Problem 54 - Chemistry and Chemical Reactivity

Question

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), is a compound with very irritating vapors; it 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 \(\mathrm{Cl}_{2}\). \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K=0.045\) at \(375^{\circ} \mathrm{C}\) (a) Suppose \(6.70 \mathrm{g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(1.00-\mathrm{L}\). flask and then 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

Part (a) has a higher dissociation compared to part (b) due to excess Cl₂ in part (b) shifting equilibrium left.

Step 1: Calculate Initial Concentrations

Start with part (a). Convert the mass of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) to moles using its molar mass, which is approximately \( 134.96 \, \text{g/mol} \). For a 6.70 g sample:\[ \text{Moles of } \mathrm{SO}_{2} \mathrm{Cl}_{2} = \frac{6.70 \, \text{g}}{134.96 \, \text{g/mol}} \approx 0.0497 \, \text{mol} \]Since the flask is 1.00 L, the initial concentration \([\mathrm{SO}_{2} \mathrm{Cl}_{2}]_0\) is:\[ [\mathrm{SO}_{2} \mathrm{Cl}_{2}]_0 = \frac{0.0497 \, \text{mol}}{1.00 \, \text{L}} = 0.0497 \, \text{M} \] Initially, \([\mathrm{SO}_{2}]_0\) and \([\mathrm{Cl}_{2}]_0\) are 0 M.

Step 2: Set up the ICE Table

Construct an ICE (Initial, Change, Equilibrium) table reflecting initial concentrations, changes that occur as equilibrium is achieved, and final equilibrium concentrations:\[\begin{array}{c|ccc}& \mathrm{SO}_{2} \mathrm{Cl}_{2}(\text{g}) & \mathrm{SO}_{2}(\text{g}) & \mathrm{Cl}_{2}(\text{g}) \\hline\text{Initial (M)} & 0.0497 & 0 & 0 \\text{Change (M)} & -x & +x & +x \\text{Equilibrium (M)} & 0.0497-x & x & x \\end{array} \]

Step 3: Apply the Equilibrium Constant Expression

Use the equilibrium expression for the reaction:\[ K = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]} = 0.045 \]Substitute the equilibrium concentrations from the ICE table:\[ 0.045 = \frac{x \cdot x}{0.0497-x} \]This simplifies to:\[ 0.045 = \frac{x^2}{0.0497-x} \]

Step 4: Solve the Quadratic Equation

Rearrange the equation from Step 3:\[ 0.045(0.0497-x) = x^2 \]Expanding and rearranging gives:\[ 0.0022365 - 0.045x = x^2 \]\[ x^2 + 0.045x - 0.0022365 = 0 \]Using the quadratic formula, the positive root determines \(x\):\[ x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \]Where \( a = 1, b = 0.045, c = -0.0022365 \). Solve for \(x\).

Step 5: Calculate Equilibrium Concentrations

Substituting the value of \(x\) from Step 4 into the equilibrium concentrations:- \([\mathrm{SO}_{2}] = x \)- \([\mathrm{Cl}_{2}] = x \)- \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = 0.0497 - x \)

Step 6: Determine the Fraction Dissociated

Calculate the fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated:\[ \text{Fraction dissociated} = \frac{x}{0.0497} \]

Step 7: Analyze Part (b) Initial Conditions

In part (b), you start with an additional \( \mathrm{Cl}_{2} \) concentration due to 1 atm of gas using the ideal gas law:\[ \text{Concentration of } \mathrm{Cl}_{2} = \frac{\text{1 atm}}{\text{RT}} = \frac{1}{0.0821 \times (375 + 273)} \approx 0.00342 \, \text{M} \]Set up the ICE table with these initial conditions and solve as in Steps 2-5 for equilibrium concentrations. Adjust \(x\) values.

Step 8: Compare Fractions Dissociated and Analyze

After finding the new fraction of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) dissociated in part (b), compare it to part (a).According to Le Chatelier's principle, the presence of additional \( \mathrm{Cl}_{2} \) should shift the equilibrium to the left, reducing the degree of dissociation compared to part (a).

Key Concepts

Equilibrium Constant

The equilibrium constant, expressed as \( K \), is crucial in understanding chemical equilibrium. It is derived from the balanced chemical equation and represents the ratio of the concentrations of products to reactants at equilibrium. For the reaction of sulfuryl chloride decomposing into sulfur dioxide and chlorine, the equilibrium expression is:

• \( K = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]} \)

This equation helps us determine what concentrations to expect at equilibrium. With \( K = 0.045 \) at \( 375^{\circ} \mathrm{C} \), it tells us that at this temperature, the concentration of the products is low compared to the reactants at equilibrium.
It's also important to remember that the equilibrium constant is specific to a particular reaction and a specific temperature, so any change in these factors will alter \( K \). This allows us to predict the direction a reaction will take as conditions change, which is fundamental in understanding how reactions reach and maintain equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a critical concept in chemical equilibrium. It states that if an external change is applied to a system at equilibrium, the system will adjust to minimize that change and re-establish equilibrium.

• If concentration, temperature, or pressure changes, the equilibrium will shift to counteract the change.

In the provided exercise, additional chlorine gas is introduced in part (b). According to Le Chatelier's Principle, this increase in chlorine concentration shifts the equilibrium to the left, meaning more sulfuryl chloride will remain undissociated.
This shift results in a lower degree of decomposition for sulfuryl chloride compared to the initial conditions in part (a). Le Chatelier's Principle helps us predict this behavior by considering how a system compensates for added stress. Understanding this principle gives insights into designing reactions and controlling the extent of reactions in industrial processes.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a structured approach for organizing data related to chemical reactions and changes as they reach equilibrium. It is especially helpful with equilibrium calculations, allowing us to clearly see the progression of a reaction.
To build an ICE table, follow these steps:

• Identify initial concentrations (I).
• Determine the change in concentrations (C) as the system moves towards equilibrium. These are typically expressed in terms of \( x \), the change in concentration.
• Calculate the final equilibrium concentrations (E).

For the decomposition of sulfuryl chloride, the ICE table shows:

• Initially: \([\mathrm{SO}_2 \mathrm{Cl}_2] = 0.0497 \text{ M}, [\mathrm{SO}_2] = 0 \text{ M}, [\mathrm{Cl}_2] = 0 \text{ M}\).
• Change: \(-x, +x, +x\) respectively.
• At equilibrium: \(0.0497-x, x, x\).

The ICE table provides a comprehensive method to use equilibrium equations and concentrations in solving for unknowns in chemical equilibrium reactions.

Mole Calculations

Mole calculations form a foundation for understanding chemistry, particularly in relation to concentration and reaction stoichiometry. In these exercises, converting grams to moles is a fundamental step:

• Determine moles using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
• The molar mass of sulfuryl chloride is approximately \( 134.96 \, \text{g/mol} \).

For example, with 6.70 g of \( \mathrm{SO}_2 \mathrm{Cl}_2\), we have:

• \( \approx 0.0497 \text{ mol} \) of \( \mathrm{SO}_2 \mathrm{Cl}_2 \).

Given its presence in a 1-liter flask, its initial concentration is \( 0.0497 \text{ M} \). Such mole calculations are not only essential to find initial concentrations but also to understand how much of a reactant has converted to product and what remains. Having a good grasp of these calculations is vital for success in chemical equilibrium problems.