Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

For the following reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) The initial pressure was \(\mathrm{P}_{0}\) while pressure after time ' \(\mathrm{t}\) ' was \(\mathrm{P}_{1}\). The rate constant \(\mathrm{k}\) will be (a) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{t}}}\) (b) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{\mathrm{1}}}\) (c) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{1}}\) (d) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-2 \mathrm{P}_{\mathrm{t}}}\)

Step-by-Step Solution

Answer

The correct answer is (b). \( k = \frac{2.303}{t} \log_{10} \frac{P_0}{2P_0 - P_1} \).

Step 1: Understand the Reaction

The reaction given is \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g}) \). It is a first-order reaction since one reactant decomposes into products.

Step 2: Determine Rate Law for First-Order Reactions

The rate law for a first-order reaction is \( k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]} \). Pressure is analogous to concentration in gaseous reactions, so we can use pressures in our calculations.

Step 3: Relate Initial and Final Pressures to Concentration

Initially, the pressure is \( P_0 \), which is the pressure due to reactant \( \mathrm{A} \) alone. After time \( t \), the pressure of \( \mathrm{A} \) decreases as it forms \( \mathrm{B} \) and \( \mathrm{C} \). The total final pressure \( P_1 = P_0 - [A] + [B] + [C] = 2P_0 - [A] \) assuming \([B]\) and \([C]\) have equal contributions as \([A] \).

Step 4: Apply First-Order Formula to Given Options

You need to find which equation aligns with the first-order formula \( k = \frac{2.303}{t} \log_{10} \frac{P_0}{P} \). We identify \( P \) as the effective pressure of \( \mathrm{A} \) at time \( t \), which relates directly to option (b). Thus, \( k = \frac{2.303}{t} \log_{10} \frac{P_0}{2P_0 - P_1} \).

Step 5: Check Consistency of Option with Derived Formula

Given the direct relationship from the first-order reaction and using pressure instead of concentration, the formula in option (b) is consistent with our derivation. It accounts for the change in pressure due to the formation of products \( \mathrm{B} \) and \( \mathrm{C} \) from \( \mathrm{A} \).

Key Concepts

Rate Constant Determination

Determining the rate constant is a crucial aspect of understanding chemical reactions, especially first-order reactions. For a first-order reaction involving a single reactant breaking down into products, such as the reaction \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g}) \), determining the rate constant \( k \) involves examining the change in concentration or pressure over time.

The formula for determining a first-order rate constant is given by \( k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]} \), where \([A]_0\) is the initial concentration of the reactant, and \([A]\) is the concentration at time \( t \). In gaseous reactions, instead of concentration, we can measure pressure. This leads to the corresponding formula \( k = \frac{2.303}{t} \log_{10} \frac{P_0}{P_t} \), with \( P_0 \) and \( P_t \) being the pressures at time 0 and time \( t \), respectively.

This determination is pivotal because the rate constant serves as an indicator of the speed of the reaction. It is also essential for modeling the kinetics of the reaction and predicting how the concentration of reactants decreases over time.

Gaseous Reactions

In reactions involving gases, pressure serves as a direct analog to concentration used in liquid-phase reactions due to the ideal gas equation, which relates pressure, volume, and temperature to the number of moles of a gas.

The reaction \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g}) \) is a typical gaseous reaction where each substance is in the gaseous state. As the reaction progresses, the pressure exerted by the substances changes. Initially, the pressure is fully due to reactant \( \mathrm{A} \), and as it decomposes, products \( \mathrm{B} \) and \( \mathrm{C} \) form, contributing to the total pressure of the system.

Monitoring these pressure changes over time allows researchers to study the reaction kinetics. The change in total pressure reflects the change in the number of gas molecules due to the reaction, which is a key aspect in applying the rate law to first-order reactions and thus calculating the rate constant.

Rate Law for First-Order Reactions

The rate law for first-order reactions is straightforward yet fundamental in the study of chemical kinetics. It defines how the rate at which reactants are consumed depends solely on the concentration (or pressure, in the case of gases) of one reactant.

For a first-order reaction like \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g}) \), the rate of reaction is directly proportional to the concentration of \( \mathrm{A} \). Mathematically, this is expressed as \( k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]} \), and when considering pressure, it becomes \( k = \frac{2.303}{t} \log_{10} \frac{P_0}{P_t} \).

This principle highlights that the greater the concentration or pressure of the reactant, the faster the reaction proceeds. Importantly, the first-order rate law indicates that the rate constant \( k \) is independent of temperature, concentration of products, or catalysts, focusing solely on the initial and present concentrations or pressures of the reactants. This approach elegantly simplifies understanding the kinetics of a variety of chemical and biochemical processes.