Problem 73 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following reaction, how is the rate of appear ance of the underlined product related to the rate of disappearance of the underlined reactant? \(\mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 3 \mathrm{Br}_{2}(1)\) \(+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (b) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (c) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (d) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\)

Step-by-Step Solution

Answer

The correct relationship is (d): \(\frac{d[\mathrm{Br}_{2}]}{dt} = -\frac{3}{5} \frac{d[\mathrm{Br}]}{dt}\).

Step 1: Understand the Reaction

The reaction given is \(\mathrm{BrO}_{3}^{-} + 5 \mathrm{Br} + 6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_{2} + 3 \mathrm{H}_{2}\mathrm{O}\). This equation involves the disappearance of \(\mathrm{Br}\) as a reactant and the appearance of \(\mathrm{Br}_{2}\) as a product.

Step 2: Analyze the Stoichiometry

The coefficients in the balanced equation indicate the ratios of reactants consumed and products formed. For every 5 moles of \(\mathrm{Br}\) consumed, 3 moles of \(\mathrm{Br}_{2}\) are produced.

Step 3: Relate the Rates of Reaction

The rate of disappearance of \(\mathrm{Br}\) and the rate of appearance of \(\mathrm{Br}_{2}\) are directly related to their coefficients in the balanced equation. Since 5 \(\mathrm{Br}\) produce 3 \(\mathrm{Br}_{2}\), we have the relationship:\[5 \frac{d[\mathrm{Br}]}{dt} = 3 \frac{d[\mathrm{Br}_{2}]}{dt}\].

Step 4: Solve for Rate of \(\mathrm{Br}_{2}\)

Rearrange the relationship to solve for \(\frac{d[\mathrm{Br}_{2}]}{dt}\):\[\frac{d[\mathrm{Br}_{2}]}{dt} = \frac{3}{5} \frac{d[\mathrm{Br}]}{dt}\]. The equation shows that the rate of appearance of \(\mathrm{Br}_{2}\) equals \(\frac{3}{5}\) times the rate of disappearance of \(\mathrm{Br}\).

Step 5: Determine the Correct Option

Given choices include rate relations. Based on \(\frac{d[\mathrm{Br}_{2}]}{dt} = \frac{3}{5} \frac{d[\mathrm{Br}]}{dt}\), the correct relation reflecting the balanced equation is option (d):\[\frac{d[\mathrm{Br}_{2}]}{dt} = -\frac{3}{5} \frac{d[\mathrm{Br}]}{dt}\].

Key Concepts

Rate of Reaction

The rate of reaction is a fundamental concept in chemical kinetics, measuring how quickly reactants transform into products in a chemical reaction. It is usually expressed as the change in concentration of a reactant or product per unit time. In the given reaction, the rate of disappearance of bromide ions, \([\mathrm{Br}]\), signifies how fast they are consumed. Similarly, the rate at which bromine, \([\mathrm{Br}_2]\), appears indicates the speed at which it is produced.

The mathematical representation involves derivatives, such as \(\frac{d[\mathrm{Br}]}{dt}\) and \(\frac{d[\mathrm{Br}_2]}{dt}\), which denote the rate of change over time. Understanding these rates helps us link the changes in reactant and product concentrations, illustrating how the stoichiometry of a balanced chemical equation can define these relationships. For our equation, the rate of appearance of \(\mathrm{Br}_2\) is directly proportional to a fraction of the rate of disappearance of \(\mathrm{Br}\). This linkage is crucial in predicting how concentrations evolve as the reaction progresses.

Stoichiometry

Stoichiometry is a key component in determining the relationship between the concentration changes of reactants and products. It provides the quantitative basis of chemical reactions, ensuring reactants and products are balanced. The balanced equation is essential for calculating other quantities, such as reaction rates.

In the provided chemical reaction, \(\mathrm{BrO}_3^- + 5 \mathrm{Br} + 6 \mathrm{H}^+ \rightarrow 3 \mathrm{Br}_2 + 3 \mathrm{H}_2\mathrm{O}\), stoichiometry reveals the ratios, such as the 5 moles of \(\mathrm{Br}\) corresponding to 3 moles of \(\mathrm{Br}_2\). Consequently, stoichiometry allows us to deduce that for every 5 molecules of \(\mathrm{Br}\) disappearing, 3 molecules of \(\mathrm{Br}_2\) are appearing.

This ratio translates into rate relationships, where the slope or ratio in the rate expressions is derived directly from stoichiometric coefficients. For any chemical problem, having the right stoichiometric calculations ensures accurate and meaningful interpretations of reaction dynamics.

Reaction Mechanism

The reaction mechanism is a sequence of step-by-step molecular events that serves to transform reactants into products. These sequences shed light on how a reaction proceeds at the molecular level and help explain the overall conversion process. While the provided exercise does not delve into the specifics of the reaction mechanism, understanding reaction mechanisms is vital for a full comprehension of complex reactions.

Each step within a mechanism can encompass different reaction rates, contributing to the overall rate of the principal reaction. Individual steps demonstrate interactions at a molecular level, sometimes involving intermediates that do not appear in the overall balanced equation. Thus, the observed rate laws are usually reflections of the slowest step in the mechanism, frequently termed the rate-determining step.

Knowledge of the stoichiometry and rate laws can support predictions about mechanisms, helping verify the coherence of proposed steps with experimental data. Through reaction mechanisms, we edge closer to depicting the actual pathways, offering insights impossible to gain from examination of the balanced equation alone.