Problem 57 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2} .\) Given values are \(-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=k_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=k_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Now the relation in between \(k_{1} k_{2}\) and \(k_{3}\) is (a) \(k_{1}=k_{2}=k_{3}\) (b) \(3 k_{1}=k_{2}=2 k_{3}\) (c) \(2 k_{1}=4 k_{2}=k_{3}\) (d) \(2 k_{1}=k_{2}=4 k_{3}\)

Step-by-Step Solution

Answer

The correct relationship is \(3k_1 = k_2 = 2k_3\). Option (b).

Step 1: Understand the Reaction Stoichiometry

The given chemical reaction is \(\mathrm{N}_2 \mathrm{O}_5 \rightarrow 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2\). From this, we can see that 1 mole of \(\mathrm{N}_2 \mathrm{O}_5\) produces 2 moles of \(\mathrm{NO}_2\) and \(\frac{1}{2}\) mole of \(\mathrm{O}_2\). Keep this stoichiometric relationship in mind as it will inform the relationships between the rate constants.

Step 2: Write the Rate Equations

The rate equation for the decomposition of \(\mathrm{N}_2 \mathrm{O}_5\) is given by \(-\frac{\mathrm{d}[\mathrm{N}_2 \mathrm{O}_5]}{\mathrm{dt}}=k_1[\mathrm{N}_2 \mathrm{O}_5]\). The formation rate of \(\mathrm{NO}_2\) is \(\frac{\mathrm{d}[\mathrm{NO}_2]}{\mathrm{dt}}=k_2[\mathrm{N}_2 \mathrm{O}_5]\) and for \(\mathrm{O}_2\) it's \(\frac{\mathrm{d}[\mathrm{O}_2]}{\mathrm{dt}}=k_3[\mathrm{N}_2 \mathrm{O}_5]\).

Step 3: Relate Rate Constants to Stoichiometry

The reaction shows that for every 1 mole of \(\mathrm{N}_2 \mathrm{O}_5\) consumed, 2 moles of \(\mathrm{NO}_2\) are produced. So, \(k_2\) should be twice \(k_1\), thus \(k_2 = 2k_1\). Similarly, for \(\mathrm{O}_2\), the stoichiometry tells us \(k_3\) should be half of \(k_1\) because only \(\frac{1}{2}\) mole of \(\mathrm{O}_2\) is produced per mole of \(\mathrm{N}_2 \mathrm{O}_5\), thus \(k_3 = \frac{k_1}{2}\).

Step 4: Determine the Relationship Between Constants

From step 3, we have that \(k_2 = 2k_1\) and \(k_3 = \frac{k_1}{2}\). To compare these relationships with the given options: it is necessary that \(3k_1 = k_2 = 2k_3\) as derived: \(k_2 = 2k_1 = 3k_1\) when applied consistently with \(k_3 = \frac{k_1}{2}\).

Key Concepts

Reaction Stoichiometry

When studying chemical reactions, stoichiometry is all about the quantitative relationships between reactants and products. In the decomposition reaction of \(_2 \mathrm{O}_5\) into \(_2 \mathrm{NO}_2\) and \(_2 \mathrm{O}_2\), the stoichiometry tells us how much of each product is formed from the reactant. Here, 1 mole of \(_2 \mathrm{O}_5\) breaks down to give 2 moles of \(_2 \mathrm{NO}_2\) and 0.5 moles of \(_2 \mathrm{O}_2\). This stoichiometric ratio is crucial for predicting how the reaction will proceed and is directly related to the rate at which products form. Understanding this baseline allows us to work out the relationships between rate constants.

Rate Equations

Rate equations describe how the concentration of substances involved in a chemical reaction changes over time. For the decomposition of \(_2 \mathrm{O}_5\), we have multiple rate equations reflecting how each product forms. The rate of decomposition of \(_2 \mathrm{O}_5\) itself is expressed as \(-\frac{d[\mathrm{N}_2 \mathrm{O}_5]}{dt}=k_1[\mathrm{N}_2 \mathrm{O}_5]\), showing it depends on its own concentration. The formation of \(_2 \mathrm{NO}_2\) is described by \(\frac{d[\mathrm{NO}_2]}{dt}=k_2[\mathrm{N}_2 \mathrm{O}_5]\), while the formation of \(_2 \mathrm{O}_2\) obeys: \(\frac{d[\mathrm{O}_2]}{dt}=k_3[\mathrm{N}_2 \mathrm{O}_5]\). These equations are integral as they allow us to determine how quickly products appear once we know the concentration of the decomposing reactant.

Rate Constants

Rate constants, represented typically as \(k_1, k_2, \) and \(k_3\), are specific to each step in a reaction and link the rate equations to the concentration of reactants. These constants depend on factors like temperature and catalyst presence but remain consistent for given conditions. In the decomposition of \(_2 \mathrm{O}_5\), understanding how the rate constants relate to one another is vital. Here, \(k_2\) is twice \(k_1\) because the stoichiometry shows each mole of \(_2 \mathrm{O}_5\) forms 2 moles of \(_2 \mathrm{NO}_2\). Similarly, \(k_3\) is half \(k_1\) due to the 0.5-mole formation ratio of oxygen. Thus, our calculated relationships yield that \(3k_1 = k_2 = 2k_3\), fitting one of the given choices in the exercise.

Decomposition Reactions

In chemistry, decomposition reactions involve a single compound breaking into two or more simpler substances. This type of reaction is exemplified by the breakdown of \(_2 \mathrm{O}_5\) into \(_2 \mathrm{NO}_2\) and \(_2 \mathrm{O}_2\). Often driven by heating or by light, these reactions are critical in various natural and industrial processes. During the decomposition, the reactant's bonds break, forming new bonds to create the products. Analyzing decomposition allows us to determine crucial aspects like how a reaction progresses and what conditions optimize speed and efficiency. Such knowledge assists in fields ranging from chemical manufacturing to understanding natural phenomena.