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Sulfuryl chloride is used in organic chemistry as a chlorinating agent. At moderately high temperatures it decomposes as follows: $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ with \(K_{c}=0.045\) at \(650 \mathrm{~K}\). 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 \(650 \mathrm{~K}\). What are the equilibrium concentrations of all of the species? b. 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
a) [SOâ‚‚Clâ‚‚] = 0.0121 M, [SOâ‚‚] = [Clâ‚‚] = 0.049 M; b) 80.1% decomposed; c) Adding Clâ‚‚ decreases decomposition.

Step by step solution

01

Calculate Initial Concentration of SOâ‚‚Clâ‚‚

First, calculate the number of moles of SOâ‚‚Clâ‚‚ using the given mass and its molar mass. The molar mass of SOâ‚‚Clâ‚‚ is approximately 134.97 g/mol.\[\text{Moles of } \mathrm{SO}_2\mathrm{Cl}_2 = \frac{8.25 \text{ g}}{134.97 \text{ g/mol}} = 0.0611 \text{ mol}\]Since this is placed in a 1.00 L reaction vessel, the initial concentration of SOâ‚‚Clâ‚‚ is:\[[\mathrm{SO}_2\mathrm{Cl}_2]_0 = 0.0611 \text{ M}\]
02

Set Up the Equilibrium Expression

The equilibrium expression for the reaction is based on the given equilibrium constant:\[K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\]We know that at equilibrium:- If x is the amount of SOâ‚‚Clâ‚‚ that decomposes, then:- \([\mathrm{SO}_2\mathrm{Cl}_2]\) at equilibrium = \(0.0611 - x\)- \([\mathrm{SO}_2]\) and \([\mathrm{Cl}_2]\) at equilibrium = x.
03

Substitute and Solve the Equilibrium Equation

Substitute the values into the equilibrium expression:\[K_c = \frac{x \cdot x}{0.0611 - x} = 0.045\]This simplifies to:\[x^2 = 0.045 (0.0611 - x)\]\[x^2 + 0.045x - 0.0027495 = 0\]Use the quadratic formula to solve for x:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plugging in the values:- a = 1, b = 0.045, c = -0.0027495,We find,\[x \approx 0.049 \text{ M}\]
04

Calculate Equilibrium Concentrations

Using the value of x, calculate the equilibrium concentrations:\[[\mathrm{SO}_2\mathrm{Cl}_2]] = 0.0611 - 0.049 = 0.0121 \text{ M}\]\[[\mathrm{SO}_2] = [\mathrm{Cl}_2] = 0.049 \text{ M}\]
05

Determine Fraction Decomposed

Using the initial and equilibrium values, calculate the fraction decomposed:\[\text{Fraction decomposed} = \frac{x}{[\mathrm{SO}_2\mathrm{Cl}_2]_0} = \frac{0.049}{0.0611} \approx 0.801\]So, approximately 80.1% of SOâ‚‚Clâ‚‚ has decomposed.
06

Predict Effect of Adding Clâ‚‚ on Decomposition

Adding Clâ‚‚ to the system shifts the equilibrium to the left according to Le Chatelier's Principle (to decrease the increase in \([\mathrm{Cl}_2]\)). This results in a decrease in the fraction of SOâ‚‚Clâ‚‚ that has decomposed, as more SOâ‚‚ and Clâ‚‚ recombine to form SOâ‚‚Clâ‚‚.

