Problem 32 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\), is \(3.0 \times 10^{-5}\) \(\mathrm{sec}^{-1}\). If the rate is \(2.40 \times 10^{-5}\) mol litre \(^{-1} \mathrm{sec}^{-1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in mol litre \(^{-1}\) ) is [2000S] (a) \(1.4\) (b) \(1.2\) (c) \(0.04\) (d) \(0.8\)

Step-by-Step Solution

Answer

The concentration of \(\mathrm{N}_2 \mathrm{O}_5\) is \(0.8\,\text{mol L}^{-1}\) (option d).

Step 1: Identify the Rate Law

For the given reaction, we know it follows first-order kinetics based on the units of the rate constant, which are \( ext{sec}^{-1} \). Therefore, the rate law is given by \( ext{Rate} = k[ ext{N}_2 ext{O}_5] \).

Step 2: Substitute Given Values

Substitute the given values into the rate law equation. We know the rate \(( ext{Rate} = 2.40 \times 10^{-5} ext{ mol L}^{-1} ext{ sec}^{-1} )\) and the rate constant \(( k = 3.0 \times 10^{-5} ext{ sec}^{-1} )\). Thus, we have: \[ 2.40 \times 10^{-5} = 3.0 \times 10^{-5}[ ext{N}_2 ext{O}_5] \].

Step 3: Solve for Concentration

Rearrange the equation to find the concentration of \( ext{N}_2 ext{O}_5 \): \[ [ ext{N}_2 ext{O}_5] = \frac{2.40 \times 10^{-5}}{3.0 \times 10^{-5}} \].

Step 4: Calculate the Result

Calculate the result of the division: \[ [ ext{N}_2 ext{O}_5] = \frac{2.40}{3.0} \], which simplifies to \([\text{N}_2\text{O}_5] = 0.8 ext{ mol L}^{-1}\).

Key Concepts

First-Order Reaction

In the world of chemical kinetics, reactions can occur at different orders, leading to variations in how their rates depend on the concentration of the reactants. A first-order reaction is where the rate of reaction is directly proportional to the concentration of a single reactant. This means, in a practical sense, if you double the concentration of this reactant, the rate of reaction will also double. In first-order reactions, the rate is described by the equation:\[ \text{Rate} = k[A] \]Here, \(k\) is the rate constant and \([A]\) is the concentration of the reactant. Understanding this helps in predicting how fast the reaction will proceed when you know the concentration of the reactant.

Rate Law

The rate law is a mathematical description of how the rate of a chemical reaction depends on the concentration of its reactants. For our first-order reaction scenario, the rate law takes on a specific form, simplifying the relationship between variables.The rate law expression for a first-order reaction like the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) is given by:\[ \text{Rate} = k[\mathrm{N}_2\mathrm{O}_5] \]Here, \(k\) is the rate constant, characterizing the speed of the reaction at a given temperature. The concentration of \(\mathrm{N}_2\mathrm{O}_5\) is raised to the power of one, indicating its first-order nature. This directly implies that the rate changes linearly with variations in \([\mathrm{N}_2\mathrm{O}_5]\). This principle is vital for determining reaction dynamics and how effectively the process can occur under different conditions.

Rate Constant

The rate constant, denoted as \(k\), is a crucial part of the rate law and encompasses the effects of temperature and other environmental conditions on the reaction rate. It is specific to each reaction and is determined experimentally. For first-order reactions, the units of \(k\) are \(\text{sec}^{-1}\), signifying its relationship with time. Knowing the value of the rate constant is essential for making accurate predictions about how a reaction proceeds over time. For example, if the rate constant for the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) is given as \(3.0 \times 10^{-5}\ \text{sec}^{-1}\), it informs us that under specified conditions, the reaction progresses at a specific pace.

Concentration Calculation

A critical application of understanding first-order reactions and rate laws is performing concentration calculations. These calculations enable us to find out the unknown concentration of a reactant in the reaction mixture from known variables. Starting with the rate law:\[ \text{Rate} = k[\mathrm{N}_2\mathrm{O}_5] \]We can rearrange to solve for the concentration of \(\mathrm{N}_2\mathrm{O}_5\):\[ [\mathrm{N}_2\mathrm{O}_5] = \frac{\text{Rate}}{k} \]By substituting the known rate \(2.40 \times 10^{-5}\ \text{mol L}^{-1} \text{ sec}^{-1}\) and rate constant \(3.0 \times 10^{-5}\ \text{sec}^{-1}\), we find:\[ [\mathrm{N}_2\mathrm{O}_5] = \frac{2.40 \times 10^{-5}}{3.0 \times 10^{-5}} = 0.8\ \text{mol L}^{-1}\]This calculation demonstrates how we apply theoretical concepts to solve practical problems, crucial for managing chemical processes and predicting outcomes in reactions.