Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

A substance reacts according to first-order kinetics. The rate constant for the reaction is \(1 \times\) \(10^{-2} \mathrm{sec}^{1} .\) Its initial concentration is IM. Its initial rate is (a) \(2 \times 10^{2} \mathrm{Ms}^{-1}\) (b) \(1 \times 10^{2} \mathrm{Ms}^{-1}\) (c) \(1 \times 10^{-2} \mathrm{Ms}^{-1}\) (d) \(2 \times 10^{-2} \mathrm{Ms}^{-1}\)

Step-by-Step Solution

Answer

The initial rate is \( 1 \times 10^{-2} \, \mathrm{Ms}^{-1} \), so the answer is (c).

Step 1: Understand the Problem

We need to calculate the initial rate of a reaction that follows first-order kinetics. We have been given the rate constant and the initial concentration.

Step 2: Review First-Order Kinetics Formula

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

Step 3: Insert Given Values

We are given \( k = 1 \times 10^{-2} \mathrm{s}^{-1} \) and the initial concentration \( [A] = 1 \mathrm{M} \). Substitute these values into the first-order rate equation.

Step 4: Perform the Calculation

The rate is calculated as \( \text{Rate} = (1 \times 10^{-2} \mathrm{s}^{-1})(1 \mathrm{M}) = 1 \times 10^{-2} \, \mathrm{M} \cdot \mathrm{s}^{-1} \).

Step 5: Compare with Given Options

Compare the calculated initial rate \( 1 \times 10^{-2} \, \mathrm{M} \cdot \mathrm{s}^{-1} \) with the options provided: (a) \( 2 \times 10^{2} \, \mathrm{Ms}^{-1} \) (b) \( 1 \times 10^{2} \, \mathrm{Ms}^{-1} \) (c) \( 1 \times 10^{-2} \, \mathrm{Ms}^{-1} \) (d) \( 2 \times 10^{-2} \, \mathrm{Ms}^{-1} \) The correct option is (c) \( 1 \times 10^{-2} \, \mathrm{Ms}^{-1} \).

Key Concepts

Rate Constant

In the realm of chemical kinetics, the rate constant is a crucial component that helps us understand the speed at which a reaction occurs. It is symbolized by the letter "k" and is a proportionality factor that links the reaction rate to the concentrations of reactants. The rate constant is specific to each chemical reaction and remains constant as long as the temperature and pressure do not change.

In this exercise, the rate constant is given as \(1 \times 10^{-2} \mathrm{s}^{-1}\). This indicates that the reaction exhibits first-order kinetics, meaning the rate of reaction is directly proportional to the concentration of one reactant. Understanding the concept of the rate constant allows us to predict how fast a reaction will proceed under given conditions.

• Units: For first-order reactions, the units of the rate constant are \(\mathrm{s}^{-1}\)
• Temperature sensitivity: The value of "k" changes with temperature since it is dependent on the rate of collisions and the energy with which particles collide.

Rate Law

A rate law expresses the relationship between the concentration of reactants and the reaction rate. In simple terms, it gives us a formula to calculate how fast a reaction goes using the concentrations of the reactants involved. For a first-order reaction, the rate law can be represented as follows:

\[ \text{Rate} = k[A] \\]where "\([A]\)" is the concentration of the reactant, and "k" is the rate constant for the reaction implied.

• This linear relationship means if the concentration of \(A\) doubles, the rate will also double.
• In our case, since \(A\) is initially \(1 \mathrm{M}\), the rate can be calculated directly by substituting the values into the formula given.

Understanding the rate law allows you to determine the effect of varying reactant concentrations on the overall reaction rate, providing insights into the reaction mechanics.

Initial Concentration

The initial concentration refers to the amount of a reactant present at the start of the reaction. It plays a key role in kinetic calculations as it's an important determinant of the initial rate of the reaction. In this exercise, the initial concentration of the reactant is given as \(1 \mathrm{M}\) (molarity), meaning at the beginning of the reaction, every liter of solution contains one mole of the reactant.

Initial concentrations provide a baseline for assessing how much a reaction changes over time. In first-order kinetics, even if you have more of the reactant initially, given the linear nature of the rate law, it will simply adjust the initial reaction rate accordingly. Therefore, the initial concentration is directly incorporated into the rate law formula, impacting the computation of the rate.

Reaction Rate Calculation

Calculating the reaction rate involves using the rate law to find how quickly the concentration of a reactant changes over time. For first-order kinetics, the calculation is quite straightforward. Using the rate law formula \( \text{Rate} = k[A] \), you can insert the known values for the rate constant and the initial concentration to find the initial rate.

For the provided exercise:

• The rate constant \( k \) is \(1 \times 10^{-2} \mathrm{s}^{-1}\)
• The initial concentration \( [A] \) is \(1 \mathrm{M}\)
• Thus, the rate calculation becomes as straightforward as it gets: \( \text{Rate} = (1 \times 10^{-2} \mathrm{s}^{-1})\times (1 \mathrm{M}) = 1 \times 10^{-2} \, \mathrm{M} \cdot \mathrm{s}^{-1}\)

This calculation tells you how fast the reactant is being used up initially. Having performed the computation, we can confidently identify, from the options given, that the reaction's initial rate corresponds to \(1 \times 10^{-2} \, \mathrm{M} \cdot \mathrm{s}^{-1}\), matching answer (c) from the given choices.