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Sulfuryl chloride is used in organic chemistry as a chlorinating agent. At moderately high temperatures it decomposes as follows: $$ \begin{array}{r} \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \\ \text { with } K_{c}=0.045 \text { at } 650 \mathrm{~K} . \end{array} $$ a. A sample of \(8.25 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel and heated to \(650 \mathrm{~K}\). b. What are the equilibrium concentrations of all of the species? What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has decomposed? c. If \(5 \mathrm{~g}\) of chlorine is inserted into the reaction vessel, what qualitative effect would this have on the fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that has decomposed?

Short Answer

Expert verified
At equilibrium, \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] \approx 0.0117\, \mathrm{M}\), \([\mathrm{SO}_2] = [\mathrm{Cl}_2] \approx 0.049\, \mathrm{M}\); 80.7% decomposed. Adding \(\mathrm{Cl}_2\) decreases decomposition.

Step by step solution

01

Calculate Initial Concentrations

First, we need to determine the initial concentration of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \). We have an initial mass of \( 8.25 \text{ g} \). The molar mass of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) is calculated as follows: - Sulfur (S): 32.07 g/mol- Oxygen (O, 2 atoms): 16.00 g/mol \( \times 2 = 32.00 \text{ g/mol} \)- Chlorine (Cl, 2 atoms): 35.45 g/mol \( \times 2 = 70.90 \text{ g/mol} \)Thus, the molar mass is \( 32.07 + 32.00 + 70.90 = 135.97 \text{ g/mol} \). Hence, the initial moles of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) are:\[ n = \frac{8.25}{135.97} \approx 0.0607 \text{ mol} \]In a 1-liter vessel, the initial concentration \([\mathrm{SO}_{2} \mathrm{Cl}_{2}]_0 = 0.0607 \text{ M}\). Initially, the concentrations of \( \mathrm{SO}_2 \) and \( \mathrm{Cl}_2 \) are zero.
02

Set Up the ICE Table

The reaction \( \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g) \) follows an equilibrium approach. Using the ICE (Initial, Change, Equilibrium) table, we define changes in concentration:- Initial: \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.0607 \text{ M} \), \([\mathrm{SO}_2] = 0 \), \([\mathrm{Cl}_2] = 0 \)- Change: \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = -x \), \([\mathrm{SO}_2] = +x \), \([\mathrm{Cl}_2] = +x \)- Equilibrium: \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.0607 - x \), \([\mathrm{SO}_2] = x \), \([\mathrm{Cl}_2] = x \)
03

Apply the Equilibrium Constant Expression

The equilibrium constant expression for the reaction is:\[ K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]} \]Substituting the equilibrium concentrations, we get:\[ 0.045 = \frac{x^2}{0.0607 - x} \]This equation gives us an expression to solve for \( x \).
04

Solve for x

Rearranging the equation, we have:\[ 0.045 (0.0607 - x) = x^2 \]\[ 0.0027315 - 0.045x = x^2 \]Rearrange into quadratic form:\[ x^2 + 0.045x - 0.0027315 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 0.045 \), and \( c = -0.0027315 \):\[ x = \frac{-0.045 \pm \sqrt{(0.045)^2 + 4 \cdot 1 \cdot 0.0027315}}{2 \cdot 1} \]Calculate \( x \approx 0.049 \text{ M} \).
05

Calculate Equilibrium Concentrations

Substitute \( x = 0.049 \text{ M} \) back into the equilibrium expressions:- \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.0607 - 0.049 \approx 0.0117 \text{ M} \)- \([\mathrm{SO}_2] = 0.049 \text{ M} \)- \([\mathrm{Cl}_2] = 0.049 \text{ M} \)
06

Calculate Fraction Decomposed

To find the fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) that has decomposed, use the initial and equilibrium concentrations:Fraction decomposed = \( \frac{\text{Initial moles} - \text{Equilibrium moles}}{\text{Initial moles}} = \frac{0.0607 - 0.0117}{0.0607} \approx 0.807 \)Thus, about 80.7% of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) has decomposed.
07

Effect of Adding Chlorine

Adding 5 g of \( \mathrm{Cl}_2 \) will increase the concentration of chlorine, thus shifting the equilibrium to the left according to Le Chatelier's Principle to form more \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \). Therefore, the fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) that has decomposed will decrease.

