Problem 130 - A Complete Resource Book in Chemistry for JEE Main

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) \(313 \mathrm{~mm} \mathrm{Hg}\)

Step-by-Step Solution

Answer

The initial pressure of \(\mathrm{A}\) is \(313 \, \text{mm Hg}\) (option d).

Step 1: Identify the Reaction and Data

The decomposition reaction is \(\mathrm{A} \rightarrow \mathrm{B} + 2\mathrm{C}\). The net change in pressure is given by a 96 mm Hg increase after 5 minutes and 20 seconds. The rate constant \(k\) is \(5.2 \times 10^{-4} \, \text{s}^{-1}\).

Step 2: Understand the Pressure Increase Equation

For the decomposition \(\mathrm{A} \rightarrow \mathrm{B} + 2\mathrm{C}\), the total increase in pressure (\(\Delta P_T\)) is due to an increase in three moles of gas for every mole of \(\mathrm{A}\) decomposing. Hence, \(\Delta P_T = 3x\), where \(x\) is the pressure due to decomposed \(\mathrm{A}\). Here, \(\Delta P_T = 96 \, \text{mm Hg}\).

Step 3: Calculate Decomposed Pressure Change

Since \(\Delta P_T = 3x = 96 \, \text{mm Hg}\), solve for \(x\) giving \(x = \frac{96}{3} = 32 \, \text{mm Hg}\). This means 32 mm Hg of \(\mathrm{A}\) has decomposed.

Step 4: Apply First Order Kinetics Equation

Use the first order reaction formula: \(k = \frac{1}{t} \ln \left( \frac{P_0}{P_0 - x} \right)\), where \(t = 5\, \text{minutes} + 20 \, \text{seconds} = 320 \, \text{s}\).

Step 5: Rearrange to Find Initial Pressure

Rearrange the first order equation: \(P_0 = x \frac{e^{kt}}{e^{kt} - 1}\). Substitute \(x = 32 \, \text{mm Hg}\), \(k = 5.2 \times 10^{-4} \, \text{s}^{-1}\), and \(t = 320 \, \text{s}\).

Step 6: Calculate and Compare Results

Plug the values into the equation: \(P_0 = 32 \frac{e^{(5.2 \times 10^{-4} \times 320)}}{e^{(5.2 \times 10^{-4} \times 320)} - 1}\). Solve this numerically to find \(P_0 \approx 313 \, \text{mm Hg}\). This matches option (d).

Key Concepts

First Order Reactions

In chemical kinetics, a first order reaction is a type of reaction where the rate is directly proportional to the concentration of one reactant. This means that if the concentration of the reactant doubles, the rate of reaction also doubles.

First order reactions follow the equation:

• Rate = k[A]

where

• k is the rate constant
• [A] is the concentration of the reactant

A key feature of first order reactions is that they have a constant half-life, regardless of the starting concentration. This makes them predictable over time. To calculate the concentration of a reactant at any given time, we often use the formula:

• ln([A]_0/[A]) = kt

where:

• [A]_0 is the initial concentration
• [A] is the concentration at time t
• t is the time elapsed

Decomposition Reactions

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. These reactions require an input of energy that could be in the form of heat, light, or electricity. In our exercise example, the decomposition reaction involves:

• A (g) → B (g) + 2C (g)

This reaction involves the breakdown of one molecule of A into three molecules in total (1 of B and 2 of C).

Characteristics of decomposition reactions include:

• Being endothermic - they absorb energy
• The resulting products are often in different physical states
• A change in pressure if occurring in a gaseous state, due to the increase in number of gas molecules

Pressure Change

Pressure change in a chemical reaction involving gases is an important indicator of the progression of the reaction. When a gaseous reactant decomposes, the number of gas molecules often changes, which directly affects the pressure.

In the decomposition reaction

• A (g) → B (g) + 2C (g), for example, the total number of molecules increases, leading to an increase in pressure.

Based on the given exercise, the total pressure increase observed is 96 mm Hg, which indicates the formation of additional gas molecules from the decomposition process.

• The equation ΔP_T = 3x is used, where x is the pressure due to decomposed A.
• Here, ΔP_T = 96 mm Hg corresponds to a decomposition of A leading to this pressure increase.

Rate Constant

The rate constant, denoted as k, is a crucial value in chemical kinetics. It is a proportionality factor in the rate equation of a chemical reaction. For first order reactions, the rate constant has the units of s^-1, reflecting its dependence solely on time.

The rate constant provides valuable information such as:

• The speed of the reaction - a higher k indicates a faster reaction.
• It is unaffected by the concentration of substances but can vary with temperature.
• For our exercise, the rate constant is 5.2 × 10^-4 s^-1, which is used in the calculation of initial pressure.

Using the rate constant with the first order kinetics equation allows us to find important details like the initial pressure or concentration of the reactant in question.