Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)

Step-by-Step Solution

Answer

The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\).

Step 1: Identify the rate law

For a first-order reaction, the rate law is expressed as \( ext{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

Step 2: Substitute given values into rate law

We know the rate constant \( k = 3 \times 10^{-6} \text{ s}^{-1} \) and the initial concentration \( [A] = 0.10 \text{ M} \). Substitute these values into the rate law: \( ext{Rate} = (3 \times 10^{-6} \text{ s}^{-1})(0.10 \text{ M}) \).

Step 3: Calculate the initial rate

Multiply the rate constant by the initial concentration: \( ext{Rate} = 3 \times 10^{-7} \text{ M s}^{-1} \).

Step 4: Determine the correct answer

Compare the calculated initial rate with the provided options. The correct answer is (a) \( 3 \times 10^{-7} \text{ M s}^{-1} \).

Key Concepts

Rate Constant

In a first-order reaction, the concept of the rate constant, commonly denoted as \( k \), is pivotal as it helps in understanding how fast a reaction proceeds. The rate constant has unique units depending on the order of the reaction, and for a first-order reaction, it is simply measured in s\( ^{-1} \). When given a rate constant of \( 3 \times 10^{-6} \) per second, this means that the reaction’s progress is somewhat slow, due to the small value.

• Rate constant determines the speed of reaction.
• Specific to each reaction; can't compare between different reactions.

As a constant, it remains the same throughout the reaction, provided the temperature is constant. If you change the temperature, the rate constant will generally change, often increasing with higher temperatures.

Initial Concentration

The initial concentration, symbolized as \([A]_0\), represents the amount of reactant present at the very beginning of a reaction. In this scenario, it is \(0.10 \text{ M}\). Understanding this concentration is crucial because it directly impacts the reaction rate, especially in a first-order reaction where

• The reaction rate is directly proportional to \([A]_0\).
• An increase in the initial concentration results in an increased initial reaction rate.

It’s also important to note that initial concentration helps in monitoring how much of the reactant is consumed as the reaction progresses. For practical purposes, this allows chemists to predict how long the reaction may take or when a given amount of products will be formed.

Rate Law

The rate law of a chemical reaction is an equation that links the reaction rate with the concentrations of its reactants. For first-order reactions, the rate law is expressed as \( \text{Rate} = k[A] \). This simplicity means that the rate of the reaction is directly proportional to the concentration of one reactant.
A pivotal point is that the reaction rate depends linearly on the concentration of the reactant:

• Rate increases if concentration increases.
• Helps determine the reaction mechanism.

Thus, it assists in predicting how alterations in the reactant concentration can affect the speed of the reaction—an essential tool for chemists in the lab.

Reaction Rate Calculation

Calculating the reaction rate for a reaction is crucial for understanding how the reaction progresses over time. In the problem given, we calculate the initial rate by substituting available data into the rate law \( \text{Rate} = k[A] \). With the provided rate constant \( k = 3 \times 10^{-6} \text{ s}^{-1} \) and the initial concentration \( [A] = 0.10 \text{ M} \), the rate calculation becomes straightforward:

• Substitute the known values: \( \text{Rate} = (3 \times 10^{-6} \text{ s}^{-1})(0.10 \text{ M}) \)
• Perform the multiplication to find the rate: \( 3 \times 10^{-7} \text{ M s}^{-1} \)

This equation tells us that at the start of the reaction, the rate is \( 3 \times 10^{-7} \text{ M s}^{-1} \), providing insight into how the initial conditions dictate the rate at which products are formed. Understanding this concept helps in planning and controlling reaction conditions effectively.