Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

In a first-order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times 10^{4}\) sec. The rate constant of reaction in sec \({ }^{-1}\) is (a) \(2 \times 10^{4}\) (b) \(3.45 \times 10^{-5}\) (c) \(1.386 \times 10^{-4}\) (d) \(2 \times 10^{4}\)

Step-by-Step Solution

Answer

The rate constant is \(1.386 \times 10^{-4} \text{ sec}^{-1}\), which corresponds to option (c).

Step 1: Understand the Problem

We are dealing with a first-order reaction, which means the rate of reaction depends linearly on the concentration of the reactant. The initial concentration is 800 mol/dm³, and it decreases to 50 mol/dm³ over a time period of 2 × 10⁴ seconds. We need to determine the rate constant, k.

Step 2: Use the First-Order Reaction Formula

For a first-order reaction, the relationship between concentration and time is given by the formula: \[ ext{ln} rac{[A]_0}{[A]} = kt \]Where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time t, and k is the rate constant.

Step 3: Substitute Known Values

Substitute the given values into the formula: \([A]_0 = 800 \) mol/dm³, \([A] = 50 \) mol/dm³, and \(t = 2 \times 10^4 \) seconds.Thus, we have:\[ \text{ln} \frac{800}{50} = k \times 2 \times 10^4 \]

Step 4: Solve for the Natural Logarithm

Calculate the natural logarithm:\[ \text{ln} \frac{800}{50} = \text{ln} 16 \]Since \( \text{ln} 16 \approx 2.7726 \), we can substitute this value back into the equation.

Step 5: Calculate the Rate Constant k

Now substitute \( \text{ln} 16 \) into the equation and solve for k:\[ 2.7726 = k \times 2 \times 10^4 \]Thus, \[ k = \frac{2.7726}{2 \times 10^4} \approx 1.386 \times 10^{-4} \text{ sec}^{-1} \]

Step 6: Select the Correct Answer

The calculated rate constant matches option (c) \(1.386 \times 10^{-4} \text{ sec}^{-1}\).

Key Concepts

Rate Constant

In the context of chemical reactions, the rate constant is a crucial value that provides insight into the speed at which a reaction proceeds. For first-order reactions, the rate constant is denoted as \( k \) and has the units of time, often written as \( ext{sec}^{-1} \). This indicates how many reactions occur per second, contributing to how quickly the concentration of the reactants decreases.

In a first-order reaction, changes in concentration over time are mathematically modeled by the formula:

• \( ext{ln} \frac{[A]_0}{[A]} = kt \)

Here, \([A]_0\) represents the initial concentration, \([A]\) the concentration at a later time \( t \), and \( k \) the rate constant.

Given this formula, determining the rate constant involves calculating how fast the reactant concentration falls over time. For students, grasping this identification of speed through \( k \) helps in understanding reaction dynamics more vividly.

Natural Logarithm

The natural logarithm, denoted as \( \text{ln} \), is an essential mathematical function in determining the rate constant of first-order reactions. Unlike regular logarithms, natural logarithms provide a way to express processes that change continuously, which is typical in chemical reactions.

Natural logs are described as the logarithms to the base \( e \), where \( e \approx 2.718 \). In the context of first-order reactions, these logs are used to compare the initial concentration of a reactant to its concentration after some time has passed:

• \( ext{ln} \frac{[A]_0}{[A]} \)

This formula makes it convenient to link changes in concentration directly to time, helping to unravel the behavior of a reaction within a specific timeframe. By using natural logarithms, chemists can easily compute essential variables like the rate constant, crucial for predicting reaction kinetics.

Reaction Concentration

In a first-order reaction, reaction concentration reflects how the quantity of a reactant decreases over time. For first-order kinetics, this decrease is proportional to the remaining concentration, meaning the rate at which reactants disappear is directly linked to how much of the reactant is left.

The initial concentration \([A]_0\) is the amount of reactant present at the beginning of the observation. During the reaction, this concentration will drop until there is none remaining or until the reaction reaches completion. To determine the dynamics of this decrease, first-order reaction formulas use both the initial and post-time \( t \) concentrations:

• \([A]_0\)
• \([A]\)

Calculating concentrations over time via these indicators enables chemists to understand the exact amount reacted, which in turn allows them to compute the rate constant. This fundamental understanding is critical for effectively monitoring and optimizing reaction conditions.

Kinetics

Kinetics is the branch of chemistry focused on the rates of chemical reactions. It analyzes how different variables like concentration, temperature, and catalysts affect the speed at which reactions occur. For first-order reactions, models link time directly to concentration changes, providing a clear picture of how reactant levels decrease.

First-order kinetics is characterized by a constant half-life, the time required for half of the reactant concentration to reduce. The reaction rate, governed by the rate constant \( k \), depends directly on the concentration of one reactant, simplifying many complexities found in other reaction types.

• Illustrated by the equation: \( ext{ln} \frac{[A]_0}{[A]} = kt \)

Through this equation, understanding the rate of reactions becomes manageable, aiding chemists in predicting concentration changes over time. This predictability is essential for fields that require precise chemical control, such as pharmaceuticals and environmental sciences.