Question

A gaseous compound decomposes on heating as per the following equation: \(\mathrm{A}(\mathrm{g}) \longrightarrow B(\mathrm{~g})+2 \mathrm{C}(\mathrm{g}) .\) After 5 minutes and 20 seconds, the pressure increases by \(96 \mathrm{~mm} \mathrm{Hg}\). If the rate constant for this first order reaction is \(5.2 \times 10^{-4} \mathrm{~s}^{-1}\), the initial pressure of \(\mathrm{A}\) is (a) \(226 \mathrm{~mm} \mathrm{Hg}\) (b) \(37.6 \mathrm{~mm} \mathrm{Hg}\) (c) \(616 \mathrm{~mm} \mathrm{Hg}\) (d) \(313 \mathrm{~mm} \mathrm{Hg}\)

Short answer

None of the provided options is correct; the initial pressure of A is approximately 208.8 mm Hg.

Step-by-step solution

Understand the Reaction

The decomposition reaction is \[\mathrm{A}(\mathrm{g}) \longrightarrow B(\mathrm{g})+2 \mathrm{C}(\mathrm{g}).\] This means for every molecule of \(\mathrm{A}\) that decomposes, there is an increase in two molecules of \(\mathrm{C}\) and one molecule of \(\mathrm{B}\). Therefore, the total increase in gaseous molecules is three when one molecule of \(\mathrm{A}\) decomposes.

Express Pressure Increase

The total pressure increase of \(96 \mathrm{~mm} \mathrm{Hg}\) is due to the production of \(B\) and \(C\). Since the stoichiometry of the reaction shows an increase of \(3\) moles of gas for each mole of \(\mathrm{A}\) that decomposes, the pressure for \(\mathrm{A}\) that decomposed must be one third of the total pressure increase, or \(96/3 = 32 \mathrm{~mm} \mathrm{Hg}\).

Calculate Time in Seconds

Convert the time of the reaction from minutes and seconds to seconds only: \(5\) minutes is \(300\) seconds, so Total time = \(300 + 20 = 320\) seconds.

Use the First Order Rate Equation

The first order rate equation is: \[\ln \left(\frac{P_0}{P_0 - P_{\text{increase}}}\right) = k \cdot t.\]Where \(P_0\) is the initial pressure of \(\mathrm{A}\), \(P_{\text{increase}}\) is \(32 \mathrm{~mm} \mathrm{Hg}\), \(k\) is \(5.2 \times 10^{-4} \mathrm{~s}^{-1}\), and \(t\) is \(320\) seconds.

Plug Values into the Equation

Solve for \(P_0\):\[\ln \left(\frac{P_0}{P_0 - 32}\right) = (5.2 \times 10^{-4}) \times 320.\]Compute the right-hand side first:\[5.2 \times 10^{-4} \times 320 = 0.1664.\]

Solve for Initial Pressure, P_0

Rewriting the equation:\[\ln \left(\frac{P_0}{P_0 - 32}\right) = 0.1664.\]Taking the exponential on both sides, we find:\[\frac{P_0}{P_0 - 32} = e^{0.1664}.\]\[e^{0.1664} \approx 1.1813.\]Substituting and solving \[P_0 \approx 1.1813 \times (P_0 - 32).\]Expand and simplify this: \[P_0 \approx 1.1813 P_0 - 37.8496.\]Solving for \(P_0\) gives:\[0.1813 P_0 = 37.8496,\]\[P_0 \approx \frac{37.8496}{0.1813} \approx 208.8 \text{ mm Hg}.\]

Key concepts

Gas Decomposition

In chemistry, gas decomposition refers to a process where a gaseous compound breaks down into simpler gases. In this specific reaction, the compound \( ext{A(g)}\) decomposes into two gases: \(\text{B(g)}\) and \(\text{2C(g)}\). This means that each molecule of \(\text{A}\) loses its original structure and results in the formation of three new molecules - one of \(\text{B}\) and two of \(\text{C}\). The decomposition increases the total number of gaseous molecules present in the system. The decomposition is driven by heating, which provides the necessary energy to break the chemical bonds in compound \(\text{A}\). As the bonds are broken, the atoms rearrange to form the two new compounds, \(\text{B}\) and \(\text{C}\). This increase in the number of gas molecules affects the pressure, resulting in an increase in the total pressure of the system as observed in the exercise.

Rate Constant Calculation

The rate constant, often symbolized by \(k\), plays a critical role in determining the speed of the reaction for a first-order reaction. For these types of reactions, the rate constant is a unique value that is dependent on various factors like temperature and the nature of the reactants. In this exercise, the rate constant \(k\) is provided as \(5.2 \times 10^{-4} \text{ s}^{-1}\).For a first-order decomposition reaction, we use the equation:\[\ln \left(\frac{P_0}{P_0 - P_{\text{increase}}}\right) = k \cdot t\]Where:

• \(P_0\) is the initial pressure of the gaseous substance \(\text{A}\)
• \(P_{\text{increase}}\) is the increase in pressure due to decomposition
• \(t\) is the time over which the reaction occurs, here \(320\) seconds

The equation stems from the integrated rate law for first-order reactions, which shows a logarithmic relationship between the initial and current pressures (or concentrations) over time. Calculating the rate constant helps predict how quickly a particular concentration of \(\text{A}\) will decompose over time, which can be crucial for understanding the dynamics of the reaction in practical scenarios.

Stoichiometry in Gaseous Reactions

Stoichiometry is the study of the quantitative relationships between the amounts of reactants used and products formed by a chemical reaction. For gaseous reactions, like the decomposition reaction \(\text{A(g)} \rightarrow \text{B(g)} + 2 \text{C(g)}\), stoichiometry plays an important role in calculating changes in pressure and the amounts of reactants and products.In this exercise, the stoichiometric coefficients indicate that one mole of \(\text{A}\) produces one mole of \(\text{B}\) and two moles of \(\text{C}\). This results in a total increase of three moles of gas from every mole of \(\text{A}\) that decomposes. This is why the increase in total pressure due to products \(\text{B}\) and \(\text{C}\) is more than the decrease in pressure due to the consumption of \(\text{A}\).Understanding stoichiometry is essential to accurately calculate the amount of each substance involved, predict the amounts of products formed, and infer other related changes like pressure variations in gaseous reactions.