Problem 59 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) according to the equation : \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) [1991 - 6 Marks] is a first order reaction. After \(30 \mathrm{~min}\). from the start of the decomposition in a closed vessel, the total pressure developed is found to be \(284.5\) \(\mathrm{mm}\) of \(\mathrm{Hg}\) and on complete decomposition, the total pressure is \(584.5\) \(\mathrm{mm}\) of \(\mathrm{Hg}\). Calculate the rate constant of the reaction.

Step-by-Step Solution

Answer

The rate constant \(k\) is calculated using the formula for a first-order reaction with the given pressures.

Step 1: Understand the Reaction

The reaction begins with the decomposition of \(2 \ \mathrm{N}_{2} \mathrm{O}_{5}\). For each mole of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that decomposes, one mole of \(\mathrm{O}_{2}\) and two moles of \(\mathrm{NO}_{2}\) are produced, totaling three moles of gas products per mole of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposing.

Step 2: Analyze Initial and Final Pressures

The total pressure initially is due only to the undecomposed \(\mathrm{N}_{2} \mathrm{O}_{5}\). At the start, consider the pressure due to \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be \(P_0\). At completion, when all \(\mathrm{N}_{2} \mathrm{O}_{5}\) is decomposed, pressure is \(584.5 \ mm \ Hg\). This pressure corresponds to the product gases.

Step 3: Calculate Partial Pressures

The change in pressure from the initial to when \(\mathrm{N}_{2} \mathrm{O}_{5}\) is partially decomposed after 30 minutes is given by \( \Delta P = 284.5 \ mm \ Hg - P_0\). This change in pressure arises from newly formed \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\).

Step 4: Calculate Initial Partial Pressure

From stoichiometry and the given total pressures, if \(P_0\) is the initial pressure due to \(\mathrm{N}_{2} \mathrm{O}_{5}\), the pressure difference is \(584.5 \ mm \ Hg - P_0\). Since the products exert 3/2 times the pressure drop of reacted \(\mathrm{N}_{2} \mathrm{O}_{5}\), find \(P_0 = \frac{2}{3}(584.5) = 389.67 \ mm \ Hg\).

Step 5: Calculate Rate Constant for First Order Reaction

For a first-order reaction, the rate constant \(k\) is given by the equation:\[k = \frac{2.303}{t} \log \frac{P_{\text{final}}}{P_{\text{final}} - (P_{\text{observed}} - P_0)}\]Substitute \(t = 30 \text{ minutes} = 1800 \text{ seconds}\), \(P_{\text{final}} = 584.5\), \(P_{\text{observed}} = 284.5\), and \(P_0 = 389.67\) into the equation.\[k = \frac{2.303}{1800} \log \left( \frac{584.5}{584.5 - (284.5 - 389.67)} \right)\]Calculate to find the rate constant.

Key Concepts

Rate Constant Calculation

In a first-order reaction, the rate constant (\( k \)) is an essential parameter that measures the speed at which a reaction proceeds. The decomposition of \( \mathrm{N}_{2} \mathrm{O}_{5} \) is a classic example of such a reaction. Here, we aim to determine \( k \) using pressures and the known formula for first-order reactions. The equation used is:

• \[ k = \frac{2.303}{t} \log \frac{P_{\text{final}}}{P_{\text{final}} - (P_{\text{observed}} - P_0)} \]

Where:

• \( t \) is the time taken (in seconds),
• \( P_{\text{final}} \) is the pressure when the reaction is complete, and
• \( P_0 \) represents the initial pressure of \( \mathrm{N}_{2} \mathrm{O}_{5} \).

This formula accounts for changes in pressure as the reaction progresses. By converting time to seconds and using the observed pressure values, \( k \) can be determined. This method provides a quantitative understanding of how quickly \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposes in a closed system. Understanding this concept is fundamental for anyone studying chemical kinetics.

Stoichiometry in Decomposition Reactions

Stoichiometry describes the quantitative relationships in chemical reactions. For the decomposition of \( \mathrm{N}_{2} \mathrm{O}_{5} \), stoichiometry dictates the amounts of products formed from reactants. According to the balanced equation:

• \( 2\ \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4\ \mathrm{NO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \)

One mole of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposes to produce three moles of gas (two \( \mathrm{NO}_{2} \) and one \( \mathrm{O}_{2} \)), which affects the total pressure in the vessel. This stoichiometric relationship allows for the calculation of initial pressures based on known total pressures at various stages of the reaction. Understanding stoichiometry is crucial to predict how reactions will progress and to calculate the amounts of products formed from given reactants. It also helps in deciding the proportions of reactants needed for reactions to proceed completely.

Gaseous Pressure Changes

The decomposition of gases often involves changes in pressure, particularly in a closed system. For the reaction involving \( \mathrm{N}_{2} \mathrm{O}_{5} \), pressure changes are key indicators of reaction progress. Initially, the pressure is purely due to undecomposed \( \mathrm{N}_{2} \mathrm{O}_{5} \), but as decomposition occurs, the pressure reflects the sum of partial pressures from \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \).The total pressure increases because more moles of gas are produced than are consumed. Monitoring these changes provides a practical way to measure the reaction's extent and to verify the completion of the reaction. By understanding gaseous pressure changes, we gain insight into how chemical reactions can change the properties of a system, especially in scenarios where volume remains constant. This principle is a fundamental concept in physical chemistry that can apply in various industrial and laboratory contexts.