Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a first order reaction \(A(\mathrm{~g}) \rightarrow 2 B(\mathrm{~g})+C(\mathrm{~g})\) at constant volume and \(300 \mathrm{~K}\), the total pressure at the beginning \((t=0)\) and at time \(t\) are \(P_{0}\)and \(P_{p}\) respectively. Initially, only \(A\) is present with concentration \([A]_{0}\), and \(t_{1 / 3}\) is the time required for the partial pressure of \(A\) to reach \(1 / 3^{\text {rd }}\) of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases) [Adv. 2018]

Step-by-Step Solution

Answer

The correct total pressure at \( t = t_{1/3} \) is \( \frac{7}{3} P_0 \).

Step 1: Identify the reaction dynamics

The reaction is first-order with respect to A: \( A(g) \rightarrow 2B(g) + C(g) \). Initially, only \( A \) is present.

Step 2: Express initial conditions

At \( t=0 \), the total pressure is \( P_0 \). This pressure is solely due to \( A \), thus \( P_A = P_0 \).

Step 3: Express the pressure change as the reaction proceeds

If \( x \) is the fraction of \( A \) that has reacted at time \( t \), then the pressure due to \( A \) remaining is \( P_A = P_0(1-x) \). From the stoichiometry, \( 2x \) moles of \( B \) and \( x \) moles of \( C \) are formed.

Step 4: Total pressure expression

The total pressure \( P_t \) at time \( t \) equals the sum of pressures due to \( A \), \( B \), and \( C \). Thus: \[ P_t = P_0(1-x) + 2P_0x + P_0x = P_0 + 2P_0x. \]

Step 5: Determine \( t_{1/3} \) conditions

At \( t = t_{1/3} \), the pressure of \( A \) is \( \frac{P_0}{3} \), which means that \( 2/3 \) of \( A \) has reacted (i.e., \( x = \frac{2}{3} \)).

Step 6: Calculate the new total pressure at \( t_{1/3} \)

Substitute \( x = \frac{2}{3} \) into the total pressure expression: \[ P_{t_{1/3}} = P_0 + 2P_0x = P_0 + 2P_0 \left( \frac{2}{3} \right) = P_0(1 + \frac{4}{3}) = \frac{7}{3}P_0. \]

Step 7: Determine the result

Repeating the calculations confirms that at \( t = t_{1/3} \), the total pressure \( P_t = \frac{7}{3} P_0 \), verifying this condition for the options provided.

Key Concepts

Reaction Dynamics

In a first order reaction, the rate of reaction depends only on the concentration of a single reactant. This means the reaction rate for our example, where a gaseous substance A decomposes to form products B and C, is directly proportional to the concentration of A. As time progresses, the concentration of A decreases, leading to a change in reaction rate.

For this particular first order reaction, initially only A is present. The reaction can be represented as:

• \[ A(g) \rightarrow 2B(g) + C(g) \]

At the start, the system contains only the reactant, which gradually converts into products over time. Dynamics in this reaction involve tracking how A breaks down into B and C, and how pressures of these gases change accordingly.

Stoichiometry

Stoichiometry in this reaction refers to the quantitative relationship between reactants and products. Here, 1 mole of A produces 2 moles of B and 1 mole of C. This relationship can be represented by the balanced chemical equation we studied earlier.

• Stoichiometrically, the formation of products is always tied to the consumption of the reactant.
• If \( x \) is the fraction of A that has reacted, then 2x moles of B and x moles of C are formed.

Understanding this relationship helps us calculate changes in pressure as the reaction proceeds from the initial state to any subsequent point in time. Thus, using stoichiometry, the change in quantities of each gas can be traced, giving a comprehensive picture of the reaction progress.

Ideal Gases

In this exercise, we're dealing with gases presumed to behave ideally. An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. The ideal gas law, given by \[PV = nRT \] is essential for understanding how gases will behave under different conditions of temperature (T), pressure (P), and volume (V).

• Here, the gases A, B, and C are each treated as ideal gases.
• This assumption simplifies calculations, allowing us to use direct proportionality between pressure and number of moles.

Such calculations are integral to chemistry and help predict gas behavior in reactions, as we analyze using the changes in pressures throughout the reaction's progression.

Partial Pressure

Partial pressure is a term used when describing the pressure contributed by a single gas in a mixture. Each component of a gas mixture exerts its own pressure, which is termed its partial pressure. The total pressure of the system is the sum of all partial pressures.

• Initially, the total pressure is due solely to A, hence, the partial pressure of A is \( P_0 \).
• As the reaction proceeds, the partial pressure of A decreases, and the partial pressures of B and C increase due to their formation.

When time \( t = t_{1/3} \), the partial pressure of A becomes \( \frac{1}{3}P_0 \), indicating that \( 2/3 \) of A has reacted. This change directly impacts the total system pressure and serves as a basis for calculating the time-dependence of this reaction's pressure dynamics.