Question

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B},\) what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Short answer

(a) For a generic second-order reaction, plotting \(\frac{1}{[A]_t}\) against time (\(t\)) will yield a straight line. (b) The slope of the straight line is the rate constant, \(k\). (c) The half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

Step-by-step solution

Part (a): Determine the y-axis quantity for the graph

We start by writing the rate equation for a second-order reaction: \(rate = k \times [A]^2\) Where \(k\) is the rate constant, and \([A]\) represents the concentration of reactant A. We are asked to find a quantity that will yield a straight line when graphed against time. To do this, we can convert the rate equation into an integrated rate law. For a second-order reaction, the integrated rate law is: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) Where \([A]_t\) is the concentration of A at a specific time, \([A]_0\) is the initial concentration of A, and \(t\) is time. Here, if we plot \(\frac{1}{[A]_t}\) against time (\(t\)), we will get a straight line.

Part (b): Find the slope of the straight line

To find the slope of the straight line from part (a), we can rewrite the integrated rate law in the form of a linear equation: \(\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\) This equation is in the form of \(y = mx + c\), where \(y = \frac{1}{[A]_t}\), \(x = t\), \(m = k\), and \(c = \frac{1}{[A]_0}\). The slope of the straight line, or the coefficient of \(t\), is the rate constant, \(k\).

Part (c): Comparison of half-lives between first-order and second-order reactions

The half-life, \(t_{1/2}\), is the amount of time it takes for the concentration of the reactant to reduce to half its initial value. The half-life of a reaction depends on the order of the reaction. For a first-order reaction, the half-life is given by: \(t_{1/2}^{(1)} = \frac{0.693}{k}\) Here, we can see that the half-life is constant and does not depend on the initial concentration. For a second-order reaction, the half-life is given by: \(t_{1/2}^{(2)} = \frac{1}{k[A]_0}\) In this case, the half-life depends on the initial concentration of the reactant. As the initial concentration increases, the half-life becomes smaller. In summary, the half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

Key concepts

Understanding the Rate Equation in Second-Order Reactions

In chemical kinetics, the rate equation is a mathematical expression that describes the speed of a reaction in terms of the concentration of reactants. For a second-order reaction like \( \mathrm{A} \longrightarrow \mathrm{B} \), the rate equation is expressed as:\[ \text{rate} = k \times [A]^2 \]Here, \( [A] \) represents the concentration of reactant A, and \( k \) is the rate constant, a measure of the reaction's speed. The superscript '2' in \( [A]^2 \) indicates that the reaction rate is proportional to the square of the concentration of A. This means that even a small change in \([A]\) can noticeably affect the reaction rate.The rate equation serves as a foundational concept for understanding how the concentration of substances in a reaction influences the speed at which the reaction proceeds. This quadratic relationship is specific to second-order reactions, making them distinct from zero and first-order reactions.

Exploring the Integrated Rate Law for Second-Order Reactions

The integrated rate law provides insight into the concentration changes of reactants over time in a chemical reaction. For a second-order reaction, the integrated rate law is a pivotal tool and is given by:\[ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \]In this equation, \([A]_t\) is the concentration of reactant A at time \(t\), and \([A]_0\) is the initial concentration. Rearranging this equation, we can write it in a linear form:\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]This format is akin to the equation of a straight line \( y = mx + c \), where:

• \( y = \frac{1}{[A]_t} \)
• \( x = t \)
• \( m = k \) (the slope)
• \( c = \frac{1}{[A]_0} \) (the y-intercept)

By plotting \( \frac{1}{[A]_t} \) versus time, we obtain a straight line. The slope of this line (the rate constant \(k\)) reflects how quickly the reaction proceeds. This visualization helps us track how concentrations decrease over time, providing a clearer picture of reaction dynamics.

Comparing Half-life in First-Order and Second-Order Reactions

The half-life of a reaction is a crucial concept in kinetics. It represents the time it takes for the concentration of a reactant to drop to half its initial value. The behavior of half-life varies significantly between different reaction orders.For a first-order reaction, the half-life is constant and does not depend on the initial concentration of the reactant. It is calculated as:\[ t_{1/2}^{(1)} = \frac{0.693}{k} \]This constancy makes first-order reactions easy to predict, as the time required for half of the reactant to react remains the same throughout the process.In contrast, for a second-order reaction, the half-life is dependent on the initial concentration, represented by:\[ t_{1/2}^{(2)} = \frac{1}{k[A]_0} \]As the initial concentration \([A]_0\) increases, the half-life decreases. This dependency highlights a key distinction: second-order reactions have varying half-lives based on starting concentrations. Therefore, each successive half-cycle takes less time as the reactant concentration decreases. Understanding these distinctions helps in predicting the behavior and duration of chemical reactions more accurately.