Question

The decomposition of dinitrogen pentaoxide $$ \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) It has been found experimentally that the decomposition is \(20.5 \%\) complete in 13.0 hours at 298 K. Calculate the rate constant and the half-life at 298 K.

Short answer

Rate constant: \( k = 0.016 h^{-1} \), half-life: \( t_{1/2} = 43.3 \) hours.

Step-by-step solution

Understand the Problem

We have a first-order decomposition reaction and need to find the rate constant \( k \) and the half-life for the reaction. It is given that the reaction is 20.5% complete in 13.0 hours at 298 K.

Use the First-Order Rate Law

For a first-order reaction, we use the rate law: \[ k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]}\right) \] where \( t \) is time, \( [A]_0 \) is the initial concentration, and \( [A] \) is the concentration at time \( t \). Since 20.5% of the reactant has decomposed, \( [A] = 0.795[A]_0 \).

Key concepts

First-Order Reaction

In chemical kinetics, a first-order reaction is characterized by its rate being directly proportional to the concentration of a single reactant. This means that as the concentration of the reactant decreases, the rate of reaction decreases at a similar rate. The mathematical representation of a first-order reaction is given by the rate law equation: \[\text{Rate} = k[A] \]Where:

• \( k \) is the rate constant, a unique value characteristic of the reaction under specific conditions.
• \([A]\) is the concentration of the reactant.

This representation indicates that if you double the concentration of the reactant, the rate also doubles. Such reactions follow an exponential decay pattern, which is reflected in their calculations.

Rate Constant Calculation

The rate constant \( k \) is a crucial value in the study of reaction kinetics because it provides insight into the speed of the reaction. For first-order reactions, it can be calculated using the integrated rate law:\[ k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]}\right) \]Where:

• \( t \) is the given time period, here 13.0 hours.
• \([A]_0\) is the initial concentration of the reactant.
• \([A]\) is the concentration of the reactant at time \( t \).

For the exercise, \( [A] = 0.795[A]_0 \) because 20.5% of the reactant is used up. Substitute these values into the rate law to find \( k \). Understanding the behavior of \( k \) helps predict how a reaction progresses under different conditions.

Half-Life Determination

In first-order kinetics, the half-life \( t_{1/2} \) is the time required for the concentration of the reactant to decrease to half of its initial value. One of the interesting properties of a first-order reaction is that its half-life is constant and does not depend on the initial concentration. The equation relating the half-life to the rate constant is:\[t_{1/2} = \frac{0.693}{k}\]This relationship makes calculating half-life straightforward once \( k \) is determined. Since the half-life is constant, it is especially useful in predicting how long a reaction will take to proceed through various stages, regardless of how much of the reactant is present at the start.