Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in chloroform was followed by measuring the volume of \(\mathrm{O}_{2}\) gas evolved : \(2 \mathrm{~N}_{2} \mathrm{O}_{5}\left(\mathrm{CCl}_{4}\right) \stackrel{2}{\longrightarrow} 2 \mathrm{~N}_{2} \mathrm{O}_{4}\left(\mathrm{CCl}_{4}\right)+\mathrm{O}_{2}(g) .\) The maximum volume of \(\mathrm{O}_{2}\) gas obtained was \(100 \mathrm{~cm}^{3}\). In 500 minutes, \(90 \cdot \mathrm{cm}^{3}\) of \(\mathrm{O}_{2}\) were evolved. The first order rate constant (in \(\min ^{-1}\) ) for the disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is : (a) \(\frac{2.303}{500}\) (b) \(\frac{2.303}{500} \log \frac{100}{90}\) (c) \(\frac{2.303}{500} \log \frac{90}{100}\) (d) \(\frac{100}{10 \times 500}\)

Short answer

(b) \(\frac{2.303}{500} \log \frac{100}{90}\)

Step-by-step solution

Understand the chemical reaction and data given

The given reaction is the decomposition of dinitrogen pentoxide into dinitrogen tetroxide and oxygen gas. We are given the total evolved oxygen gas volume as 100 cm³ and the volume of oxygen gas evolved in 500 minutes as 90 cm³.

Apply the integrated rate law for a first order reaction

For a first-order reaction, the integrated rate law is ln([A]_0 / [A]) = kt, where [A]_0 and [A] represent the initial concentration and the concentration at time t, respectively, and k is the first-order rate constant.

Calculate initial and remaining concentrations of N2O5

The initial concentration of N2O5 can be represented by the volume of O2 it can produce, which is 100 cm³. After 500 minutes, 90 cm³ have evolved, indicating that 10 cm³ of O2 is yet to evolve. This implies 10 cm³ of N2O5 is remaining.

Express the concentration terms as fractions

We can express the concentration of N2O5 at time t relative to its initial concentration as a ratio of the volumes of O2 since 1 mole of N2O5 yields 0.5 moles of O2. For the initial concentration [A]_0 = 100 cm³ and the remaining concentration [A] = (100 - 90) cm³ = 10 cm³.

Insert values into the integrated rate equation and solve for k

ln([A]_0 / [A]) = kt becomes ln(100 / 10) = k * 500 min. Solving for k gives k = (1/500) * ln(10). The natural log ln(10) is approximately 2.303, so the equation reduces to k = (2.303 / 500).

Choose the correct answer

Among the given options, option (b) matches the final expression for the rate constant k. Thus, the first-order rate constant for the reaction is (2.303 / 500) * log(100 / 90) which simplifies to just (2.303 / 500) since log(100 / 10) is 1.

Key concepts

Chemical Kinetics

Chemical kinetics is the study of how quickly chemical reactions occur and what factors affect these rates. It's a key topic in chemistry because it helps scientists and engineers understand and control reactions, from industrial processes to biological systems.

At the core of kinetics is the reaction rate, the speed at which reactants transform into products. This rate can be influenced by various parameters such as temperature, concentration of reactants, and the presence of catalysts. By analyzing the kinetics of a reaction, one can determine the reaction order, which indicates how the rate is affected by the concentration of reactants.

For instance, in the textbook exercise provided, the decomposition of dinitrogen pentoxide is analyzed. Knowing that this reaction is a first-order reaction allows us to predict how the concentration of N2O5 will change over time and to calculate the reaction's rate constant, which is a quantitative measure of how fast the reaction proceeds.

Integrated Rate Law

The integrated rate law is a mathematical expression that relates the concentrations of reactants to time, allowing us to determine the rate constant and the concentrations of reactants at any given time during a reaction.

For first-order reactions, the integrated rate law is presented as \( \text{ln}(\frac{[A]_0}{[A]}) = kt \), where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time t, and k is the rate constant. This relationship is the foundation for solving problems like the decomposition of dinitrogen pentoxide in the exercise.

By manipulating this law, we were able to use the given volumes of oxygen gas to calculate the remaining concentration of N2O5, which in turn provided the information needed to find the rate constant, a crucial piece of data for predicting reaction behavior over time.

Reaction Decomposition

Reaction decomposition is a type of chemical reaction in which a single compound breaks down into two or more products. It's commonly a unimolecular process, meaning that the rate of reaction depends on the concentration of the single reactant.

In our textbook exercise, the reactant dinitrogen pentoxide decomposes to form dinitrogen tetroxide and oxygen gas. This particular decomposition is essential because it shows a direct relationship between the concentration of the reactant and the volume of gas produced. By understanding the stoichiometry and the order of the reaction, we can make use of the kinetic data like gas volumes to monitor the reaction progress and to calculate the kinetic parameters such as the rate constant.

Concentration and Rate Relationship

Understanding the relationship between concentration and rate is crucial for controlling chemical reactions. In a first-order reaction, the rate is directly proportional to the concentration of the reactant. As the reactant is consumed, the rate of the reaction decreases.

In the provided example, by measuring the volume of oxygen gas generated, which is directly related to the concentration of N2O5, we gained insights into the reaction's progress. This proportional relationship between concentration and rate enables us to use simple measures, like gas volume, to make complex determinations about reaction rates and mechanisms.

By applying this knowledge, students and researchers can design experiments to measure reaction rates, determine rate laws, and ultimately optimize reactions for a variety of applications, from industrial synthesis to pharmaceuticals production.