x \) and \( [\mathrm{I}] = 2x \).
04

Substitute into Equilibrium Expression

Substitute the expressions for concentrations into the equilibrium expression: \( 3.76 \times 10^{-3} = \frac{(2x)^2}{0.00854 - x} \).
05

Solve the Quadratic Equation

Rearrange the expression to form a quadratic equation: \( 4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x) \). Solving this gives \( x = 0.0023 \).
06

Calculate Equilibrium Concentrations

Find equilibrium concentrations using \( x = 0.0023 \): \( [\mathrm{I}_2] = 0.00854 - 0.0023 = 0.00624 \text{ M} \) and \( [\mathrm{I}] = 2 \times 0.0023 = 0.0046 \text{ M} \).

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Key Concepts

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

Dissociation Reaction
A dissociation reaction involves breaking a compound into smaller particles, typically atoms or simpler molecules. In this exercise, iodine molecules (\( \mathrm{I}_2 \)) dissociate into iodine atoms (\( 2 \mathrm{I} \)). This process is important because it changes how the atoms or molecules interact and form new arrangements, influencing the overall chemical behavior.
Understanding dissociation reactions is crucial in fields like chemistry and biology. They often involve energy absorption and are reversible. When you know the dissociation reaction, you can predict the concentrations of the products and reactants at different stages, especially at equilibrium.
Iodine Molecules
Iodine molecules (\( \mathrm{I}_2 \)) are diatomic, meaning they consist of two iodine atoms bonded together. At high temperatures, such as 1000 K, they tend to dissociate into individual iodine atoms. This transformation is reversible, as seen in the chemical equation:
\[ \mathrm{I}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \]
In the example, 0.105 mol of iodine molecules is initially placed in a flask. Knowing the initial concentration helps us track how much of the iodine dissociates over time. Understanding iodine's behavior at different temperatures is key in industrial and laboratory settings.
Equilibrium Expression
The equilibrium expression quantifies the balance between reactants and products in a reversible reaction at equilibrium. For the dissociation of iodine, the equilibrium constant (\( K_c \)) is defined as:
\[ K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]
This formula relates the concentrations of iodine atoms and iodine molecules at equilibrium. The given \( K_c \) value, \( 3.76 \times 10^{-3} \), dictates the extent to which the reaction favors products (iodine atoms) or reactants (iodine molecules).
By substituting the equilibrium concentrations into the equilibrium expression, we can solve for unknowns, helping to determine how much of each species is present at equilibrium.
Quadratic Equation
A quadratic equation appears when solving for the concentration changes during a reaction at equilibrium. In our iodine dissociation example, the change in concentration, denoted as \( x \), helps form the equation:
\[ 4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x) \]
We rearrange it to solve for \( x \), allowing us to find how much iodine dissociates. Quadratic equations often arise in equilibrium problems and can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, solving for \( x \) helps us calculate the equilibrium concentrations of the dissociated iodine atoms and remaining iodine molecules precisely.

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Most popular questions from this chapter

The ammonia complex of trimethylborane, \(\left(\mathrm{NH}_{3}\right) \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3},\) dissociates at \(100^{\circ} \mathrm{C}\) to its components with \(K_{\mathrm{p}}=4.62\) (when the pressures are in atmospheres). \(\left(\mathrm{NH}_{3}\right) \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3}(\mathrm{g}) \quad \rightleftarrows \quad \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g})\) If \(\mathrm{NH}_{3}\) is changed to some other molecule, the equilibrium constant is different. For \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{P}\right] \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3} \quad \quad K_{\mathrm{p}}=0.128\) For \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right] \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3} \quad \quad K_{\mathrm{p}}=0.472\) (a) If you begin an experiment by placing 0.010 mol of each complex in a flask, which would have the largest partial pressure of \(\mathbf{B}\left(\mathrm{CH}_{3}\right)_{3}\) at \(100^{\circ} \mathrm{C} ?\) (b) If \(0.73 \mathrm{g}(0.010 \mathrm{mol})\) of \(\left(\mathrm{NH}_{3}\right) \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3}\) is placed in a \(100 .\) -mL. flask and heated to \(100^{\circ} \mathrm{C},\) what is the partial pressure of each gas in the equilibrium mixture, and what is the total pressure? What is the percent dissociation of \(\left(\mathrm{NH}_{3}\right) \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3} ?\)

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

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=\) $$ 5.0 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L}, \text { and } $$ \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilib- rium? If not, which way will the reaction proceed to reach equilibrium? $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) K_{\mathrm{c}}=279 $$

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature, it decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\). \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{c}=0045 \mathrm{at} 375^{\circ} \mathrm{C}\) (a) A \(1.00-1 .\) flask containing \(6.70 \mathrm{g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is heated to \(375^{\circ} \mathrm{C}\). What is the concentration of each of the compounds in the system when equilibrium is achieved? What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (b) What are the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{Cl}_{2}\) at equilibrium in the \(1.00-\mathrm{L}\). flask at \(375^{\circ} \mathrm{C}\) if you begin with a mixture of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(6.70 \mathrm{g})\) and \(\mathrm{Cl}_{2}\) \((1.00 \mathrm{atm}) ?\) What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (c) Compare the fractions of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in parts (a) and (b). Do they agree with your expectations based on Le Chatelier's principle?

At \(1800 \mathrm{K},\) oxygen dissociates very slightly into its atoms. $$ \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{p}}=1.2 \times 10^{-10} $$ If you place 1.0 mol of \(\mathrm{O}_{2}\) in a \(10 .\).L. vessel and heat it to \(1800 \mathrm{K}\), how many \(\mathrm{O}\) atoms are present in the flask?
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