Question

(a) Nitrogen dioxide, \(\mathrm{NO}_{2}\), decomposes at \(6.5\) \(\mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\) by the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\). Determine the rate of formation of \(\mathrm{O}_{2}\). (b) What is the unique rate of the reaction?

Short answer

The rate of formation of \(\mathrm{O}_{2}\) is \(3.25 \mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\). The unique rate of the reaction is \(6.5 \mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\).

Step-by-step solution

Identify the reaction and stoichiometry

Understand that the given reaction is \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})\). The stoichiometry is 2:2:1 for \(\mathrm{NO}_{2}\):\(\mathrm{NO}\):\(\mathrm{O}_{2}\), respectively.

Calculate the rate of formation of \(\mathrm{O}_{2}\)

Since the decomposition rate of \(\mathrm{NO}_{2}\) is given as \(6.5 \mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\), and the stoichiometric ratio of \(\mathrm{NO}_{2}\) to \(\mathrm{O}_{2}\) is 2:1, divide the rate of decomposition of \(\mathrm{NO}_{2}\) by 2 to get the formation rate of \(\mathrm{O}_{2}\): Rate of formation of \(\mathrm{O}_{2}\) = \(\frac{6.5 \mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}}{2}\).

Calculate the unique rate of the reaction

The unique rate of reaction is the rate of disappearance of reactants or the rate of appearance of products divided by their stoichiometric coefficients. In this case, it's the same as the rate of decomposition of \(\mathrm{NO}_{2}\) because of the 1:1 ratio considering the coefficients. Therefore, the unique rate equals the given rate of \(6.5 \mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\).

Key concepts

Stoichiometry and Reaction Rates

Understanding stoichiometry is crucial when studying the kinetics of chemical reactions. Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict the amount of substances consumed and produced during a reaction.

When it comes to reaction rates, stoichiometry comes into play as it helps us relate the rates at which reactants are used up to the rates at which products are formed. For example, in the decomposition of nitrogen dioxide (NO2), the stoichiometric ratio indicates that for every two molecules of NO2 that react, one molecule of oxygen (O2) is produced. This informs us about the relative speeds at which these species participate in the reaction and allows us to calculate the rate of formation of O2 based on the rate at which NO2 decomposes.

Remember: The reaction rate is not just about how fast a substance reacts; it's about how quickly the reactant concentrations change over time, relative to the balanced chemical equation. Therefore, balancing equations is not just a formal requirement but a practical necessity to understand the kinetics of the involved species.

Rate of Formation

The rate of formation of a product in a chemical reaction is the speed at which the product is created from the reactants. It's directly tied to the stoichiometry of the reaction because it considers the amount of a substance produced per unit time.

For example, in the given exercise, the rate of formation of oxygen gas (O2) from the decomposition of nitrogen dioxide (NO2) is a key point of interest. Using the stoichiometric ratio between NO2 and O2 (2:1), we can infer that for every doubling in the rate of decomposition of NO2, O2 is formed at half that rate. Thus, determining the rate of formation of O2 involves dividing the rate at which NO2 is disappearing by the stoichiometric coefficient of NO2 in the production of O2.

In practical terms, if we have the rate of NO2 decomposition, we simply apply the stoichiometric ratio to find how fast O2 appears in the reaction mixture. This can help chemists understand the dynamics of the reaction and predict the concentration of products at different times.

Nitrogen Dioxide Decomposition

The decomposition of nitrogen dioxide (NO2) into nitric oxide (NO) and oxygen (O2) is a classic example used to illustrate reaction kinetics and stoichiometry. This reversible reaction happens at a certain rate, which can change depending on temperature, pressure, and the presence of a catalyst.

In the context of the exercise, we focus on the decomposition rate given for NO2, which sets the stage for calculating other rates. The reaction proceeding at a rate of 6.5 mmol L^-1 s^-1 tells us how rapidly the concentration of NO2 falls over time. Decomposition reactions like this are common in the atmosphere and are critical to understanding environmental chemistry, such as the formation of smog and other pollution-related phenomena.

The quantitative study of the kinetics of this reaction helps scientists and engineers control and predict the behavior of NO2 under various conditions, thereby informing strategies for pollution reduction and environmental protection.

Unique Rate of Reaction

The unique rate of a reaction is a standardized way of reporting reaction rates irrespective of the reactant or product being considered. It is defined as the rate of change of concentration of any species involved in the reaction, divided by its stoichiometric coefficient in the balanced chemical equation.

In this exercise, this concept helps us understand the overall rate of the reaction. Since the stoichiometric coefficients of NO2 and NO are both 2, their rates will be identical. However, for O2 with a stoichiometric coefficient of 1, the rate would appear different if not adjusted for stoichiometry. To find this unique rate, we use the rate of disappearance of NO2 (or the rate of appearance of O2) divided by their respective coefficients. This provides us with a consistent measure that applies across all reactants and products, simplifying comparability and calculations in reaction kinetics.

Understanding the unique rate is key to grasping the big picture of how fast a reaction proceeds and is especially useful when comparing different reactions or different conditions for the same reaction.