Problem 169 - A Complete Resource Book in Chemistry for JEE Main

Question

If a is the initial concentration of reactant and \((a-x)\) is the remaining concentration after time "t' in a first order reaction of rate constant \(\mathrm{k}_{1}\), then which of the following relations is /are correct? (a) \(k_{1}=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)\) (b) \(x=a\left(1-c^{k_{1} t}\right)\) (c) \(t_{1 / 2}=\frac{1.414}{k_{1}}\) (d) \(t_{a v}=\frac{1}{k_{1}}\)

Step-by-Step Solution

Answer

Correct relations: (a) and (d).

Step 1: Understanding First Order Kinetics

In first order kinetics, the rate of reaction is proportional to the concentration of one reactant. The integrated rate law for a first order reaction is given as:\[ rac{d[A]}{dt} = -k_1[A] \]which can be rearranged and integrated to:\[ ext{ln}rac{[A]}{[A]_0} = -k_1 t \]where \([A]_0 = a\) is the initial concentration and \([A] = a-x\) is the concentration at time \(t\).

Step 2: Convert to Logarithmic Form

The integrated rate equation can also be expressed in terms of common logarithm, which is often more practical:\[ k_1 = \frac{2.303}{t}\log\left(\frac{a}{a-x}\right) \]This equation corresponds to option (a) and is correct for a first order reaction.

Step 3: Evaluating Option (b)

Option (b) suggests a different reaction form: \( x = a(1 - c^{k_1 t}) \). This format is not consistent with first order reaction kinetics, where the exponential decay is in the form \(e^{-k_1 t}\), not involving another base \(c\). Thus, option (b) is incorrect.

Step 4: Checking Half-life Equation

The half-life \(t_{1/2}\) for a first order reaction is given by:\[ t_{1/2} = \frac{0.693}{k_1} \]Option (c) states: \( t_{1/2} = \frac{1.414}{k_1} \),which is incorrect because \(1.414 eq 0.693\). Thus, option (c) is incorrect.

Step 5: Evaluating Option (d)

Option (d) gives the definition for average time : \( t_{av} = \frac{1}{k_1} \). In the context of first order reactions, this equation refers to the mean lifetime of a reactant molecule, which is indeed expressed as \( \frac{1}{k_1} \). Hence, option (d) is correct.

Key Concepts

Rate Constant

In chemical kinetics, the rate constant, denoted by \( k_1 \), is a crucial factor that gives us the relationship between the concentration of a reactant and the rate of a chemical reaction. For a first order reaction, the rate constant is essential in determining how fast or slow the reaction proceeds.
It essentially tells us the proportionality between the decrease in concentration of the reactant and time.
The formula to find the rate constant for a first order reaction can be expressed as:

• \( k_1 = \frac{2.303}{t} \log \left(\frac{a}{a-x}\right) \)

This formula allows you to calculate the rate constant if you know the initial concentration of the reactant \( a \), the concentration remaining after time \( t \) which is \( a-x \,\) and the time \( t \). It incorporates logarithms to transition from the natural base \( e \) to base 10, which is often more intuitive for calculations in many practical problems.

Integrated Rate Law

The integrated rate law is vital for understanding how the concentration of reactants in a reaction changes over time. For first order reactions, the general form can be written as follows:

• \( \ln \left( \frac{[A]}{[A]_0} \right) = -k_1 t \)

Here, \( [A]_0 \) is the initial concentration of the reactant, and \( [A] \) is the concentration at time \( t \. \) The equation is derived from integrating the first order rate law \( \frac{d[A]}{dt} = -k_1 [A] \).
The integrated rate law for a first order reaction is especially useful because it helps in calculating the concentration of reactant remaining at any point in time. By converting logarithms from the natural base to base 10, we have another expression:

• \( k_1 = \frac{2.303}{t} \log \left(\frac{a}{a-x}\right) \)

This yields a practical equation that aligns with empirical observations in chemical labs.

Half-life

The half-life of a reaction is the time required for the concentration of a reactant to reduce to half of its initial value. For first order reactions, calculating the half-life is simplified through a unique relationship with the rate constant, given as:

• \( t_{1/2} = \frac{0.693}{k_1} \)

This constant fraction, 0.693, arises because of the exponential nature of first order reactions. The half-life is independent of the initial concentration of the reactant, making it a particularly useful concept.
It's a critical measurement because it gives insight into how quickly a reaction progresses and helps in modeling the decay of substances in various fields, from pharmacology to nuclear physics.

Mean Lifetime

The mean lifetime, often symbolized as \( t_{av} \), refers to the average time a molecule of reactant exists before it reacts. For first order reactions, the mean lifetime is elegantly simple:

• \( t_{av} = \frac{1}{k_1} \)

This equation shows that the mean lifetime is inversely related to the rate constant \( k_1 \. \) A higher rate constant results in a shorter lifetime, revealing that the substance reacts more quickly.
The mean lifetime is important as it provides a statistical sense of the reaction duration, helping chemists and engineers design reactors and processes effectively. It translates complex reaction kinetics into manageable data when planning and conducting chemical experiments or industrial processes.