Problem 54 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At constant temperature and volume, \(X\) decomposes as [2005 - 4 Marks] \(2 \mathrm{X}(\mathrm{g}) \rightarrow 3 \mathrm{Y}(\mathrm{g})+2 \mathrm{Z}(\mathrm{g}) ; P_{X}\) is the partial pressure of \(X\) \begin{tabular}{|c|c|c|} \hline Observation No. & Time (in minute) & \(P_{X}\) (in mm of \(\mathrm{Hg}\) ) \\\ \hline 1 & 0 & 800 \\ \hline 2 & 100 & 400 \\ \hline 3 & 200 & 200 \\ \hline \end{tabular} (i) What is the order of reaction with respect to \(X ?\) (ii) Find the rate constant. (iii) Find the time for \(75 \%\) completion of the reaction. (iv) Find the total pressure when pressure of \(X\) is \(700 \mathrm{~mm}\) of \(\mathrm{Hg}\).

Step-by-Step Solution

Answer

(i) First-order. (ii) 0.00693 min⁻¹. (iii) 200 minutes. (iv) 950 mmHg.

Step 1: Determine Reaction Order

The reaction follows the form \( aX \rightarrow bY + cZ \). The data provided shows the partial pressure of \( X \) vs. time. To find the order of the reaction, we note that the pressure of \( X \) halves over each 100-minute interval (e.g., from 800 to 400 then 400 to 200). This indicates a first-order reaction because the half-life is constant.

Step 2: Calculate Rate Constant (k)

For a first-order reaction, we use the formula \( k = \frac{1}{t} \ln\left(\frac{P_{0}}{P}\right) \). Using the data, \( P_0 = 800 \) mmHg and \( P = 400 \) mmHg at \( t = 100 \) minutes. Therefore, \( k = \frac{1}{100} \ln\left(\frac{800}{400}\right) = \frac{1}{100} \ln(2) = \frac{0.693}{100} = 0.00693 \) min\(^{-1}\).

Step 3: Calculate Time for 75% Completion

A 75% completion means 25% of \( X \) remains, so \( P = 0.25 \times 800 = 200 \) mmHg. Using the first-order rate equation \( t = \frac{1}{k} \ln\left(\frac{P_0}{P}\right) \), we find \( t = \frac{1}{0.00693} \times \ln\left(\frac{800}{200}\right) = 200 \) minutes.

Step 4: Calculate Total Pressure When \( P_X = 700 \) mmHg

Initially, the pressure is 800 mmHg. When \( P_X = 700 \), \( \ X \) has decomposed by 100 mmHg. Using stoichiometry \( 2X \rightarrow 3Y + 2Z \), 100 mmHg of \( X \) decomposing produces \( \frac{3}{2} imes 100 = 150 \) mmHg of \( Y \) and 100 mmHg of \( Z \). Thus, total pressure = 700 mmHg (\( X \)) + 150 mmHg (\( Y \)) + 100 mmHg (\( Z \)) = 950 mmHg.

Key Concepts

Reaction Order

The reaction order tells us how the rate of a chemical reaction depends on the concentration of the reactants. For the reaction involving the decomposition of substance \( X \), we examine how the partial pressure of \( X \) changes over time. By observing that the pressure of \( X \) halves every 100 minutes (from 800 mmHg to 400 mmHg, and from 400 mmHg to 200 mmHg), we determine that the reaction is first-order. This consistent half-life signals that the reaction rate is directly proportional to the concentration of \( X \). To put it simply, the order of one means if you double the concentration of \( X \), the rate also doubles.

Rate Constant

The rate constant \( k \) is a proportionality factor in the rate equation of a chemical reaction. It provides a link between the reaction rate and the concentrations of reactants. For a first-order reaction, like our example, the rate constant can be calculated using the formula: \[ k = \frac{1}{t} \ln\left(\frac{P_{0}}{P}\right) \]where \( P_{0} \) is the initial partial pressure and \( P \) is the pressure at time \( t \). In this case, \( P_{0} = 800 \) mmHg and \( P = 400 \) mmHg at \( t = 100 \) minutes. Plugging in these values, we find that \( k = \frac{0.693}{100} \approx 0.00693 \) min\(^{-1}\). The rate constant gives a sense of the speed of the reaction and is crucial in predicting how fast a reaction will proceed.

Partial Pressure

Partial pressure refers to the pressure exerted by a single component of a mixture of gases. In our decomposition reaction, we focus on how the partial pressure of substance \( X \) changes over time. Partial pressure is directly related to the amount of the gaseous substance present in the system. As the reaction progresses:

• The partial pressure of \( X \) decreases, indicating its consumption in the reaction.
• The partial pressures of \( Y \) and \( Z \), the products, increase due to their formation.

Understanding partial pressures is essential when dealing with reactions in closed containers where volume and temperature remain constant. It allows calculation of the total pressure in the system using the individual pressures of each gas involved.

Stoichiometry

Stoichiometry involves the quantitative relationships or ratios in which substances participate in chemical reactions. The balanced chemical equation used in our problem states: \[ 2\mathrm{X}(\mathrm{g}) \rightarrow 3\mathrm{Y}(\mathrm{g}) + 2\mathrm{Z}(\mathrm{g}) \]This tells us that for every 2 moles (or any equivalent measure such as mmHg in gases) of \( X \) that decomposes, 3 moles of \( Y \) and 2 moles of \( Z \) are produced. In the calculation when \( P_X \) decreases from 800 mmHg to 700 mmHg:

• 100 mmHg of \( X \) is consumed.
• This results in the production of \( \frac{3}{2} \times 100 = 150 \) mmHg of \( Y \) and 100 mmHg of \( Z \).

Stoichiometry is crucial because it helps us calculate how much of each product is formed and how the total pressure changes as the reaction proceeds.