Problem 83 - Chemistry: Principles and Reactions

Question

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), to sulfur dioxide and chlorine gases is a first-order reaction. $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At a certain temperature, the half-life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(7.5 \times 10^{2} \mathrm{~min}\). Consider a sealed flask with \(122.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (a) How long will it take to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in the sealed flask to \(45.0 \mathrm{~g}\) ? (b) If the decomposition is stopped after \(29.0 \mathrm{~h}\), what volume of \(\mathrm{Cl}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm is produced?

Step-by-Step Solution

Answer

Question: Based on the given information, find (a) the time needed for the decomposition of SO2Cl2 to reduce its amount to 45.0 g, and (b) the volume of Cl2 produced after 29.0 h. Solution: (a) First, we determine the rate constant (k) using the half-life formula and find the molar mass of SO2Cl2. After converting the given masses to moles, we use the integrated rate law formula to find the time needed for the SO2Cl2 to decompose to 45.0 g. (b) Next, we determine the moles of SO2Cl2 remaining after 29.0 h using the integrated rate law formula and find the moles of Cl2 produced. We then use the ideal gas law to find the volume of Cl2 produced under the given pressure and temperature conditions.

Step 1: (a) Calculate the time needed for decomposition

To calculate this part, we will use the integrated rate law for first-order reactions and the half-life to find the time needed for the SO2Cl2 to decompose to 45.0 g. The first-order integrated rate law is given by: $$ \ln{\frac{N_o}{N_t}} = kt $$ where \(N_o\) is initial concentration of SO2Cl2, \(N_t\) is concentration after time t, k is the rate constant, and t is time. We first need to find the rate constant (k) using the half-life formula: $$ k = \frac{\ln{2}}{t_{1/2}} = \frac{\ln{2}}{7.5\times10^2\mathrm{~min}} $$ Now we can solve for time by reorganizing the rate formula: $$ t = \frac{\ln{\frac{N_o}{N_t}}}{k} $$ We need to convert the mass to moles by using the molar mass of SO2Cl2: $$ \text{Molar mass of SO2Cl2=}\\ \text{S:} 32.07\mathrm{~g/mol} + \text{2}(\text{O:} 16.00\mathrm{~g/mol}) + \text{2}(\text{Cl:} 35.45\mathrm{~g/mol})=\\ 136.97\mathrm{~g/mol} $$ Find the initial moles (\(N_o\)) and final moles (\(N_t\)) of SO2Cl2: $$ N_o = \frac{122.0\mathrm{~g}}{136.97\mathrm{~g/mol}} \\ N_t = \frac{45.0\mathrm{~g}}{136.97\mathrm{~g/mol}} $$ And finally, plug the values into the formula to find the time (t): $$ t = \frac{\ln{\frac{N_o}{N_t}}}{k} $$

Step 2: (b) Calculate the volume of produced Cl2

After determining the time needed for the decomposition, we can find the moles of produced \(\mathrm{Cl}_{2}\). The stoichiometry of the reaction is 1:1, which means that for each mole of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) that decomposes, 1 mole of \(\mathrm{Cl}_{2}\) is produced. Use the given stopped time, 29.0 hr, and convert it to minutes: $$ 29.0\mathrm{~h} = 29.0\mathrm{~h}\times\frac{60\mathrm{~min}}{1\mathrm{~h}} = 1740\mathrm{~min} $$ Now, use the integrated rate law to find the moles of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) remaining at this time: $$ N_t = N_o \exp{(-kt)} = N_o \exp{(-\frac{\ln{2}}{7.5\times10^2\mathrm{~min}} \times 1740\mathrm{~min})} $$ To get the amount of \(\mathrm{Cl}_{2}\) produced, subtract \(N_t\) from \(N_o\): $$ \Delta N=\text{moles of Cl}_2 = N_o - N_t $$ Now we have the number of moles, and we can determine the volume of \(\mathrm{Cl}_{2}\) by using the ideal gas law (PV = nRT). Since we are given the pressure and temperature, we can determine the volume like this: $$ V = \frac{nRT}{P} = \frac{\Delta N \times (0.08206\frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{K}\cdot\mathrm{mol}} )(27^{\circ}\mathrm{C} + 273.15\mathrm{K})}{1.00\mathrm{~atm}} $$

Key Concepts

Sulfuryl Chloride Decomposition

Sulfuryl chloride, (\(\mathrm{SO}_2\mathrm{Cl}_2\)), decomposes into sulfur dioxide (\(\mathrm{SO}_2\)) and chlorine gas (\(\mathrm{Cl}_2\)), with a simple chemical reaction: \[\mathrm{SO}_2\mathrm{Cl}_2(g) \rightarrow \mathrm{SO}_2(g) + \mathrm{Cl}_2(g)\] This reaction is classified as a first-order reaction. First-order reactions depend on the concentration of a single substance, meaning their rate is directly proportional to the substance concentration. This characteristic simplifies calculations related to how fast a reaction proceeds and how long it takes for a specific amount of reactant to decompose.

Half-Life Calculation

Half-life is an essential concept used to describe the time taken for half of the initial amount of a substance to undergo a reaction. For first-order reactions, the half-life \((t_{1/2})\) is constant and can be calculated by the formula: \[ t_{1/2} = \frac{\ln{2}}{k} \]where

• \( k \) is the rate constant.

This means the half-life does not depend on the initial concentration of the substance in a first-order reaction.In our exercise, the half-life of sulfuryl chloride is given as \(7.5 \times 10^2 \mathrm{~min}\). With this value, we can determine the rate constant, which in turn helps us predict the time required for any amount of sulfuryl chloride to decompose to a different specified amount.

Integrated Rate Law

The integrated rate law for first-order reactions is crucial for determining the amount of reactant left after a certain time or to establish the time required for a given amount of reaction. The formula is:\[ \ln{\frac{N_o}{N_t}} = kt \]where

• \( N_o \) is the initial concentration,
• \( N_t \) is the remaining concentration after time \( t \),
• \( k \) is the rate constant,
• \( t \) is the time.

Using this equation, we can solve for any unknown variable, given the other parameters. For instance, to calculate the time taken for sulfuryl chloride to decompose from 122.0 g to 45.0 g, we first convert these amounts to moles, then apply this formula to find the time \( t \). The rate constant \( k \) is derived using the half-life, ensuring all units align for consistent calculations.

Ideal Gas Law

The ideal gas law is a fundamental equation that relates the amount of gas to its pressure, volume, and temperature. It is given as: \[ PV = nRT \] where

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (\(0.08206 \frac{\mathrm{L}\,\mathrm{atm}}{\mathrm{K}\,\mathrm{mol}}\)),
• \( T \) is the temperature in Kelvin.

This equation allows us to calculate the volume of chlorine gas produced from the decomposition reaction, based on the number of moles of chlorine generated and the given reaction conditions of pressure and temperature. For our specific scenario, the reaction is stopped after \(29.0\) hours, and the temperature and pressure are \(27^{\circ} \mathrm{C}\) and \(1.00\) atm respectively.