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) \(\frac{1}{t} \ln \frac{2 P_{3}}{P_{3}-P_{2}}\) (c) \(\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{3}}{\mathrm{P}_{3}-\mathrm{P}_{2}}\) (d) \(\frac{1}{t} \ln \frac{P_{3}}{2 P_{3}-P_{2}}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Reaction

This reaction is a first-order reaction, where substance \( A \) decomposes to form \( B \) and \( C \). Using the expression for a first-order reaction, we will calculate the reaction constant \( k \).

Step 2: Identify Initial Variables

- At time \( t = 0 \), let the initial pressure of \( A \) be \( P_0 \).- As time progresses, \( A \) decomposes, and the pressure will change accordingly.

Step 3: Apply Pressure Changes at Time t

At time \( t' \), the total pressure is \( P_2 \). Since one molecule of \( A \) forms one molecule each of \( B \) and \( C \), assume \( x \) as the pressure of \( A \) that has reacted. Hence, the total pressure at \( t' \) is \( P_0 + 2x = P_2 \).

Step 4: Consider Pressure at Completion

At completion (\( t = \infty \)), all \( A \) has decomposed, so the total pressure is \( P_3 \). Thus, \( P_0 + x = P_3 \), because all of \( A \) is converted to \( B \) and \( C \).

Step 5: Derive x in Terms of Pressures

From Step 3, we have \( x = P_2 - P_3 \).

Step 6: Expression for k in First Order Kinetics

The reaction constant \( k \) for first-order kinetics is \( k = \frac{1}{t} \, \ln{\left( \frac{P_0}{P_0 - x} \right)} \).

Step 7: Substitute Values Into k

Using \( P_0 = P_3 - x \) and the expression from Step 5, substitute these into the equation for \( k \):\[ k = \frac{1}{t} \, \ln{\left( \frac{P_3}{P_3 - (P_2 - P_3)} \right)} \]This simplifies to:\[ k = \frac{1}{t} \, \ln{\left( \frac{P_3}{P_3 - P_2} \right)} \].

Step 8: Comparing Options

The correct option corresponds to:\[ \text{Option (c) } \frac{1}{t} \ln \frac{P_{3}}{P_{3} - P_{2}} \]

Key Concepts

Reaction Kinetics

In chemistry, reaction kinetics deals with understanding the rates of chemical reactions. It provides insights into how quickly reactants convert into products. When examining reaction kinetics, it's vital to comprehend the concept of reaction order.

A first-order reaction signifies that the rate of the reaction is directly proportional to the concentration of one reactant. In simpler terms, if you double the amount of the reactant, the rate of reaction also doubles. Such reactions usually follow an exponential decay pattern over time.

• First-order reactions have a specific rate constant, denoted as \(k\).
• The rate of reaction can be represented by the equation: \(\text{Rate} = k[A]\), where \[A\] is the concentration of the reactant.

Understanding these concepts is crucial because they form the basis for deriving equations to calculate the rate constants, making it easier to predict the behavior of chemical processes.

Pressure Change in Reactions

Chemical reactions involving gases often result in changes in pressure. This change occurs because gases occupy volume and exert pressure on their container's walls. As reactions progress, converting reactants into products, the number of gas molecules changes, impacting the pressure.

For the decomposition reaction \(\text{A(g)} \rightarrow \text{B(g) + C(g)}\), the pressure at any given moment depends on the mole ratio of reactants to products. A helpful concept in these reactions is using the relationship between initial and final pressures:

• The pressure of individual components contributes to total pressure.
• Pressure changes can directly relate to the stoichiometry of the balanced equation of the reaction.

At the start of the reaction, the initial pressure, denoted as \[P_0\], decreases as \[A\] decomposes. By the end, the total pressure, \[P_3\], includes contributions from both \[B\] and \[C\]. Analyzing these pressure changes helps us determine how far a reaction has proceeded.

Decomposition Reactions

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. In gaseous reactions like \(\text{A(g)} \rightarrow \text{B(g) + C(g)}\), the decomposition typically involves pressure changes as the number of gas molecules changes.

Such reactions are essential in reaction kinetics, as they often proceed through defined pathways and can be catalyzed or influenced by external factors such as temperature or pressure. Key characteristics include:

• The reaction initiates with a single substance, often resulting in a product gas mixture.
• Decomposition is energetically significant, often requiring specific conditions to commence.

Understanding the nature of decomposition reactions allows chemists to manipulate reaction conditions effectively, aiming to optimize product yields or influence reaction rates.

Rate Constant Calculation

The rate constant \(k\) for a reaction is a crucial parameter in kinetics and indicates how fast a reaction will proceed. For first-order reactions, \(k\) can be calculated using the initial and total pressures of reactants and products.

This parameter is vital for designing reaction processes and predicting how long it will take to achieve a particular level of decomposition. To calculate \(k\) for a given scenario, one uses the equation derived from first-order kinetics:

• The formula \[ k = \frac{1}{t} \ln{\left( \frac{P_3}{P_3 - P_2} \right)} \] incorporates time \[t\] and pressures \[P_2\] and \[P_3\]. These represent intermediate and total pressures.
• The natural logarithm part of the equation accounts for the exponential nature of first-order reactions.

Accurately calculating \(k\) helps chemists ensure optimal conditions are met to accomplish efficient and cost-effective chemical reactions.