Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

Observe the following reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{C}\) the rate of formation of is \(2.2 \times 10^{-3}\) mol. \(L^{-1}\). What is the value of \(-\mathrm{d}[\mathrm{A}] / \mathrm{dt}\left(\right.\) in \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}\right)\) ? (a) \(2.2 \times 10^{-3}\) (b) \(1.1 \times 10^{-3}\) (c) \(4.4 \times 10^{-3}\) (d) \(5.5 \times 10^{-3}\)

Step-by-Step Solution

Answer

The value of \(-d[A]/dt\) is \(4.4 \times 10^{-3}\) mol L\(^{-1}\) min\(^{-1}\).

Step 1: Understand the Reaction

We are given a chemical reaction: \(2 \mathrm{~A} + \mathrm{B} \rightarrow \mathrm{C}\). The reaction tells us that for every two moles of \(\mathrm{A}\) that react, one mole of \(\mathrm{C}\) is produced.

Step 2: Determine Given Rate

The rate of formation of \(\mathrm{C}\) is \(2.2 \times 10^{-3}\) mol \(L^{-1}\) min\(^{-1}\). This means that \(\mathrm{C}\) is forming at this rate.

Step 3: Relate the Rate of Formation of C to Rate of Consumption of A

The rate of reaction is given by the stoichiometry of the balanced equation. For \(2\mathrm{~A} + \mathrm{B} \rightarrow \mathrm{C}\), for every 1 mole of \(\mathrm{C}\) produced, 2 moles of \(\mathrm{A}\) are consumed. Therefore, the rate of disappearance of \([\mathrm{A}]\) is twice the rate of formation of \(\mathrm{C}\).

Step 4: Calculate -d[A]/dt

Since the rate of formation of \([\mathrm{C}]\) is \(2.2 \times 10^{-3}\) mol \(L^{-1}\) min\(^{-1}\), the rate of disappearance of \([\mathrm{A}]\) is \(2 \times 2.2 \times 10^{-3} = 4.4 \times 10^{-3} \) mol \(L^{-1}\) min\(^{-1}\).

Key Concepts

Rate of Reaction

In a chemical reaction, the rate of reaction measures how quickly reactants are converted into products. It is expressed as the change in concentration of a reactant or product over time. This concept is key for understanding how a reaction progresses. The rate can vary based on factors such as temperature, concentration, and presence of catalysts. In our exercise, the rate of formation of compound \( C \) is \( 2.2 \times 10^{-3} \) mol \( L^{-1} \) min\(^{-1}\), showing how quickly \( C \) is being produced during the reaction.

• **Reactant Concentration**: If there are more molecules available, the rate typically increases due to more frequent collisions.
• **Temperature**: Raising the temperature usually speeds up reactions as molecules move faster and collide more often.
• **Catalysts**: These substances increase the reaction rate without being consumed in the process.

These aspects help chemists control reactions for desired outcomes.

Stoichiometry

Stoichiometry is a fundamental aspect of chemistry that describes the proportions of elements and compounds involved in a reaction. It is used to calculate the quantities of reactants needed and products formed. In the exercise provided, the balanced equation \(2 \text{A} + \text{B} \rightarrow \text{C} \) shows the stoichiometric relationship between reactants \( \text{A} \), \( \text{B} \), and product \( \text{C} \).

• **Mole Ratios**: The coefficients in balanced equations indicate the mole ratios of the reactants and products.
• **Mole-to-Mole Conversions**: These ratios are used to calculate how much of each reactant is needed to produce a desired amount of product.

In our example, for every 1 mole of \( C \), 2 moles of \( A \) are consumed. This relationship helps determine the rate of \(-d[A]/dt\) showing that \( A \) is consumed twice as fast as \( C \) is formed.

Chemical Equilibrium

Chemical equilibrium occurs in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of all reactants and products remain constant over time. However, our original reaction is not in equilibrium. Hence, the rate of reaction matters more than equilibrium in this context.

• **Dynamic Equilibrium**: Even though no visible changes occur, molecular processes continue at equal rates in both directions.
• **Le Chatelier's Principle**: If a change occurs in conditions (like pressure or concentration), the system will adjust to restore equilibrium.

Understanding chemical equilibrium is crucial for reactions where equilibrium needs to be established or maintained. However, for our scenario, we focus on stoichiometry and the rate of reactions as the system hasn't reached equilibrium.