Problem 151 - A Complete Resource Book in Chemistry for JEE Main

Question

For a first order reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\), the total pressure of \(\mathrm{A}+\mathrm{B}+\mathrm{C}\) at time ' \(\mathrm{t}^{\prime}\) and \(\infty\) are \(\mathrm{P}_{2}\) and \(\mathrm{P}_{3}\) respectively. The constant \(\mathrm{k}\) of the reaction is (a) \(\frac{1}{t} \ln \frac{P_{3}}{2\left(P_{3}-P_{2}\right)}\) (b) \(\frac{1}{t} \ln \frac{2 P_{3}}{P_{1}-P_{2}}\) (c) \(\frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\) (d) \(\frac{1}{t} \ln \frac{P_{3}}{2 P_{3}-P_{2}}\)

Step-by-Step Solution

Answer

The correct option is (c) \(\frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\).

Step 1: Write down the Reaction Scheme

The given first order reaction is \(\mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)}\). Since it's a first-order reaction, the rate depends only on the concentration of \(\mathrm{A}\).

Step 2: Analyze the Pressure Changes

Initially, at \(t=0\), the pressure is \(P_A\), the partial pressure of \(\mathrm{A}\). As the reaction proceeds, \(\mathrm{A}\) will transform into \(\mathrm{B}\) and \(\mathrm{C}\). If \(\mathrm{x}\) moles of \(\mathrm{A}\) react, they will create an equal amount of \(\mathrm{B}\) and \(\mathrm{C}\). Thus, the total pressure at any time \(t\) will be \(P_{t} = P_{A} + x + x = P_{A} + 2x\).

Step 3: Express Final and Intermediate Pressures

At time \(t\), the total pressure is given as \(P_2 = P_{A} + 2x\). At infinite time, when the reaction is complete, the total pressure will be \(P_3 = P_A + 2(P_A) = 2P_A\) assuming \(P_A\) moles of \(\mathrm{A}\) converted completely into \(\mathrm{B} \) and \(\mathrm{C}\).

Step 4: Determine the Change in Concentration

Since \(P_3 = 2P_A\), and \(P_2 = P_A + 2x\), it implies that when \(t = \infty\), \(x = P_A\). We use \(P_3\) for reference, then the change in pressure \(P_3 - P_2 = 2P_A - (P_A + 2x)\). Replacing \(x = P_A - (P_3 - P_2)/2\), we have \(P_A = P_3-x\).

Step 5: Calculate the Rate Constant \(k\)

The rate constant for a first-order reaction is given by the formula \(k = \frac{1}{t} \ln \frac{[A]_0}{[A]}\). Using pressures, this can be translated to \(k = \frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\).

Step 6: Choose the Correct Option

Review the options to find the expression for \(k\) that matches with the derived formula. Option (c), \(\frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\), matches the formula derived for the rate constant \(k\).

Key Concepts

reaction rate constant

In chemical kinetics, the reaction rate constant (often denoted as \( k \)) is a fundamental factor for the rate equation of a reaction. For a first-order reaction, the rate of the reaction is directly proportional to the concentration of a single reactant, specifically the reactant that is disappearing as the reaction proceeds. Here, we're dealing with the reaction \( \mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)} \) where the rate is dependent solely on the concentration of \( \mathrm{A} \), the reactant. The rate constant \( k \) uniquely defines how quickly a reaction proceeds under a given set of conditions. It's derived from the formula for first-order kinetics: \[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]In this scenario, pressures can be used to represent concentrations, allowing us to rewrite this equation in terms of observable pressures. This is useful in gas-phase reactions where partial pressures can easily be measured. The rate constant is found by relating initial and final partial pressures, which gives us an expression: \[ k = \frac{1}{t} \ln \left( \frac{P_{3}}{P_{3}-P_{2}} \right) \]This aligns with the fact that \( P_{3} \) represents the total pressure at reaction completion and \( P_{3} - P_{2} \) reflects the pressure change as the reaction develops.

pressure changes in reactions

Pressure changes in chemical reactions, especially in gas-phase reactions, provide insight into the progression of the reaction. For the given first-order reaction \( \mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)} \), the changes in pressure over time reveal information about the reactants and products involved. At the beginning, the full pressure is attributed to \( \mathrm{A(g)} \) alone. As time passes and \( \mathrm{A} \) converts into \( \mathrm{B} \) and \( \mathrm{C} \), the total pressure increases because two moles of gas replace each disappearing mole of \( \mathrm{A} \).

• Initially, the pressure, \( P_A \), reflects only \( \mathrm{A} \).
• As \( \mathrm{A} \) changes into \( \mathrm{B} \) and \( \mathrm{C} \), the combined pressure jumps to \( P_2 = P_A + 2x \), where \( x \) is the mole change over time.
• At completion, the reaction reaches \( P_3 = 2P_A \), showing total pressure resulting from complete conversion of \( \mathrm{A} \).

Tracking pressure adjustments enables us to calculate how much reactant was used and to determine other quantities such as the rate constant.

chemical kinetics

Chemical kinetics explores the rates of chemical processes and the factors affecting them – concentration, temperature, and catalysts, to name a few. It provides an understanding of how fast reactions progress and which variables modify these rates. For the reaction \( \mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)} \), we are examining a first-order reaction. This implies that the rate is proportional to the concentration of only one species, \( \mathrm{A} \). When concentrations are difficult to measure directly, like in many gaseous reactions, pressure changes provide a direct measure of reaction progress.The key aspects of chemical kinetics in this context include:

• Understanding concentration changes and how they relate to observable components, such as pressure.
• Using the rate law, \( \text{Rate} = k[A] \), to derive suitable expressions for quantifying reaction rates and constants through proxy measures like pressure.
• Appreciating the complete reaction mechanism, informed by kinetic data, that predicts how products form from reactants.

Chemical kinetics informs us not only about the speed of reactions but also offers valuable insights into optimizing reaction conditions for desired outcomes.