91Ó°ÊÓ

In chemistry, understanding gaseous decomposition reactions is crucial for analyzing how substances break down over time. This specific reaction involves the decomposition of a gaseous compound, denoted as \( \mathrm{A(g)} \), into other gaseous products \( \mathrm{B(g)} \) and \( \mathrm{2C(g)} \). This type of reaction is common in many real-world environments, such as the breaking down of pollutants or the combustion process.

During gaseous decomposition, a single reactant undergoes a breakdown into simpler substances. This reaction increases the total number of gas molecules, leading to an increase in pressure. It's important to note the stoichiometry: one mole of \( \mathrm{A} \) produces one mole of \( \mathrm{B} \) and two moles of \( \mathrm{C} \), thus changing the overall pressure in a measurable way. Understanding these changes is critical to determining reaction rates and the dynamics within a sealed environment.

Pressure change is an essential factor when dealing with gaseous reactions, especially when assessing a reaction's progress. In this exercise, the pressure increase is a direct result of the decomposition of compound \( \mathrm{A(g)} \), leading to the generation of products \( \mathrm{B} \) and \( \mathrm{C} \).

The initial challenge involves linking the pressure increase to the volumes of the products formed. Since the products \( \mathrm{B} \) and \( \mathrm{2C} \) contribute a total pressure increase of \( 96\, \mathrm{mmHg} \), we need to break this down:

• \( \mathrm{B} \) contributes \( x \), and \( \mathrm{2C} \) contributes \( 2x \)
• The equation becomes \( x + 2x = 96\, \mathrm{mmHg} \)
• Solving gives \( 3x = 96\, \mathrm{mmHg} \) which simplifies to \( x = 32\, \mathrm{mmHg} \)

The contribution of the total pressure change due to the emerging gases helps us deduce some critical parameters, such as the rate of reaction and the nature of the reactant's conversion.

The rate constant is a key parameter in understanding the speed of a chemical reaction, particularly in first-order reactions. For this exercise, the given rate constant \( k \) is \( 5.2 \times 10^{-4} \, \mathrm{s}^{-1} \), which indicates how quickly the gaseous compound \( \mathrm{A(g)} \) decomposes.

In a first-order reaction, the rate is directly proportional to the concentration of the single reactant. This relationship is mathematically represented using the formula:
\[ k = \frac{1}{t} \ln \left(\frac{P_0}{P}\right)\]
This equation integrates the rate constant \( k \), the initial pressure \( P_0 \), and the pressure \( P \) at time \( t \). By substituting known values, we can analyze the reaction kinetics and predict how rapidly the decomposition occurs under the given conditions.

Calculating the initial pressure of a reacting gas is a crucial aspect of analyzing gaseous decomposition. To find \( P_0 \), the initial pressure of \( \mathrm{A} \), we use the known rate constant and pressure dynamics.

In this exercise, after decomposing \( \mathrm{A} \) over \( 320 \) seconds (5 minutes and 20 seconds), we observe a pressure increase of \( 96\, \mathrm{mmHg} \) due to the formation of \( \mathrm{B} \) and \( \mathrm{C} \). The pressure decrease of \( \mathrm{A} \) equates to the presence of \( x = 32\, \mathrm{mmHg} \). Substituting this into the rate equation:

\[5.2 \times 10^{-4} = \frac{1}{320} \ln \left(\frac{P_0}{P_0 - 32}\right)\]
Solving the logarithmic equation gives the initial pressure \( P_0 = 313\, \mathrm{mmHg} \). This approach demonstrates how understanding the relationship between pressure changes and reaction kinetics allows for the determination of important reactant metrics in a straightforward way.