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Key Concepts

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

Chlorination Reaction
Chlorination reactions are a fascinating part of chemical transformations where chlorine is introduced into organic molecules. In the case of sulfuryl chloride \(\text{SO}_2\text{Cl}_2\), it's not only a chlorination agent but also undergoes decomposition at high temperatures. The reaction:
  • \( \mathrm{SO}_2\mathrm{Cl}_2(g) \rightleftharpoons \mathrm{SO}_2(g) + \mathrm{Cl}_2(g) \).
This balances the generation of chlorine gas \( \mathrm{Cl}_2 \) and sulfur dioxide \( \mathrm{SO}_2 \) from sulfuryl chloride. Chlorination reactions typically lead to the substitution or addition of chlorine atoms into organic compounds, critical in building complex molecules. This particular reaction represents a dynamic equilibrium where both forward and reverse reactions occur at equal rates, allowing the system to maintain balance without any net change in concentration of reactants and products.
Le Chatelier's Principle
Le Chatelier's Principle provides essential insights into how a chemical equilibrium reacts to changes in concentration, temperature, or pressure. If a system at equilibrium experiences a change, it will adjust to mitigate that change while forming a new equilibrium.

In our reaction involving sulfuryl chloride, increasing chlorine levels shifts the equilibrium. According to Le Chatelier's Principle:
  • If we add more \( \text{Cl}_2 \) to the reaction vessel, the equilibrium will shift to the left, favoring the formation of \( \text{SO}_2\text{Cl}_2 \) to offset the disturbance.
  • This shift will decrease the fraction of \( \text{SO}_2\text{Cl}_2 \) decomposition, as the system balances the concentration of reactants and products to find a new equilibrium position.
Understanding this principle is crucial in predicting how chemical systems respond to changes, allowing for effective control and optimization in industrial chemical processes.
Equilibrium Constant
The equilibrium constant \( K_c \) plays a vital role in understanding how far an equilibrium reaction has proceeded. It quantifies the ratio of concentrations of products to reactants at equilibrium, giving a measure of reaction direction and extent.

For the sulfuryl chloride decomposition, the expression is:
  • \( K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]} \)
Here, with \( K_c = 0.045 \) at 650 K, it indicates that under these conditions, the reaction does not heavily favor products, as the \( K_c \) is less than one.

Solving the equilibrium constant equation involves setting up an ICE (Initial, Change, Equilibrium) table to determine changes in concentrations as the system approaches equilibrium. Knowing \( K_c \) allows predictions of concentrations at equilibrium, guiding chemical production and studying reactions under different conditions more efficiently.

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Most popular questions from this chapter

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals 49 at \(230^{\circ} \mathrm{C}\). If \(0.400 \mathrm{~mol}\) each of phosphorus trichloride and chlorine are added to a 4.0-L reaction vessel, what is the equilibrium composition of the mixture at \(230^{\circ} \mathrm{C} ?\)

A graduate student places 0.272 mol of \(\mathrm{PCl}_{3}(g)\) and \(8.56 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{PCl}_{5}(g)\) into a 0.718 - \(\mathrm{L}\) flask at a certain temperature. \(\mathrm{PCl}_{5}(g)\) is known to decompose as follows: $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ After the reaction attains equilibrium, the student finds that the flask contains \(2.51 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{Cl}_{2} .\) Calculate the equilibrium constant \(K_{c}\) for the reaction at this temperature. a.0.114 b.\(8.51 \times 10^{-2}\) c.0.157 d.\(8.88 \times 10^{4}\) e.\(2.40 \times 10^{4}\)

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 \(150^{\circ} \mathrm{C} ?\)

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 \(2.000 \mathrm{~mol} \mathrm{Br}_{2}\) ?

A 2.0-L reaction flask initially contains \(0.010 \mathrm{~mol}\) \(\begin{array}{lllll}\mathrm{CO}, & 0.80 \mathrm{~mol} \mathrm{H}_{2} \text { , and } & 0.50 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH} & \text { (methanol). }\end{array}\) If this mixture is brought in contact with a zinc oxidechromium(III) oxide catalyst, the equilibrium $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ is attained. The equilibrium constant \(K_{c}\) for this reaction at \(300^{\circ} \mathrm{C}\) is \(1.1 \times 10^{-2}\). What is the direction of reaction (forward or reverse) as the mixture attains equilibrium?

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