Problem 40 - Chemistry Structure and Properties

Question

This reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}:\) $$\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)$$ The rate constant for the reaction at a certain temperature is 0.053\(/ \mathrm{s}.\) \begin{equation} \begin{array}{l}{\text { a. Calculate the rate of the reaction when }\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.055 \mathrm{M} \text { . }} \\\ {\text { b. What is the rate of the reaction at the concentration indicated in part }} \\ {\text { a if the reaction is second order? Zero order? (Assume the same } n u-} \\ {\text { merical value for the rate constant with the appropriate units.) }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. The rate for a first-order reaction is 0.002915 \( M/s \). b. For a second-order reaction, the rate is 0.000159625 \( M^{-1}s^{-1} \); for a zero-order reaction, the rate is 0.053 \( M/s \).

Step 1: Write down the rate law for a first-order reaction

For a first-order reaction, the rate law is given by the formula: \( Rate = k \left[ \mathrm{N}_2\mathrm{O}_5 \right] \), where \( k \) is the rate constant and \( \left[ \mathrm{N}_2\mathrm{O}_5 \right] \) is the concentration of \( \mathrm{N}_2\mathrm{O}_5 \).

Step 2: Calculate the rate for a first-order reaction

Substitute the given values into the rate law. \( k = 0.053 \, /s \) and \( \left[ \mathrm{N}_2\mathrm{O}_5 \right] = 0.055 \, M \). So the rate is \( 0.053 \times 0.055 \).

Step 3: Write down the rate law for a second-order reaction

For a second-order reaction, the rate law is given by the formula: \( Rate = k \left[ \mathrm{N}_2\mathrm{O}_5 \right]^2 \).

Step 4: Calculate the rate for a second-order reaction

Using the same rate constant value, we assume \( k = 0.053 \, M^{-1}s^{-1} \). The rate is \( 0.053 \times (0.055)^2 \).

Step 5: Write down the rate law for a zero-order reaction

For a zero-order reaction, the rate law is given by the formula: \( Rate = k \), indicating that the rate does not depend on the concentration of reactants.

Step 6: Calculate the rate for a zero-order reaction

The rate is simply the value of the rate constant, which is \( 0.053 \, M/s \), since we assume the same numerical value with appropriate units.

Key Concepts

First-Order Reaction

Understanding first-order reactions is essential for students tackling chemical kinetics. In a first-order reaction, the rate at which the reaction occurs is directly proportional to the concentration of a single reactant. This type of reaction follows the rate law:
\[ \text{Rate} = k[\text{A}] \]
where \( k \) is the rate constant and \( [\text{A}] \) is the concentration of the reactant. An intuitive way to grasp this is by considering that if you double the concentration of \( \text{A} \), the reaction rate will also double. This is a fundamental concept in chemistry; as reactants are consumed over time, the rate of the reaction decreases proportionally.
For the exercise given, with the reaction:
\[ \text{N}_2\text{O}_5(g) \rightarrow \text{NO}_3(g) + \text{NO}_2(g) \]
and a rate constant \( k \) of 0.053 \( \text{/s} \), calculating the rate when \( [\text{N}_2\text{O}_5] = 0.055 \text{ M} \) involves simply plugging into our rate law:
\[ \text{Rate} = (0.053 \text{/s}) \times (0.055 \text{ M}) = 0.002915 \text{ M/s} \]
This step-by-step solution aids in visualizing how the rate changes with different concentrations, reinforcing the proportional relationship of a first-order reaction.

Second-Order Reaction

In contrast to first-order reactions, second-order reactions are influenced by the concentration of either two reactants or the square of a single reactant's concentration. The rate law for a second-order reaction is expressed as:
\[ \text{Rate} = k[\text{A}]^2 \]
or when involving two reactants, \( \text{A} \) and \( \text{B} \):
\[ \text{Rate} = k[\text{A}][\text{B}] \]
In this scenario, if the concentration of \( \text{A} \) is doubled, the rate would increase by four times, showcasing the quadratic relationship. This highlights the sensitivity of reaction rates to concentration changes in second-order kinetics.
Applying this to the same exercise, the question also considers the rate if the reaction of \( \text{N}_2\text{O}_5 \) were second-order. We then use the given rate constant with adjusted unit \( k = 0.053 \text{ M}^{-1}\text{s}^{-1} \), and calculate:
\[ \text{Rate} = 0.053 \times (0.055)^2 = 0.000159025 \text{ M/s} \]
Students often find second-order reactions trickier to understand due to the squared concentration term. By providing examples where they can directly calculate and compare rates, educators can greatly aid students in comprehending these concepts.

Rate Constant

The rate constant, denoted as \( k \), is an imperative factor in the kinetics of chemical reactions. Depending on the order of the reaction, its units will vary. For first-order reactions, the unit is inverse seconds (\( \text{s}^{-1} \)), while for second-order reactions, it is mole inverse liters inverse seconds (\( \text{M}^{-1}\text{s}^{-1} \)). The value of the rate constant is indicative of the speed of the reaction; a larger \( k \) value corresponds to a faster reaction under the same conditions.

It is also temperature-dependent and can be calculated using the Arrhenius equation. In educational settings, it is crucial to emphasize that the rate constant is not changed by varying reactant concentrations; it is intrinsic to the reaction at a given temperature.
As seen in the exercise, the rate constant was utilized in different scenarios to calculate reaction rates for first-order, second-order, and even zero-order behaviors, underlining its versatility in kinetics. The exercise also provides an opportunity to reinforce the concept that assuming the same numerical value for \( k \) across different reaction orders is hypothetical and typically used for comparison purposes only in educational exercises. In reality, the rate constant would need to be experimentally determined for each specific reaction order.