Problem 87 - Chemistry: The Central Science

Question

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M} ?\)

Step-by-Step Solution

Answer

The reaction is second-order in \(\mathrm{NO}_{2}\), based on the given rate constant's units. To find the time it takes for the concentration to decrease from \(0.100 \mathrm{M}\) to \(0.025 \mathrm{M}\), use the integrated second-order rate law: \( \frac{1}{[\mathrm{A(t)}]} - \frac{1}{[\mathrm{A(0)}]} = kt\). Plug in the given concentrations and rate constant, then solve for \(t\), which is approximately \(95.24 \ \mathrm{s}\).

Step 1: Determine the reaction order from the rate constant's units

The units for the given rate constant \(k\) affects the reaction order. As the given units are \( \mathrm{M}^{-1} \mathrm{~s}^{-1} \), this indicates that the rate law, \( rate = k [\mathrm{NO}_{2}]^{n}\), has the same units as the rate law for a second-order reaction. Thus, the reaction is second-order in \(\mathrm{NO}_{2}\).

Step 2: Write the second-order integrated rate law

For a second-order reaction, we can write the integrated rate law as: \( \frac{1}{[\mathrm{A(t)}]} - \frac{1}{[\mathrm{A(0)}]} = kt\), where \(\mathrm{A(t)}\) is the concentration of \(\mathrm{NO}_{2}\) at time \(t\), \(\mathrm{A(0)}\) is the initial concentration of \(\mathrm{NO}_{2}\), and \(k\) is the rate constant.

Step 3: Use initial and final concentrations to solve for time

We are given the initial concentration of \(\mathrm{NO}_{2}\), \(\mathrm{A(0)} = 0.100 \mathrm{M}\), and we want to know how long it would take for the concentration to decrease to \(\mathrm{A(t)} = 0.025 \mathrm{M}\). Plugging these values into the integrated second-order rate law, we get: \( \frac{1}{0.025 \mathrm{M}} - \frac{1}{0.100 \mathrm{M}} = (0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1})t \) Now, solve for \(t\).

Step 4: Solve for time t

Rearrange the integrated rate law equation to isolate \(t\) : \( t = \frac{\frac{1}{0.025 \mathrm{M}} - \frac{1}{0.100 \mathrm{M}}}{0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}} \) Calculate \(t\): \( t \approx 95.24 \ \mathrm{s} \) It will take approximately \(95.24 \ \mathrm{s}\) for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.100 \mathrm{M}\) to \(0.025 \mathrm{M}\).

Key Concepts

Integrated Rate Law

The integrated rate law is an equation that links the concentration of reactants to time and allows us to calculate how concentrations change as a reaction proceeds. For second-order reactions, the integrated rate law is particularly important. It helps to track the changing concentration over time, providing insights into how long a reaction will take to reach a certain state.

For a second-order reaction, like the one with nitrogen dioxide ( ext{NO}_2), the integrated rate law is expressed as:

• \( \frac{1}{[\text{A(t)}]} - \frac{1}{[\text{A(0)}]} = kt \)

This formula is derived from the ordinary differential form of the rate law, and it allows you to find the concentration \([\text{A(t)}]\) at any given time \(t\) if you know the initial concentration \([\text{A(0)}]\) and the rate constant \(k\).

In simple terms, it tells us that for a second-order reaction, the change in the reciprocal of the concentration over time is proportional to the product of the rate constant and time. This is very useful when you need to calculate how long it will take for a reaction to proceed to a certain point.

Reaction Rate

The reaction rate of a chemical reaction indicates how fast reactants are converted into products. In essence, it measures the change in concentration of a reactant or product per unit time. This depends on factors like concentration, temperature, and the presence of a catalyst, but it's also directly related to the order of the reaction and the rate constant.

In our context of second-order reactions, the rate can be defined by the equation:

• \( \text{rate} = k [\text{NO}_2]^2 \)

where \(k\) is the rate constant and \([\text{NO}_2]\) is the concentration of nitrogen dioxide. As the concentration of \(\text{NO}_2\) affects the reaction rate, a higher concentration leads to an increase in the reaction rate. This is due to the fact that in a second-order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant.

Understanding the reaction rate is crucial for predicting how long a process will take, which is essential information for laboratory settings and industrial processes.

Reaction Order

Reaction order refers to the sum of the powers of the concentration terms in the rate law equation. It provides vital information about the relationship between the concentration of reactants and the rate of reaction. In the specific case of nitrogen dioxide ( ext{NO}_2), we have a second-order reaction.

Recognizing the reaction order is straightforward once the units of the rate constant are provided. For our reaction, the rate constant has units \(\text{M}^{-1} \text{s}^{-1}\), characteristic of second-order reactions. This means the rate of reaction is dependent on the square of \([\text{NO}_2]\), written as:

• \( \text{rate} = k [\text{NO}_2]^2 \)

Knowing the reaction order is essential not only for applying the correct integrated rate law but also for understanding the underlying reaction mechanism. It aids in predicting how a change in concentration will affect the rate, and thus the overall progress of the reaction.