Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following reaction, \(\mathrm{A} \longrightarrow\) products This reaction is completed in 100 minutes. The rate constant of this reaction at \(\mathrm{t}_{1}=10 \mathrm{~min}\) is \(10^{-2} \mathrm{~min}^{-1}\). What is the rate constant (in \(\min ^{-1}\) ) at \(\mathrm{t}_{2}=20 \mathrm{~min} ?\) (a) \(2 \times 10^{-2}\) (b) \(10^{-2}\) (c) \(5 \times 10^{-3}\) (d) \(0.1\)

Step-by-Step Solution

Answer

The rate constant at \( t_2 = 20 \text{ min} \) is \( 10^{-2} \text{ min}^{-1} \), option (b).

Step 1: Understand the Reaction Type

Since the rate constant for the reaction is given, assume the reaction follows first-order kinetics unless stated otherwise. In a first-order reaction, the rate constant remains constant over time.

Step 2: Analyze Rate Constant Behavior

In a first-order reaction, the rate constant (k) does not change over time as it is independent of the concentration of the reactants. This means that for any time period, the value of k is the same.

Step 3: Apply First-Order Kinetics

The rate constant at \( t_1 = 10 \text{ min} \) is given as \( 10^{-2} \text{ min}^{-1} \). For a first-order reaction, the rate constant at all other times, such as \( t_2 = 20 \text{ min} \), will be the same since it is independent of time.

Step 4: Conclude the Rate Constant at t2

Since the rate constant does not change in a first-order reaction, at \( t_2 = 20 \text{ min} \), the rate constant will also be \( 10^{-2} \text{ min}^{-1} \).

Key Concepts

Rate Constant

The rate constant is a crucial factor in understanding how quickly a chemical reaction proceeds. It is typically denoted by the symbol \( k \) and is intrinsic to the reaction type. The rate constant links the reaction rate to the concentrations of the reactants involved through the rate law equation.
For a first-order reaction, the rate is directly proportional to the concentration of one reactant. Hence, the rate law is expressed as:

• \( ext{Rate} = k [ ext{A}] \)

Here, \( [ ext{A}] \) represents the concentration of reactant A.
Understanding the rate constant is essential for predicting how fast reactions occur and for determining reaction mechanisms. In the context of first-order reactions, the rate constant remains consistent, indicating that no matter when you measure it, the rate constant is unchanging over time.

Reaction Kinetics

Reaction kinetics is a branch of chemistry that studies the speed or rate of a chemical reaction and the factors affecting it. It delves into understanding how reaction rate is influenced by various conditions like temperature, pressure, concentration, and the presence of catalysts.
The primary aim of studying reaction kinetics is to gain insights into the pathway from reactants to products. It considers how molecular interactions transform one or more substances into different entities.
In our scenario, where a first-order reaction is taking place, reaction kinetics emphasizes how just one reactant primarily governs the rate of reaction. Such insights can help chemists manipulate conditions to control the speed of reactions, whether they need to slow down a reaction, for safety reasons, or speed it up, to increase yield or efficiency.

First Order Kinetics

First-order kinetics describes a reaction where the rate is proportional to the concentration of a single reactant. These reactions are characterized by a linear relationship on a naturally plotted logarithmic graph between concentration and time.
The mathematical expression for first-order kinetics is given by:

• \( ext{ln}[ ext{A}]_t = -kt + ext{ln}[ ext{A}]_0 \)

Where \( [ ext{A}]_t \) is the concentration of A at time \( t \), \( k \) is the rate constant, and \( [ ext{A}]_0 \) is the initial concentration.
Notably, in first-order reactions, the rate constant \( k \) is constant and independent of the concentrations of reactants or the passage of time. This property is crucial because it simplifies calculations and allows chemists to predict the behavior of reactions across various time intervals. In the exercise provided, the rate at \( t_2 = 20 \) min remains \( 10^{-2} \text{ min}^{-1} \), just as it was at \( t_1 = 10 \) min.