Problem 72 - Chemistry and Chemical Reactivity

Question

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ with a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0-L. flask, what is the partial pressure of each reactant and product after 245 minutes? What is the partial pressure of each reactant after 12 hours?

Step-by-Step Solution

Answer

After 245 minutes, each product is 12.5 mmHg, and \\(SO_{2}Cl_{2}\\) is 12.5 mmHg. After 12 hours, \\(SO_{2}Cl_{2}\\) is 2.79 mmHg.

Step 1: Understand the First-Order Reaction

A first-order reaction has a rate that depends linearly on the concentration (or pressure) of one reactant. For such reactions, the half-life is constant and given as 245 minutes.

Step 2: Calculate the Rate Constant (k)

The half-life of a first-order reaction is related to the rate constant by the formula:\[t_{1/2} = \frac{0.693}{k}\]We know the half-life, \(t_{1/2} = 245\) minutes. Therefore, solving for \(k\), we have:\[k = \frac{0.693}{245} = 0.00283 \, \text{min}^{-1}\]

Step 3: Determine Remaining \\(SO_{2}Cl_{2}\\) after 245 Minutes

Since the reaction is first-order, we can use the formula for a first-order decay:\[P = P_0 e^{-kt}\]where \(P_0 = 25\) mmHg is the initial pressure, and \(t = 245\) minutes. The pressure \(P\) of \(SO_2Cl_2\) after 245 minutes is:\[P = 25 e^{-0.00283 \times 245} = 12.5 \, \text{mmHg}\]

Step 4: Determine Products' Partial Pressure after 245 Minutes

Initially, the decomposition hasn't occurred, so all the pressure comes from \(SO_2Cl_2\). After 245 minutes, the amount decomposed is given by the initial pressure minus the remaining pressure, or:\[\Delta P = 25\, \text{mmHg} - 12.5\, \text{mmHg} = 12.5\, \text{mmHg}\]This change in pressure corresponds to the formation of \(SO_2\) and \(Cl_2\), each of which has partial pressures equal to \(\Delta P\). Thus, the pressure of each product is \(12.5\, \text{mmHg}\).

Step 5: Determine Remaining \\(SO_{2}Cl_{2}\\) after 12 Hours

Convert 12 hours to minutes: \(12 \times 60 = 720\) minutes. Use the first-order equation again to find the remaining pressure of \(SO_{2}Cl_{2}\):\[P = 25 e^{-0.00283 \times 720} \]Calculate this to find:\[P \approx 2.79 \, \text{mmHg}\]

Key Concepts

Half-Life Calculation

Understanding the half-life measurement is crucial when dealing with first-order reactions. The half-life of a reaction is the time it takes for half of the reactant to be consumed. For first-order reactions, this value is constant, regardless of concentration or pressure, because the reaction's rate depends linearly on only one reactant.
This means that even if you start with different amounts of a substance, the time required to have half of it decomposed remains unchanged.

The formula used to calculate the half-life for a first-order reaction is:

• \( t_{1/2} = \frac{0.693}{k} \)

This equation shows the indirect relationship between the half-life and the rate constant \(k\). A longer half-life indicates a smaller rate constant, which translates to a slower reaction. In our example, with a half-life of 245 minutes, the decomposition process is relatively slow.

Rate Constant

The rate constant, \(k\), is a vital parameter in chemical kinetics, as it signifies the speed of a reaction. For first-order reactions, the rate constant is derived from the half-life using the equation \(k = \frac{0.693}{t_{1/2}}\).
Using this equation, if our reaction has a half-life of 245 minutes, we get a rate constant \(k = 0.00283 \, \text{min}^{-1}\).

Knowing the rate constant allows us to predict how quickly a reactant is being consumed and estimate the amount of product formed at any given time.

This information is essential for practical applications, such as determining how long a reaction needs to proceed to achieve desired conditions or maintaining safety standards in an industrial setting.

Partial Pressure

In gaseous reactions, partial pressure is a key concept used to describe the pressure contribution of an individual gas within a mixture. Each gas's partial pressure is related to its concentration in the mixture through Dalton's Law, which asserts that the total pressure is the sum of the individual partial pressures.

For our specific reaction, we start with a partial pressure of 25 mmHg of \(\text{SO}_{2}\text{Cl}_{2}\). After 245 minutes, given it is a first-order reaction, calculations involving the rate constant determine that \(\text{SO}_{2}\text{Cl}_{2}\) has decomposed, reaching a partial pressure of 12.5 mmHg.

• The decomposed portion results in the formation of products \(\text{SO}_{2}\) and \(\text{Cl}_{2}\).

Both products will individually possess a partial pressure equal to the amount of decomposed reactant, balancing the reaction's equilibrium.

Chemical Decomposition

Chemical decomposition, especially in gaseous reactions, involves a single compound breaking down into two or more products. In our case, \(\text{SO}_{2}\text{Cl}_{2}\) decomposes into \(\text{SO}_{2}\) and \(\text{Cl}_{2}\), which can be expressed as:

• \( \text{SO}_{2}\text{Cl}_{2}(\text{g}) \rightarrow \text{SO}_{2}(\text{g})+\text{Cl}_{2}(\text{g}) \)

This type of decomposition often requires energy input, usually in the form of heat, to break chemical bonds within the molecules.

Moreover, because the decomposition of \(\text{SO}_{2}\text{Cl}_{2}\) is a first-order process, the rate is solely dependent on the concentration of \(\text{SO}_{2}\text{Cl}_{2}\), which simplifies calculations and predictions regarding the progress of the reaction over time.

Understanding decomposition reactions is crucial in fields like atmospheric chemistry and pollution control, where such processes affect environmental conditions.

Gaseous Reactions

Gaseous reactions, such as the one involving \(\text{SO}_{2}\text{Cl}_{2}\), are critical in many industrial and natural processes. These reactions involve the transformation of gaseous reactants into gaseous products with often significant implications.In our example, the decomposition occurs entirely in a gaseous state, meaning it can be analyzed and measured based on changes in partial pressures alone. This property makes gaseous reactions particularly suitable for studies of physical chemistry, where pressure and volume are major variables.

Additionally, knowing the behavior of gaseous reactions allows chemists to manipulate reaction conditions such as temperature and pressure to optimize yields and efficiencies. This knowledge is applied in fields ranging from chemical manufacturing to environmental science, illustrating the vast importance of mastering gaseous reaction dynamics.