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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 (\( K_c \)) is a mathematical expression that helps chemists understand how far a reaction will proceed before reaching equilibrium. An equilibrium occurs when the rate at which the reactants are converted into products equals the rate at which products revert back into reactants. At this point, the concentrations of the reactants and products remain constant over time.
For the reaction \(\mathrm{SO}_2\mathrm{Cl}_2(g) \rightleftharpoons \mathrm{SO}_2(g) + \mathrm{Cl}_2(g)\), the equilibrium constant \( K_c \) is given by:
  • \( K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]} \)
Here, brackets represent the concentrations of each gas in the mixture. \( K_c \) is dimensionless and varies with temperature, for this reaction conducted at 650 \mathrm{~K}, \( K_c = 0.045\). This small value of the equilibrium constant indicates that, at equilibrium, the concentration of reactants (\(\mathrm{SO}_2\mathrm{Cl}_2\)) is relatively higher compared to the concentration of products (\(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\)). This means the reaction favors the reactants at this temperature.
Le Chatelier's Principle
Le Chatelier's Principle is a handy guideline that predicts how a system at equilibrium responds to disturbances or changes. When a change is introduced, the system will shift in the direction that counteracts this change and restores a new equilibrium. In simple terms, it's how nature keeps balance.
Consider our reaction: \(\mathrm{SO}_2\mathrm{Cl}_2(g) \rightleftharpoons \mathrm{SO}_2(g) + \mathrm{Cl}_2(g)\). Let's say we add more chlorine gas (\(\mathrm{Cl}_2\)), according to Le Chatelier's Principle:
  • The system will try to reduce this change by favoring the reverse reaction.
  • This means more \(\mathrm{SO}_2\mathrm{Cl}_2\) will be formed from \(\mathrm{SO}_2\) and \(\mathrm{Cl}_2\) gas.
  • The end result is a decrease in the decomposition rate of \(\mathrm{SO}_2\mathrm{Cl}_2\).
Thus, when the chlorine is added, the fraction of \(\mathrm{SO}_2\mathrm{Cl}_2\) that has decomposed decreases, demonstrating how shifts occur to minimize external stress (added chlorine in this case).
Reaction Quotient
The reaction quotient, represented as \( Q_c \), is a snapshot of the ratio of concentrations of products to reactants at any point in time—not just at equilibrium. It's like pressing pause in the middle of a movie!To calculate \( Q_c \), we use the same expression as \( K_c \):
  • \( Q_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]} \)
If \( Q_c < K_c, \) the reaction will proceed in the forward direction to reach equilibrium, producing more products.
If \( Q_c > K_c, \) the reaction proceeds in the reverse direction, turning products back into reactants.
And if \( Q_c = K_c, \) the system is at equilibrium, so no net change occurs.
Understanding \( Q_c \) allows chemists to predict which way a reaction will "shift" to achieve equilibrium, making it an essential tool in forecasting reaction behavior under varying conditions.

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Most popular questions from this chapter

Suppose sulfur dioxide reacts with oxygen at \(25^{\circ} \mathrm{C}\). $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ The equilibrium constant \(K_{c}\) equals \(8.0 \times 10^{35}\) at this temperature. From the magnitude of \(K_{c}\), do you think this reaction occurs to any great extent when equilibrium is reached at room temperature? If an equilibrium mixture is \(1.0 \mathrm{M} \mathrm{SO}_{3}\) and has equal concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), what is the concentration of \(\mathrm{SO}_{2}\) in the mixture? Does this result agree with what you expect from the magnitude of \(K_{c}\) ?

The equilibrium constant \(K_{c}\) for the equation $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ at \(900^{\circ} \mathrm{C}\) is \(27.8\). What is the value of \(K_{c}\) for the following equation? $$ \frac{1}{2} \mathrm{CS}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$

The following reaction is important in the manufacture of sulfuric acid. $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ At \(900 \mathrm{~K}, 0.0216 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) and \(0.0148 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are sealed in a 1.00-L reaction vessel. When equilibrium is reached, the concentration of \(\mathrm{SO}_{3}\) is determined to be \(0.0175 \mathrm{M}\). Calculate \(K_{c}\) for this reaction.

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. \(\left[\mathrm{CH}_{4}\right]=1.26 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.32 M\) \(\left[\mathrm{CS}_{2}\right]=1.43 M,\left[\mathrm{H}_{2}\right]=1.12 M\) b. \(\left[\mathrm{CH}_{4}\right]=1.25 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.52 \mathrm{M}\), \(\left[\mathrm{CS}_{2}\right]=1.15 M,\left[\mathrm{H}_{2}\right]=1.73 M\) c. \(\left[\mathrm{CH}_{4}\right]=1.20 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.31 \mathrm{M}\), \(\left[\mathrm{CS}_{2}\right]=1.15 M,\left[\mathrm{H}_{2}\right]=1.85 M\) d. \(\left[\mathrm{CH}_{4}\right]=1.56 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.43 M\), \(\left[\mathrm{CS}_{2}\right]=1.23 M,\left[\mathrm{H}_{2}\right]=1.91 M\)

List four ways in which the yield of ammonia in the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \Delta H^{\circ}<0 $$ can be improved for a given amount of \(\mathrm{H}_{2} .\) Explain the principle behind each way.

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