Question

For a first order reaction a. Plot between 't' and \(\log _{10}(a-X)\) will be a parabola. b. \(d x / d t=k(a-x)\) c. \(\mathrm{K}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{a}}{\mathrm{a}-\mathrm{X}}\). d. \(\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\mathrm{a}-\mathrm{X}_{1}}{\mathrm{a}-\mathrm{X}_{2}}\)

Short answer

Options b, c, and d are correct; option a is incorrect for a first order reaction.

Step-by-step solution

Analyze Option a

For a first order reaction, the plot of time \( t \) against \( \log_{10}(a-X) \) is linear, not parabolic. The relationship is described by the equation \( \log_{10}(a-X) = \log_{10} a - \frac{k}{2.303}t \) where \( k \) is the rate constant. Thus, this statement is incorrect.

Analyze Option b

The given differential equation \( \frac{dx}{dt} = k(a-x) \) correctly represents a first order reaction, where \( x \) is the concentration of the product at time \( t \), \( a \) is the initial concentration of the reactant, and \( k \) is the rate constant. This differential equation is correct for a first order reaction.

Analyze Option c

The given expression for the rate constant \( K \) of a first-order reaction is correct: \( K = \frac{2.303}{t} \log_{10}\frac{a}{a-X} \). This is derived from integrating the first order rate law \( \frac{dx}{dt} = k(a-x) \) and expresses the relationship between the concentration of reactant, time, and rate constant.

Analyze Option d

The time difference \( t_2 - t_1 \) for the change in concentration from \( a-X_1 \) to \( a-X_2 \) for a first-order reaction can be described by the equation \( t_2 - t_1 = \frac{2.303}{k} \log_{10}\frac{a-X_1}{a-X_2} \). This is correct and is derived by rearranging the integrated rate law for a first-order reaction.

Key concepts

Understanding Differential Equations in First Order Reactions

In the world of chemistry, understanding the speed of a reaction is crucial. A key tool in this area is the differential equation. For first order reactions, a common representation is \( \frac{dx}{dt} = k(a-x) \). This equation tells us the rate at which the concentration of a reactant \( a \) changes over time to produce a product \( x \).

• \( dx \) is a tiny change in concentration.
• \( dt \) is a small change in time.
• \( k \) is the rate constant, which we'll explore further.
• \( a \) is the initial concentration of the reactant.

This equation highlights the direct relationship between the rate of reaction and the concentration of the reactant that is yet to react. The term \((a - x)\) signals that as the reactant concentration decreases, the reaction speed also changes.

The Rate Constant \( k \) and Its Importance

The rate constant, denoted as \( k \), is a pivotal component in understanding first order reactions. It is a measure of how quickly a reaction proceeds.

• The larger the value of \( k \), the faster the reaction.
• For first order reactions, \( k \) is consistent and doesn’t depend on the concentration of reactants.

An equation that helps us calculate \( k \) for first order reactions is \( K = \frac{2.303}{t} \log_{10}\frac{a}{a-X} \). Here, \( t \) is the time, \( a \) is the initial concentration, and \( X \) is the concentration reacted. This formula stems from rearranging the integrated rate law and helps scientists and students understand the speed of these reactions.

Exploring the Integrated Rate Law

Integrated rate laws take the differential equation of a reaction and simplify it to a more usable form. For first order reactions, the differential equation \( \frac{dx}{dt} = k(a-x) \) transforms into the integrated form \( \log_{10}(a-X) = \log_{10} a - \frac{k}{2.303}t \). This expression allows us to describe how the concentration of reactants changes over time using logarithms.
To visualize this, we can plot the logarithm of the concentration against time.

• The plot yields a straight line for first order reactions, indicating a constant rate of reaction over time.
• This linear relationship is essential for predicting reaction behavior and calculating \( k \).

Understanding the Concentration-Time Relationship

In chemical kinetics, predicting how the concentration of reactants varies over time is crucial for first order reactions. The relationship can be captured by the integrated rate law.

• The equation \( \log_{10}(a-X) = \log_{10} a - \frac{k}{2.303}t \) describes this relationship.
• The changing concentration directly influences the slope of the line when plotted over time, demonstrating how the reaction progresses.
• Given two different times \( t_1 \) and \( t_2 \), the change in concentration can be expressed using \( t_2 - t_1 = \frac{2.303}{k} \log_{10}\frac{a-X_1}{a-X_2} \).

These concepts help chemists to model and predict how quickly a reaction will complete or reach a desired state, which is invaluable for laboratory applications and industrial processes.