Question

At a certain moment in the reaction, $$2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$$ \(\mathrm{N}_{2} \mathrm{O}_{5}\), is decomposing at a rate of \(2.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). What are the rates of formation of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ?

Short answer

The rates of formation are \(5.0 \times 10^{-6}\) mol L^-1 s^-1 for \(NO_{2}\) and \(1.25 \times 10^{-6}\) mol L^-1 s^-1 for \(O_{2}\).

Step-by-step solution

Write down the balanced chemical equation

The given reaction is already balanced: \[2 \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\]

Relate the rates of reactants and products

The stoichiometry of the reaction shows that 2 moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) produce 4 moles of \(\mathrm{NO}_{2}\) and 1 mole of \(\mathrm{O}_{2}\). Therefore, the rate of formation of \(\mathrm{NO}_{2}\) is twice the rate of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), and the rate of formation of \(\mathrm{O}_{2}\) is half the rate of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

Calculate the rate of formation of NO2

Given the rate of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(2.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\), the rate of formation of \(\mathrm{NO}_{2}\) is \(2 \times 2.5 \times 10^{-6}\) mol L^-1 s^-1, which equals \(5.0 \times 10^{-6}\) mol L^-1 s^-1.

Calculate the rate of formation of O2

Similarly, the rate of formation of \(\mathrm{O}_{2}\) is half the rate of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), which is \(\frac{2.5 \times 10^{-6}}{2}\) mol L^-1 s^-1, resulting in \(1.25 \times 10^{-6}\) mol L^-1 s^-1.

Key concepts

Reaction Stoichiometry

Understanding how substances interact in a chemical reaction is fundamental to chemistry, and reaction stoichiometry is the quantification of these interactions. For instance, suppose we have the decomposition reaction of dinitrogen pentoxide: \[2 \text{N}_2\text{O}_5 \rightarrow 4 \text{NO}_2 + \text{O}_2\].In this equation, stoichiometry explains that 2 moles of \(\text{N}_2\text{O}_5\) yield 4 moles of \(\text{NO}_2\) and 1 mole of \(\text{O}_2\). To understand the concept, envision making sandwiches: if 2 loaves of bread produce 4 sandwiches and 1 salad, then for every halving in the bread, you'd get half as many sandwiches and salads. Stoichiometry is the numerical relationship between chemical quantities in a balanced chemical equation, which can be used to predict the amount of products formed from a given amount of reactant.

Rate of Decomposition

The rate of a chemical reaction is a measure of how quickly reactants are turned into products, and specifically, the rate of decomposition gauges how rapidly a substance breaks down into simpler substances. If we consider our example of \(\text{N}_2\text{O}_5\) decomposing, its rate is expressed in moles per liter per second (mol L^-1 s^-1). The given rate in our exercise is \(2.5 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1}\).

Visualizing the Process

A good analogy is watching a sugar cube dissolve in water; the speed at which it disappears is similar to the rate of decomposition. However, it's crucial to remember that the rate of decomposition not only depends on the concentration but is also influenced by factors such as temperature and the presence of catalysts.

Rate of Formation

The flip side of the chemical reaction rate coin is the rate of formation, which charts the creation of products. It's a dynamic that gives us a clear view of how fast new substances emerge in a reaction. Referring back to our reaction, the rates of formation for \(\text{NO}_2\) and \(\text{O}_2\) are linked to the decomposition rate of \(\text{N}_2\text{O}_5\).

Understanding Proportions

These rates are proportional to the ratios outlined in the chemical equation's stoichiometry. Therefore, understanding the proportionality allows us to calculate the rate at which \(\text{NO}_2\) and \(\text{O}_2\) form from the breakdown of \(\text{N}_2\text{O}_5\) simply by applying the stoichiometric coefficients as multipliers or divisors to the rate of decomposition.

Chemical Kinetics

Chemical kinetics dives deeper into the rates of chemical reactions, studying the pathway from reactants to products and how various conditions affect this journey. It's a domain within chemistry that deals not only with speeds of reactions but also with understanding how and why reactions happen as they do. It’s an arena of study that encompasses factors such as concentration, temperature, physical state of the reactants, and the use of catalysts.

Why Kinetics Matter

Kinetics inform us about reaction mechanisms, or the 'recipe' that reactions follow, which is vital for controlling industrial chemical processes, pharmaceutical formulations, and even baking a cake efficiently! With kinetics, chemists can predict reaction behavior, allowing us to tailor conditions for desired outcomes. As students grasp kinetics, they gain the tools to understand the subtleties of how the world around us transforms through countless chemical reactions every moment.