Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For an elementary chemical reaction, \(\mathrm{A}_{2} \frac{\mathrm{k}_{\mathrm{l}}}{\mathrm{k}_{-1}} 2 \mathrm{~A}\),the expression for \(\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) is:(a) \(\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (b) \(2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (c) \(\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]+\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (d) \(2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}\)

Step-by-Step Solution

Answer

The expression is (d) \(2 k_{1}[\mathrm{A}_{2}] - 2 k_{-1}[\mathrm{A}]^2\).

Step 1: Write the Reaction

The elementary chemical reaction given is \( \mathrm{A}_{2} \leftrightarrow 2 \mathrm{~A} \), with forward rate constant \( k_{1} \) and reverse rate constant \( k_{-1} \).

Step 2: Determine Rate Laws for Forward and Reverse Reactions

The rate of the forward reaction is given by the expression: \( \text{rate}_{ ext{forward}} = k_{1}[\mathrm{A}_{2}] \). The rate of the reverse reaction is given by: \( \text{rate}_{ ext{reverse}} = k_{-1}[\mathrm{A}]^2 \).

Step 3: Write the Expression for Net Rate of Change of [A]

Using the rate expressions from Step 2, the net rate of change of \([\mathrm{A}]\) is given by:\[ \frac{d[\mathrm{A}]}{dt} = 2 k_{1}[\mathrm{A}_{2}] - 2k_{-1}[\mathrm{A}]^2 \]The factor of 2 accounts for the stoichiometry of the reaction, as 2 moles of A are produced or consumed per mole of \( \mathrm{A}_2 \) reacting.

Key Concepts

Elementary Reaction

An elementary reaction is the simplest form of a chemical process. In such reactions, the reaction occurs in a single step, directly converting reactants to products without any intermediates. The reaction given, \( \mathrm{A}_{2} \leftrightarrow 2 \mathrm{~A} \), is an example of an elementary reaction.
It involves the dissociation of a molecule \( \mathrm{A}_{2} \) into two separate \( \mathrm{A} \) atoms.
Elementary reactions have distinct characteristics:

• They can sometimes be reversible, as shown by the double arrows, meaning the reaction can proceed in both directions.
• Each elementary reaction can be characterized by specific rate constants, \( k_1 \) for the forward reaction and \( k_{-1} \) for the reverse reaction in this case.
• They follow simple rate laws as they involve direct conversion from reactants to products.

Understanding elementary reactions aids in determining the speed and mechanism of complex reactions.

Rate Law Expression

The rate law expression is a mathematical relationship that indicates the speed of a reaction in terms of the concentration of reactants. For elementary reactions, the rate law can be written directly from the stoichiometry of the reaction.
In our case, we have two expressions corresponding to the forward and reverse processes:

• Forward reaction: \( \text{rate}_{\text{forward}} = k_{1}[\mathrm{A}_{2}] \). This means that the rate of the forward reaction is proportional to the concentration of \( \mathrm{A}_{2} \).
• Reverse reaction: \( \text{rate}_{\text{reverse}} = k_{-1}[\mathrm{A}]^2 \). This shows that the reverse rate is proportional to the square of the concentration of \( \mathrm{A} \).

Rate laws are crucial in understanding how changing reactant concentrations affects reaction speed. They help in predicting how a system behaves over time.

Stoichiometry

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It tells us how much product will form from a given amount of reactant. In the context of this exercise, stoichiometry is crucial for writing rate expressions.
For the reaction \( \mathrm{A}_{2} \leftrightarrow 2 \mathrm{~A} \), each mole of \( \mathrm{A}_{2} \) generates two moles of \( \mathrm{A} \). So, when calculating the rate of change of \([\mathrm{A}]\), the stoichiometric factor comes into play:

• The expression \( \frac{d[\mathrm{A}]}{dt} = 2k_{1}[\mathrm{A}_{2}] - 2k_{-1}[\mathrm{A}]^2 \) reflects this relationship by multiplying both rate constants by 2.
• This multiplication by 2 is necessary to account for the production or consumption of two \( \mathrm{A} \) per one \( \mathrm{A}_{2} \) in the reaction.

Understanding stoichiometry helps us accurately represent how changes in reactant concentrations impact the overall reaction and predict the amounts of reactants and products over time.