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 \(7.5 \times 10^{2}\) 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?

Short answer

Question: (a) Calculate the time required to reduce the amount of SO2Cl2 to 45.0 g. (b) Calculate the volume of Cl2 produced after 29.0 hours. Answer: (a) The time required to reduce the amount of SO2Cl2 to 45.0 g is approximately 2086.9 minutes. (b) The volume of Cl2 produced after 29.0 hours is approximately 21.7 liters.

Step-by-step solution

(a) Calculate the time required to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(45.0 \: \mathrm{g}\)

First, let's find the rate constant \(k\) for the reaction using the half-life formula for first-order reactions: \(t_{1/2} = \frac{0.693}{k}\) Now we can plug in the given value of the half-life: \(k = \frac{0.693}{7.5 \times 10^{2} \: \mathrm{min}} = 9.240 \times 10^{-4} \: \mathrm{min^{-1}}\) We are given the initial amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) as \(122.0 \: \mathrm{g}\), and we need to find how long it takes to reduce the amount to \(45.0 \: \mathrm{g}\). To do this, we use the first-order kinetics equation: \(ln \frac{[A]_0}{[A]} = kt\) Where \([A]_0\) is the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), \([A]\) is the final concentration and \(t\) is the time. First, we need to convert the given masses into concentrations. The molar mass of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1 \times (32.07 \: \mathrm{g/mol}) + 2 \times (35.45 \: \mathrm{g/mol}) = 102.97 \: \mathrm{g/mol}\). Then we can calculate the concentrations: \([A]_0 = \frac{122.0 \: \mathrm{g}}{102.97 \: \mathrm{g/mol}} = 1.185 \: \mathrm{mol}\) \([A] = \frac{45.0 \: \mathrm{g}}{102.97 \: \mathrm{g/mol}} = 0.437 \: \mathrm{mol}\) Now, we can plug the concentrations and the rate constant into the first-order kinetics equation: \(ln \frac{1.185}{0.437} = (9.240 \times 10^{-4} \: \mathrm{min^{-1}})t\) Solving for \(t\): \(t = \frac{ln \frac{1.185}{0.437}}{9.240 \times 10^{-4} \: \mathrm{min^{-1}}}= 2086.9 \: \mathrm{min}\)

(b) Calculate the volume of \(\mathrm{Cl}_{2}\) produced after \(29.0 \: \mathrm{h}\)

To find the amount of \(\mathrm{Cl}_{2}\) produced, we first need to know how much \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has reacted after \(29.0 \: \mathrm{h}\). To do this, we use the first-order kinetics equation and the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\): \(ln \frac{[A]_0}{[A]} = kt\) \([A] = [A]_0e^{-kt} = 1.185 \: \mathrm{mol} \times e^{-(9.240 \times 10^{-4} \: \mathrm{min^{-1}})(29.0 \: \mathrm{h} \times 60 \: \mathrm{min/h})} = 0.305 \: \mathrm{mol}\) The amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) reacted is equal to the initial concentration minus the final concentration: \(\Delta [A] = 1.185 - 0.305 = 0.880 \: \mathrm{mol} = [Cl_{2}]\) The volume of \(\mathrm{Cl}_{2}\) produced can be found using the ideal gas law: \(PV = nRT\) Where \(P = 1.00 \: \mathrm{atm}\), \(V\) is the volume of \(\mathrm{Cl}_{2}\) in liters, \(n = 0.880 \: \mathrm{mol}\), \(R = 0.0821 \: \mathrm{(L\: atm)/(K\: mol)}\) and \(T = 27^{\circ}C + 273.15 = 300.15 \: \mathrm{K}\). Solving for \(V\): \(V = \frac{nRT}{P} = \frac{(0.880 \: \mathrm{mol})(0.0821 \: \mathrm{(L\: atm)/(K\: mol)})(300.15 \: \mathrm{K})}{1.00 \: \mathrm{atm}} = 21.7 \: \mathrm{L}\)

Key concepts

Understanding the Half-Life of a Reaction

The concept of half-life is pivotal in understanding the dynamics of a first-order reaction. The half-life of a reaction, denoted as \( t_{1/2} \), is the time required for the concentration of a reactant to decrease by half its initial value. In the context of our exercise, sulfuryl chloride (\( \text{SO}_2 \text{Cl}_2 \)) decomposes into sulfur dioxide and chlorine gases in a reaction that exhibits first-order kinetics.

This implies that the rate at which \( \text{SO}_2 \text{Cl}_2 \) decomposes is directly proportional to its concentration at any given moment. Consequently, the half-life remains constant irrespective of the initial concentration. To practically apply this concept to our problem, knowing that the half-life of sulfuryl chloride is \( 7.5 \times 10^2 \) min, we can assess how long it takes for any amount of \( \text{SO}_2 \text{Cl}_2 \) to reduce to a certain level, such as from 122.0 g to 45.0 g. Understanding the half-life concept allows us to easily determine the remaining concentration of a substance after an elapsed time period, provided that the reaction stays consistent with first-order kinetics.

Deciphering the First-Order Kinetics Equation

The first-order kinetics equation is essential for quantifying reaction rates and provides a bridge between the half-life of a substance and its rate of decomposition or reaction. Represented by the equation \( ln(\frac{[A]_0}{[A]}) = kt \), it relates the natural logarithm of the ratio between the initial concentration \( [A]_0 \) and the concentration at time \( t \), \( [A] \), to the product of the reaction rate constant \( k \) and time \( t \).

From Mass to Molar Concentrations

To use this equation effectively, as demonstrated in the exercise, we must convert the mass of the substance into its molar concentration using its molar mass. Once we have the initial and final concentrations, we can calculate the elapsed time necessary for the reaction to reach a specific concentration milestone or determine the amount of reactant remaining after a certain duration.

This calculation is beneficial in various scientific fields, including pharmacology, environmental science, and nuclear chemistry, as it helps predict how long it would take for a compound to undergo a certain degree of transformation. The simplicity of the first-order kinetics equation makes it an invaluable tool for students and professionals in these domains.

Applying the Ideal Gas Law in Reaction Analysis

The ideal gas law is a fundamental equation that relates the pressure \( P \), volume \( V \), and temperature \( T \) of an ideal gas to its amount in moles \( n \) through the equation \( PV = nRT \). Here \( R \) is the universal gas constant. When a reaction releases or consumes gases, as is the case with the decomposition of \( \text{SO}_2 \text{Cl}_2 \), the ideal gas law becomes very relevant to calculate the volume of the gases produced or consumed under specific conditions of temperature and pressure.

Volume Calculation of Gaseous Products

By knowing the amount of gas produced or consumed in moles, as well as the conditions under which the reaction takes place, we can manipulate the ideal gas law to find out other parameters such as the volume of a gas at a certain temperature and pressure. In our problem, this allows us to conclude the volume of chlorine gas produced after stopping the reaction for a set time. This concept is not only crucial for laboratory and industrial calculations but also for understanding processes in the atmosphere and aquatic systems where gases often play a role.