Problem 75 - A Complete Resource Book in Chemistry for JEE Main

Question

The reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) follows first order kinetics. The time taken for \(0.8\) mole of \(\mathrm{A}\) to produce \(0.6\) mole of is 1 hour. What is the time taken for conversion of \(0.9\) mole of \(\mathrm{A}\) to produce \(0.675 \mathrm{~mole}\) of B? (a) 2 hour (b) 1 hour (c) \(0.5\) hour (d) \(0.25\) hour

Step-by-Step Solution

Answer

1 hour. Option (b) is the answer.

Step 1: Understand First Order Kinetics

In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The rate equation can be given as \( \ln(\frac{[A_0]}{[A]}) = kt \) where \( k \) is the rate constant, \([A_0]\) is the initial concentration, and \([A]\) is the final concentration at time \( t \).

Step 2: Write the Rate Equation for Given Data

Given: Initial moles of \( \mathrm{A} \) is 0.8, final moles is 0.2 (since 0.8 - 0.6 = 0.2) after 1 hour.Using the first-order kinetics formula: \[ \ln(\frac{0.8}{0.2}) = k \cdot 1 \]Calculate the value of \( \ln(4) \).

Step 3: Calculate the Rate Constant \( k \)

\( \ln(4) \approx 1.386 \) Thus, \( k = 1.386 \) hr\(^{-1}\)

Step 4: Apply Rate Equation to New Initial Conditions

We need to find the time taken for 0.9 moles of \( \mathrm{A} \) to convert to 0.675 moles of \( \mathrm{B} \).This means \( [A]_t = 0.9 - 0.675 = 0.225 \ ext{ moles after t hours.} \)Get time \( t \) using \[ \ln(\frac{0.9}{0.225}) = 1.386t \]

Step 5: Solve for Time \( t \)

Calculate \( \ln(\frac{0.9}{0.225}) \): \[ \ln(4) \approx 1.386 \]Now substitute back, \[ 1.386 = 1.386t \].This simplifies to \( t = 1 \) hour.

Key Concepts

Rate Constant

A crucial part of understanding first-order kinetics is the concept of the rate constant, often denoted by the symbol \( k \). The rate constant is a unique value for each reaction and signifies the speed at which the reaction occurs. It is essential in calculating the time needed for a reaction to reach a certain point.
In first-order reactions, the rate constant has units of inverse time, such as hour\(^{-1}\).
This means that knowledge of the rate constant enables you to predict how both the concentration of reactants and products will evolve over time.

• The larger the rate constant, the faster the reaction progresses.
• The rate constant is independent of the concentration of reactants and products.

One can determine the rate constant using experimental data, and it becomes a pillar for predicting future outcomes of similar reactions.

Reaction Rate Equation

The reaction rate equation for first-order reactions is expressed as \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]where \([A_0]\) is the initial concentration, and \([A]\) is the concentration remaining after time \( t \).
The logarithmic form of the equation helps in simplifying calculations because it involves dividing the initial concentration by the final concentration and taking its natural logarithm.
This is useful in chemistry to understand how much time a reaction might take to achieve a certain level of completion, without having to measure time intervals repeatedly. The rate equation provides a predictive tool that simplifies dealing with complex reaction processes.

• This formula is pivotal in determining unknown variables like time \( t \) or the rate constant \( k \).
• A logarithmic form accommodates the natural decreasing nature of reactants over time.

Concentration of Reactants

The term "concentration of reactants" refers to the amount of substance in a given volume at any point in time.
Tracking how it changes is essential to understanding the progress of a reaction.
In first-order reactions, the rate of reaction depends on the initial and final concentrations of the reactants.
Concentration can be expressed in moles per liter (molarity) or simply in moles if dealing with specific volumes.

• Start by knowing both the initial and final amounts, as these are pivotal to the calculations.
• Changes in concentration provide insight into how far a reaction has progressed.

In the provided exercise, knowing the initial and remaining concentrations of reactant \( \text{A} \) enabled calculation of how long it would take to convert certain amounts into product \( \text{B} \). This showcases how vital concentration data is in applying the rate equation correctly.