Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following (a) \(\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}\) (p) Zero order (b) \(\mathrm{t}_{1 / 2}=\frac{\mathrm{a}}{2 \mathrm{k}}\) (q) First order (c) \(\mathrm{t}=\frac{1}{\mathrm{k}}\) (r) Average life (d) \(\mathrm{t}_{\frac{1}{2}}=\frac{1}{\mathrm{ak}}\) (s) Second order (t) Nuclear disintegration

Step-by-Step Solution

Answer

(a) matches (q), (b) matches (p), (c) matches (r), (d) matches (s).

Step 1: Understand the Terms

Each equation presented relates to a characteristic of a particular kinetic order of a reaction or process. Here, \( t_{1/2} \) typically refers to the half-life, \( t \) refers to the average life, \( k \) is the rate constant, and \( a \) is the initial concentration of the reactant.

Step 2: Match Equation (a)

Equation (a) \( t_{1/2} = \frac{0.693}{k} \) corresponds to first-order kinetics. This half-life formula is derived from the logarithm of 2 (0.693) for first-order reactions. Therefore, (a) matches with (q) First order.

Step 3: Match Equation (b)

Equation (b) \( t_{1/2} = \frac{a}{2k} \) is for zero-order reactions. In zero-order kinetics, the half-life is directly proportional to the initial concentration \( a \), suggesting that (b) matches with (p) Zero order.

Step 4: Match Equation (c)

Equation (c) \( t = \frac{1}{k} \) represents the formula for average life, not specific to any order but commonly encountered in contexts like radioactive decay. This indicates that (c) matches with (r) Average life.

Step 5: Match Equation (d)

Equation (d) \( t_{1/2} = \frac{1}{ak} \) is associated with second-order reactions, where the half-life is inversely proportional to the initial concentration and the rate constant. Thus, (d) matches with (s) Second order.

Key Concepts

Half-Life Equations

The concept of half-life is central to understanding reaction kinetics. It describes the time taken for half of a sample to react or decay. Various reaction orders have unique half-life equations, which reflect the underlying kinetics of the reaction.

In a first-order reaction, the half-life (\( t_{1/2} \)) is independent of the initial concentration of the reactants. This results in a consistent half-life, regardless of how much reactant is present. The formula is given by:\[ t_{1/2} = \frac{0.693}{k} \]Where \( k \) is the rate constant.

• First-order reactions: Half-life does not change with concentration.

For zero-order reactions, the half-life depends linearly on the initial concentration \( a \):\[ t_{1/2} = \frac{a}{2k} \]This means the more concentration at the start, the longer it takes for half the substance to react. Finally, second-order reactions have a half-life that inversely depends on the initial concentration and the rate constant:\[ t_{1/2} = \frac{1}{ak} \]This reveals that a higher initial concentration results in a shorter half-life. Understanding these differences is key to grasping how substances transform or decay over time.

First-Order Reactions

First-order reactions are characterized by a reaction rate that depends linearly on the concentration of a single reactant. It means that the speed of the reaction is directly proportional to how much reactant is available. These reactions have some special features.

The rate law for a first-order reaction is:\[ \text{Rate} = k[A] \]Here, \( [A] \) represents the concentration of the reactant A, and \( k \) is the rate constant. Manipulating this rate law gives the integrated form, which is a logarithmic equation that links reactant concentration over time:\[ \ln[A] = \ln[A]_0 - kt \]This demonstrates how the natural logarithm of the concentration decreases linearly as time goes on.

• Linear decrease: A straight line is seen when plotting \( \ln[A] \) against time.
• Constant half-life: Unique as the half-life stays the same no matter how much reactant you started with.

First-order kinetics are common in radioactive decay and certain chemical reactions where only one reactant is involved.

Zero-Order Reactions

In zero-order reactions, the reaction rate is constant, meaning it doesn't depend on the concentration of the reactant. This type of kinetics is quite different from first- and second-order reactions, leading to unique behavior.

The rate law for zero-order reactions is:\[ \text{Rate} = k \]Since the rate is independent of concentration, it suggests that the reaction progresses at a steady rate. Integrating the rate law gives the concentration of reactants over time:\[ [A] = [A]_0 - kt \]This indicates a linear decrease of \([A] \) as time progresses. Here are some key characteristics:

• Linear relationship: When you plot concentration versus time, youll get a straight line.
• Dependent half-life: Half-life increases with an increase in initial concentration \( a \).

Zero-order reactions can occur when a catalyst is saturated or in processes like enzyme-catalyzed reactions where substrate is abundant.

Second-Order Reactions

Second-order reactions have a reaction rate that depends on either two reactant concentrations or one concentration squared. They portray a non-linear relationship between reactant concentration and time.

The rate law for a simple second-order reaction takes the form:\[ \text{Rate} = k[A]^2 \]If the reaction is of the type \( 2A \rightarrow \) Products or \( A + B \rightarrow \) Products for different reactants contributing equally.The integrated rate law for second-order reactions can be expressed as:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]This suggests that a plot of inverse concentration vs. time will yield a straight line, with the slope equal to the rate constant \( k \).

• Non-linear kinetics: Unlike zero- and first-order, the relationship here is quadratic.
• Inverse half-life: Half-life tends to decrease with higher initial concentratons.

Second-order reactions are common in many fields, including biochemistry, and require careful analysis of concentration versus time data to understand the underlying kinetics.