Question

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 \(\mathrm{Cl}_{2}\) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.045\) at \(375^{\circ} \mathrm{C}\) (a) \(A\) 10.0-L. flask containing 6.70 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 10.0 -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 \(\mathrm{Cl}_{2}(0.10 \mathrm{atm}) ?\) 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?

Short answer

(a) 26.2% dissociation; (b) 7.9% dissociation; (c) Le Chatelier's principle predicted less dissociation in (b).

Step-by-step solution

Calculate the Initial Concentration of SO2Cl2

The initial moles of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) can be calculated from its mass. Given the molecular weight of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) is 134.97 g/mol, the initial moles are \( \frac{6.70 \text{ g}}{134.97 \text{ g/mol}} = 0.0496 \text{ mol} \). In a 10.0-L flask, the initial concentration is \( \frac{0.0496 \text{ mol}}{10.0 \text{ L}} = 0.00496 \text{ M} \).

Set up the ICE Table for Equilibrium

To find the concentrations at equilibrium, use the ICE table (Initial, Change, Equilibrium). The reaction is \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{SO}_{2} + \mathrm{Cl}_{2} \). Initially, \( [\mathrm{SO}_{2}] = 0 \) and \( [\mathrm{Cl}_{2}] = 0 \).- Initial Concentrations: \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.00496 \text{ M}\), \([\mathrm{SO}_{2}] = 0 \text{ M}, \ [\mathrm{Cl}_{2}] = 0 \text{ M}\) - Change: \(-x\) for \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \), \(+x\) for \( \mathrm{SO}_{2} \) and \( \mathrm{Cl}_{2} \)- Equilibrium: \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.00496 - x\), \([\mathrm{SO}_{2}] = x\), \([\mathrm{Cl}_{2}] = x\)

Solve for x Using the Equilibrium Constant

The equilibrium constant expression is \( K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]} \). Given \( K_c = 0.045 \), substitute the equilibrium concentrations:\[0.045 = \frac{x \cdot x}{0.00496 - x} = \frac{x^2}{0.00496 - x}\]Solve this quadratic equation for \(x\) to find \(x \approx 0.0013 \text{ M}\).

Calculate Equilibrium Concentrations

The equilibrium concentrations are:- \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.00496 - 0.0013 = 0.00366 \text{ M}\)- \([\mathrm{SO}_{2}] = 0.0013 \text{ M}\)- \([\mathrm{Cl}_{2}] = 0.0013 \text{ M}\)

Calculate the Fraction of SO2Cl2 Dissociated (a)

The fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated is \( \frac{x}{\text{Initial concentration of } \mathrm{SO}_{2} \mathrm{Cl}_{2}} = \frac{0.0013}{0.00496} \approx 0.262 \).

Determine Initial Conditions for Case with Cl2 Present (b)

Now consider the system starts with 0.00496 M \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) and \(0.10\) atm of \( \mathrm{Cl}_{2} \). Convert the pressure to concentration using \( \text{ideal gas law} \):\([\mathrm{Cl}_{2}] = \frac{0.10 \text{ atm} \times 1 \text{ mol/atm} \times 0.0821 \text{ L atm/mol K} \times 10^3 \text{ L}}{0.0821 \times 648 \text{ K}} = 0.00261 \text{ M}\).

Solve for x with Mutually Present Cl2 (b)

Apply the ICE table and solve for \(x\) just as in Step 3. However, initial \([\mathrm{Cl}_{2}] = 0.00261 \text{ M}\). Thus:\[0.045 = \frac{x \cdot (x + 0.00261)}{0.00496 - x}\]Solving yields \(x \approx 0.00039 \text{ M}\).

