Problem 49 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Which of the following represents the expression for \(3 / 4^{\text {th }}\) the life of a first-order reaction? (a) \(\frac{k}{2.303} \log \frac{4}{3}\) (b) \(\frac{2.303}{k} \log \frac{4}{3}\) (c) \(\frac{2.303}{k} \log 4\) (d) \(\frac{2.303}{k} \log 3\)

Step-by-Step Solution

Answer

The expression for \(3 / 4^{\text {th }}\) the life of a first-order reaction is \(\frac{2.303}{k} \log 4\), which corresponds to option (c).

Step 1: Understanding First-Order Reaction

The life of a first-order reaction is commonly described using its half-life formula, which is given by \( t_{1/2} = \frac{\ln(2)}{k} \) where \( k \) is the rate constant. However, to find the time for a different fraction of the reactant's life such as \( \frac{3}{4} \) life, we need to modify the equation.

Step 2: Deriving the Expression for \( \frac{3}{4} \) Life

For a first-order reaction, we use the integrated rate law \( ln \left( \frac{[A]_0}{[A]} \right) = kt \) where \( [A]_0 \) is the initial concentration and \( [A] \) is the concentration at time \( t \) . To find \( 3/4 \) life, set \( \frac{[A]}{[A]_0} = \frac{1}{4} \) (as 1/4 of the substance remains), which gives the equation \( ln(4) = kt \) or \( t = \frac{ln(4)}{k} \) .

Step 3: Converting to Common Logarithm Form

Since the options are provided in log base 10 and not ln (natural log), we'll need to convert \( ln \) to \( log \) using the conversion formula \( ln(x) = 2.303 \log(x) \) . This transforms the expression to \( t = \frac{2.303 \log(4)}{k} \) .

Step 4: Comparing with Given Options

Looking at the options, only option (c) \( \frac{2.303}{k} \log 4 \) matches the derived formula for \( \frac{3}{4} \) the life of a first-order reaction.

Key Concepts

Chemical Kinetics

Chemical kinetics is the branch of chemistry that deals with the rates of chemical reactions and the factors that affect these rates. It is a fundamental topic because it helps us understand the speed at which a reaction proceeds and what we can do to control it.

Just as a race car's speed can be influenced by its engine power or the type of tires it has, the rate of a chemical reaction can be affected by various factors such as temperature, concentration of reactants, presence of catalysts, and the nature of the reactants themselves. Kinetics plays a crucial role in many fields, from designing pharmaceuticals to understanding environmental processes which involve chemical reactions.

In the context of the exercise, the first-order reaction is one whose rate depends linearly on a single reactant's concentration. Kinetics provides us with tools such as the integrated rate laws to quantify these reaction rates over time. Through kinetics, students can predict the concentration of reactants at any point in the reaction, which has wide-ranging applications in both industry and research.

Rate Law

The rate law is a mathematical equation that relates the rate of a chemical reaction to the concentration of the reactants. It is a direct outcome of chemical kinetics study and can be written generically as: \[ Rate = k[A]^{m}[B]^{n} \]
Here, \( k \) is the rate constant, which is unique to each reaction at a given temperature, and \( m \) and \( n \) are the reaction orders with respect to reactants \( A \) and \( B \), respectively. These orders are determined experimentally and do not necessarily correspond to the stoichiometric coefficients in the balanced chemical equation.

First-order reactions are a key focus in kinetics because they have a constant half-life irrespective of initial concentration, which makes their study and mathematical description much simpler. For first-order reactions, the rate law simplifies to: \[ Rate = k[A] \] where the reaction rate is directly proportional to the concentration of \( A \). Understanding how to use and manipulate rate laws is crucial for students to predict how changes in conditions affect the reaction speed.

Half-Life of Reaction

Half-life, often represented as \( t_{1/2} \), is a concept in chemistry that refers to the time required for exactly one-half of the reactant to be consumed in a chemical reaction. For first-order reactions, the half-life is given by the equation: \[ t_{1/2} = \frac{\ln(2)}{k} \]

This relation indicates that the half-life of a first-order reaction is independent of the initial concentration of the reactant, a characteristic feature of first-order kinetics. It is an extremely useful concept because it gives us a measure of how quickly a reactant is depleted, allowing chemists to design and control reactions appropriately for their desired outcomes.

In the context of our textbook exercise, the concept of half-life is extended to calculate the time needed for the concentration of the reactant to diminish to a fraction other than one-half, in this case, representing \( \frac{3}{4} \) the life of the reaction. By grasping the basic principles of half-life, students reinforce their understanding of reaction kinetics and become adept at working with exponentially decaying processes.