Question

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{-1}\). If the rate is \(2.40 \times 10^{-5}\) mol litre \(^{-1}\) \(\mathrm{s}^{-1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in mol litre \(^{-1}\) ) is a. \(1.4\) b. \(1.2\) c. \(0.04\) d. \(0.8\)

Short answer

The concentration of \( \text{N}_2\text{O}_5 \) is \( 0.8 \) mol \(\text{L}^{-1}\). (Option d)

Step-by-step solution

Understand the Rate Law

The given reaction is a first-order reaction involving \( \text{N}_2\text{O}_5 \). For first-order reactions, the rate \( r \) is given by: \[ r = k [\text{N}_2\text{O}_5] \] where \( r \) is the rate of the reaction, \( k \) is the rate constant, and \([\text{N}_2\text{O}_5]\) is the concentration of \( \text{N}_2\text{O}_5 \).

Substitute Given Values

Substitute the given values into the rate law equation. We know that \( r = 2.40 \times 10^{-5} \) mol \(\text{L}^{-1}\) \(\text{s}^{-1}\) and \( k = 3.0 \times 10^{-5} \) \(\text{s}^{-1}\). The equation becomes: \[ 2.40 \times 10^{-5} = 3.0 \times 10^{-5} [\text{N}_2\text{O}_5] \]

Solve for Concentration

To find \([\text{N}_2\text{O}_5]\), rearrange the equation and divide both sides by \( 3.0 \times 10^{-5} \): \[ [\text{N}_2\text{O}_5] = \frac{2.40 \times 10^{-5}}{3.0 \times 10^{-5}} \] Simplifying, we find: \[ [\text{N}_2\text{O}_5] = 0.8 \] mol \(\text{L}^{-1}\).

Match the Result with Options

The calculated concentration of \( \text{N}_2\text{O}_5 \) is \( 0.8 \) mol \(\text{L}^{-1}\). This matches option (d) in the list of possible answers.

Key concepts

First-order reaction

First-order reactions are a fundamental concept in chemical kinetics. They are characterized by having a reaction rate that depends linearly on the concentration of one reactant. This means that if you double the concentration of the reactant, the rate of the reaction also doubles. In mathematical terms, this is expressed with the rate law:

• rate = k[A]

Here, "k" is the rate constant, and "[A]" is the concentration of the reactant. For our specific reaction involving \(\mathrm{N}_2\mathrm{O}_5\), the rate will directly depend on the concentration of \(\mathrm{N}_2\mathrm{O}_5\). The simplicity of first-order reactions makes them a great starting point for learning about reaction kinetics.

Rate constant

The rate constant, often denoted as "k," is a crucial parameter in the rate law equation of a reaction. It directly impacts how fast a reaction proceeds. In the context of a first-order reaction, the rate constant has units of reciprocal time, such as \(\text{s}^{-1}\).

• For example, in the reaction of \(\mathrm{N}_2\mathrm{O}_5\), the rate constant is given as \(3.0 \times 10^{-5} \, \mathrm{s}^{-1}\). This means that the reaction speed is inherently tied to this constant value, reflecting how the molecules interact and convert over time.

Knowing the rate constant allows chemists to predict the behavior of the reaction under various conditions. It's a foundational piece of information used to calculate the rate of the reaction and to explore how changing conditions might alter the speed of the reaction.

Reaction rate

The rate of a chemical reaction is a measure of how quickly the reactants are converted into products. In chemistry, understanding the reaction rate is crucial for controlling how a reaction proceeds.

• For a first-order reaction, like the one involving \(\mathrm{N}_2\mathrm{O}_5\), the rate is proportional to the concentration of the reactant: \[ \text{Rate} = k [\mathrm{N}_2\mathrm{O}_5] \]

The rate is given in units of concentration per time, which often appears as mol \(\text{L}^{-1}\text{s}^{-1}\). By using the rate law, we can determine how long it will take for a certain amount of reactant to be converted to product, which is particularly useful in industrial and lab settings where time efficiency is critical.

Chemical kinetics

Chemical kinetics is the branch of chemistry that deals with the speeds or rates at which chemical reactions occur. This field of study is essential for predicting and controlling how reactions proceed over time.

• Kinetics provides insight into the mechanisms of reactions, including how bonds are broken and formed during the process. It offers a quantitative way to understand the transformation of reactants to products.

Studying the kinetics of a reaction such as \(\mathrm{N}_2\mathrm{O}_5\) decomposition enables scientists to formulate mathematical models that predict the course of the reactions. These models help chemists in modifying conditions like concentration, temperature, or pressure to optimize the reaction.