Problem 219 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the reaction: \(\mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq}) \longrightarrow \mathrm{S}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\) The rate equation for this reaction is Rate \(=\mathrm{k}\left[\mathrm{Cl}_{2}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]\) Which of these mechanisms is/are consistent with this rate equation? [2010] (1) \(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{~S} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{Cl}^{+}+\mathrm{HS}^{-}\)(slow) \(\mathrm{Cl}^{+}+\mathrm{HS}^{-} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{S}\) (fast \()\) (2) \(\mathrm{H}_{2} \mathrm{~S} \Leftrightarrow \mathrm{H}^{+}+\mathrm{HS}^{-}\)(fast equilibrium) \(\mathrm{Cl}_{2}+\mathrm{HS}^{-} \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{H}^{+}+\mathrm{S}\) (slow) (a) 2 only (b) Both 1 and 2 (c) Neither 1 nor 2 (d) 1 only

Step-by-Step Solution

Answer

The correct answer is (b) Both 1 and 2 are consistent.

Step 1: Understanding the Given Rate Law

The rate equation provided is \( \text{Rate} = k[\text{Cl}_2][\text{H}_2 \text{S}] \). This indicates that the reaction rate is dependent on the concentration of both \( \text{Cl}_2 \) and \( \text{H}_2 \text{S} \). Each term shows that these two compounds are involved in the rate-determining step which is the slow step.

Step 2: Analyzing Mechanism 1

In mechanism 1, the slow step is \( \text{Cl}_2 + \text{H}_2 \text{S} \longrightarrow \text{H}^+ + \text{Cl}^- + \text{Cl}^+ + \text{HS}^- \). This slow step is consistent with our rate equation because both \( \text{Cl}_2 \) and \( \text{H}_2 \text{S} \) appear as reactants in this step, which matches with \( [\text{Cl}_2][\text{H}_2 \text{S}] \) in the rate law.

Step 3: Analyzing Mechanism 2

In mechanism 2, the slow step is \( \text{Cl}_2 + \text{HS}^- \longrightarrow 2 \text{Cl}^- + \text{H}^+ + \text{S} \). However, \( \text{HS}^- \) is an intermediate, not \( \text{H}_2 \text{S} \), which does not appear in the rate law directly. To be consistent, \( \text{HS}^- \) must relate back to \( \text{H}_2 \text{S} \) via a prior fast equilibrium step. The equilibrium \( \text{H}_2 \text{S} \Leftrightarrow \text{H}^+ + \text{HS}^- \) allows us to link\( [\text{HS}^-] \) to \( [\text{H}_2 \text{S}] \), making mechanism 2 consistent with the rate law.

Step 4: Conclusion

Both mechanisms 1 and 2 lead to the overall rate law with the same dependence on \( [\text{Cl}_2] \) and \( [\text{H}_2 \text{S}] \). Therefore, both mechanisms are consistent with the provided rate equation.

Key Concepts

Rate Equation

The rate equation is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. In our case, the equation is given as \( \text{Rate} = k[\text{Cl}_2][\text{H}_2 \text{S}] \). This equation indicates that the rate of reaction depends directly on the concentrations of chlorine and hydrogen sulfide.
This means higher concentrations of these substances lead to a faster reaction, provided other conditions are constant.

• \( k \) is the rate constant, which changes with temperature.
• The exponents of \([\text{Cl}_2]\) and \([\text{H}_2 \text{S}]\) are both 1, which implies a first-order dependence on each reactant.

Understanding the rate equation helps predict how the reaction progresses and what happens if either reactant concentration changes.

Rate-Determining Step

The rate-determining step in a reaction mechanism is the slowest step that controls the overall rate of the reaction. For our reaction, the slowest step determines the rate law. This step must involve all the reactants specified in the rate equation.
In Mechanism 1, the slow step \( \text{Cl}_2 + \text{H}_2 \text{S} \rightarrow \text{H}^+ + \text{Cl}^- + \text{Cl}^+ + \text{HS}^- \) includes both \( \text{Cl}_2 \) and \( \text{H}_2 \text{S} \), agreeing with \( [\text{Cl}_2][\text{H}_2 \text{S}] \) of the rate law.
While in Mechanism 2, the slow step involves an intermediate, \( \text{HS}^- \), but it is linked back to \( \text{H}_2 \text{S} \) by a prior equilibrium step, allowing it to be consistent with the rate law.
Understanding this concept is crucial because by altering this step, you directly influence the reaction speed.

Intermediates

Intermediates are species that appear in the mechanism of a reaction but are not seen in the overall balanced equation. They are formed in one elementary step and consumed in another, playing a pivotal role in connecting various stages of a reaction mechanism.
For Mechanism 2, \( \text{HS}^- \) is an intermediate. It is a product of an initial fast equilibrium stage and a reactant in the slow, rate-determining step.
These intermediates are important because they help scientists understand the pathway of a reaction, even though they do not appear in the final products. Understanding intermediates reveals how reactions progress from reactants to products over multiple steps.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the steps by which reactions occur. It is fundamental in determining the reaction mechanism, which is a detailed sequence of elementary steps describing how reactants transform into products.
Kinetics encompasses various factors, including:

• The concentration of reactants.
• The temperature at which the reaction occurs.
• Presence of catalysts, which can alter the rate by providing alternative pathways with lower activation energies.

Understanding chemical kinetics allows chemists to predict reaction behavior under different conditions, optimize conditions for desired outcomes, and elucidate the detailed mechanism of complex reactions. In this case, kinetics aids in validating the proposed mechanisms by matching them with observed data expressed through the rate equation.