Calculate Equilibrium Concentrations (b)

The equilibrium concentrations are:- \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.00496 - 0.00039 = 0.00457 \text{ M}\)- \([\mathrm{SO}_{2}] = 0.00039 \text{ M}\)- \([\mathrm{Cl}_{2}] = 0.00039 + 0.00261 = 0.003 \text{ M}\)

Calculate the Fraction of SO2Cl2 Dissociated (b)

The fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated is \( \frac{0.00039}{0.00496} \approx 0.079 \).

Compare Fractions of Dissociation (c)

In part (a), \(26.2\%\) of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociates, whereas in part (b) \(7.9\%\) dissociates. Adding \( \mathrm{Cl}_{2} \) shifts the equilibrium to the left according to Le Chatelier's principle, decreasing dissociation. This agrees with the principle's predictions.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in chemical equilibrium that helps predict how a change in conditions affects the position of equilibrium. The principle states that if a system at equilibrium is subjected to an external stress, such as a change in concentration, pressure, or temperature, the system will adjust itself to counteract the effect of the disturbance. This adjustment aims to restore equilibrium.
In the context of the given sulfuryl chloride reaction, introducing additional chloride gas (\( \mathrm{Cl}_2 \)) to the system shifts the equilibrium position. According to Le Chatelier's Principle, adding more products will shift the reaction towards the reactants. As a result, less sulfuryl chloride (\( \mathrm{SO}_2\mathrm{Cl}_2 \)) decomposes, which is why the dissociation fraction decreases with increased chloride presence.
This principle helps to qualitatively understand the behavior of equilibrium systems under different conditions, providing a guide for predicting the outcomes of changes in concentration or other factors.

Equilibrium Constant

The Equilibrium Constant (\( K_c \)) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. For the decomposition of sulfuryl chloride, the equilibrium constant at a certain temperature is given as \( K_c = 0.045 \).
Mathematically, the equilibrium constant is expressed as:
\[K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\]
This equation gives us a quantitative measure of how far the reaction proceeds before reaching equilibrium. A low \( K_c \) value, like 0.045, indicates the reaction favors the reactants (\( \mathrm{SO}_2\mathrm{Cl}_2 \)) at equilibrium. Therefore, only a small amount of reactant decomposes to form sulfur dioxide (\( \mathrm{SO}_2 \)) and chlorine (\( \mathrm{Cl}_2 \)).
Understanding the equilibrium constant provides significant information about the chemical reaction and helps to calculate equilibrium concentrations when initial conditions are known.

ICE Table

An ICE table is a straightforward tool used to simplify the calculation of concentrations at equilibrium for a chemical reaction. The term "ICE" stands for Initial, Change, and Equilibrium, and the table is structured to track these values for all species involved in the reaction.
Here's how an ICE table is constructed:

• Initial: Start by filling in the initial concentrations of the reactants and products.
• Change: Indicate the change in concentrations as the system shifts towards equilibrium (often represented by the variable \( x \)). For \( \mathrm{SO}_2\mathrm{Cl}_2 \) decomposition, this change is \( -x \) for the reactant and \( +x \) for each product.
• Equilibrium: Define the expressions for concentrations at equilibrium using the initial values and changes. These expressions help set up equations based on the equilibrium constant.

In the sulfuryl chloride exercise, the ICE table helps guide us in setting up the quadratic equation necessary to solve for \( x \), thus allowing us to determine equilibrium concentrations and the extent of dissociation.

Dissociation Fraction

The dissociation fraction measures the extent to which a reactant breaks down into its products at equilibrium. It is calculated as the ratio of the amount dissociated to the initial amount present.
In the case of sulfuryl chloride:

• The dissociation fraction in experiment (a) is about 0.262, or 26.2%, which indicates significant dissociation under the given conditions.
• For experiment (b), where additional chlorine gas is present, the dissociation fraction drops to about 0.079, or 7.9% due to the shift in equilibrium.

This concept captures the effect that different starting conditions, such as the presence of additional products, can have on the extent of reactant dissociation. Calculating the dissociation fraction provides insight into the efficiency of the reaction and the conditions needed for desired product formation.