Problem 54 - Chemistry and Chemical Reactivity

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 \(\mathrm{Cl}_{2}\). \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{c}=0045 \mathrm{at} 375^{\circ} \mathrm{C}\) (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?

Step-by-Step Solution

Answer

For (a), 95.1% dissociation. For (b), 4.4% dissociation. More SO2Cl2 dissociates without initial Cl2, consistent with Le Chatelier's principle.

Step 1: Calculate Initial Concentration of SO2Cl2

First, calculate the number of moles of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \). The molar mass of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) is \( 32.07 + 2(16.00) + 2(35.45) = 134.97\, \text{g/mol} \). Thus, moles of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} = \frac{6.70\, \text{g}}{134.97\, \text{g/mol}} \approx 0.0497\, \text{mol} \). The initial concentration is \( \frac{0.0497\, \text{mol}}{1.00\, \text{L}} = 0.0497\, \text{M} \).

Step 2: Set Up ICE Table

Set up an ICE (Initial, Change, Equilibrium) table for the reaction \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{SO}_{2} + \mathrm{Cl}_{2} \). Let initial concentration of \( \mathrm{SO}_{2} \) and \( \mathrm{Cl}_{2} \) be zero. Change in \( \mathrm{SO}_{2} \mathrm{Cl}_{2} = -x \) and change in \( \mathrm{SO}_{2} = \mathrm{Cl}_{2} = +x \).

Step 3: Apply Equilibrium Constant Expression

Use the equilibrium constant expression: \[ K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]} = 0.045 \]Plug in the equilibrium concentrations: \[ K_c = \frac{x^2}{0.0497 - x} \]Solve for \( x \).

Step 4: Solve for x and Calculate Equilibrium Concentrations

Assuming \( x \) is small, \( 0.045 = \frac{x^2}{0.0497} \) gives us \( x^2 = 0.0022365 \) and \( x = 0.0473 \). Thus, \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.0497 - 0.0473 = 0.0024\, \text{M} \), \([\mathrm{SO}_{2}] = [\mathrm{Cl}_{2}] = 0.0473\, \text{M} \).

Step 5: Calculate Fraction Dissociated for (a)

The fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated is \( \frac{x}{0.0497} = \frac{0.0473}{0.0497} \approx 0.951 \).

Step 6: Analyze Initial Conditions for Part (b)

In part (b), with \( 1.00\, \text{atm} \) of \( \mathrm{Cl}_{2} \), use \( K_c = 0.045 \). Use small \( x \) assumption: \[ K_c = \frac{(y)(1.00 + y)}{0.0497 - y} = 0.045 \]

Step 7: Solve Equilibrium Equation for Part (b)

Using the assumption that \( y \) is small: \( 0.045 = \frac{y(1.00)}{0.0497} \), \( y = 0.0022 \). Then, \([\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 0.0497 - 0.0022 = 0.0475\, \text{M} \),\([\mathrm{SO}_{2}] = 0.0022\, \text{M} \), \([\mathrm{Cl}_{2}] = 1.0022\, \text{M} \).

Step 8: Calculate Fraction Dissociated for (b)

The fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated is \( \frac{0.0022}{0.0497} \approx 0.044 \).

Step 9: Compare Fractions and Apply Le Chatelier's Principle

In (a), 95.1% dissociation, while in (b), 4.4%. More dissociation occurs in (a) due to lack of initial \( \mathrm{Cl}_{2} \), consistent with Le Chatelier's principle, which predicts less dissociation with added product.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a critical concept in understanding how reactions reach equilibrium. Whenever a change is introduced to a system at equilibrium, the system will adjust itself to counteract the effect of the change and re-establish equilibrium. In simpler terms, if you disturb a reaction in a closed system, it will "lean" to oppose that disturbance.

For example, consider a reaction where an increase in concentration of a reactant or product occurs. According to Le Chatelier’s Principle, the reaction will shift to reduce this change. If a product is increased, the reaction will move towards the reactants, while if a reactant is increased, it will progress towards products.

Using part (b) of the original exercise as an illustration, adding chlorine gas ( \( \text{Cl}_2 \)) means extra product is present initially. The system responds by shifting the equilibrium position to produce more reactant (\( \text{SO}_2 \text{Cl}_2 \)), reducing its dissociation, which aligns with Le Chatelier's Principle.

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a value that expresses the ratio of the concentration of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. This constant is specific to a given reaction at a certain temperature.

In the context of the reaction \( \text{SO}_2 \text{Cl}_2 \rightleftharpoons \text{SO}_2 + \text{Cl}_2 \), the equilibrium constant \( K_c = 0.045 \) at 375°C illuminates the balance of concentrations at equilibrium between the reagent sulfuryl chloride and its products, sulfur dioxide and chlorine gas.

Mathematically, the expression for this reaction using the equilibrium constant is:

• \( K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2 \text{Cl}_2]} \)

The value of \( K_c \) suggests that at equilibrium, there will be a relatively small concentration of products compared to the reactants. This small \( K_c \) is reflected in part (a) of the exercise where most of the reactant decomposes.

Reaction Quotient

The reaction quotient, \( Q \), serves as a snapshot indicating the progress of a reaction at any point in time compared to its equilibrium state. It's calculated in a way similar to the equilibrium constant, but it applies to concentrations that are not necessarily at equilibrium.

By evaluating \( Q \), chemists can determine whether a reaction needs to shift to the right (toward products) or to the left (toward reactants) to reach equilibrium. If \( Q = K_c \), the reaction is at equilibrium. If \( Q < K_c \), the reaction will proceed forward (producing more products), and if \( Q > K_c \), it will proceed backward (producing more reactants).

In the given problem, setting an initial \( \text{Cl}_2 \) concentration alters system dynamics initially, affecting \( Q \) and prompting system shifts according to Le Chatelier's Principle until \( Q \) aligns with \( K_c \).

Dissociation Reaction

A dissociation reaction involves the breaking apart of a compound into its components. This is a type of reversible reaction where the compound can break down into smaller molecules, which can also recombine to form the initial compound at equilibrium.

In the exercise, sulfuryl chloride (\( \text{SO}_2 \text{Cl}_2 \)) undergoes dissociation to form sulfur dioxide (\( \text{SO}_2 \)) and chlorine gas (\( \text{Cl}_2 \)). This is expressed as:

• \( \text{SO}_2 \text{Cl}_2 \rightleftharpoons \text{SO}_2 + \text{Cl}_2 \)

The extent of this dissociation is impacted by the reaction conditions and equilibrium constant.

Understanding dissociation is crucial in chemical equilibrium calculations, as it provides insights into the fraction of the compound breaking down at equilibrium. Part (a) and (b) illustrate how changes in initial conditions, such as added \( \text{Cl}_2 \), influence the fraction of \( \text{SO}_2 \text{Cl}_2 \) dissociated, with part (a) showing significant dissociation due to no initial \( \text{Cl}_2 \).