Question

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(\mathrm{~g})+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) Write rate laws for each elementary reaction in the mechanism. (d) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

Short answer

(a) The overall balanced equation for the reaction is: \( \mathrm{H}_{2}(g) + 2 \mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + 2 \mathrm{HCl}(g) \) (b) The intermediate in the mechanism is HI. (c) Rate laws for each elementary reaction are: \( rate_1 = k_1 [\mathrm{H}_{2}] [\mathrm{ICl}] \) and \( rate_2 = k_2 [\mathrm{HI}] [\mathrm{ICl}] \) (d) The rate law for the overall reaction is: \( rate = k_1 [\mathrm{H}_{2}] [\mathrm{ICl}] \)

Step-by-step solution

(a) Overall Balanced Equation

Add the two steps to get the overall balanced equation: Step 1: \( \mathrm{H}_{2}(g) + \mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g) + \mathrm{HCl}(g) \) Step 2: \( \mathrm{HI}(g) + \mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g) \) Overall: \( \mathrm{H}_{2}(g) + 2 \mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + 2 \mathrm{HCl}(g) \)

(b) Identify Intermediates

Intermediates are species that are produced in one step and consumed in a later step of the reaction mechanism. In this case, the intermediate is HI since it is produced in step 1 and consumed in step 2.

(c) Rate Laws for Each Elementary Reaction

For each elementary reaction, write the rate law using the rate constant k and the concentration of the reactants raised to their stoichiometric coefficients: Rate Law for Step 1: \( rate_1 = k_1 [\mathrm{H}_{2}] [\mathrm{ICl}] \) Rate Law for Step 2: \( rate_2 = k_2 [\mathrm{HI}] [\mathrm{ICl}] \)

(d) Rate Law for the Overall Reaction

Since step 1 is slow (rate-determining step) and step 2 is fast, the overall rate of the reaction will be determined by the rate of step 1. Therefore, the rate law for the overall reaction will be the same as the rate law for step 1: Overall Rate Law: \( rate = k_1 [\mathrm{H}_{2}] [\mathrm{ICl}] \)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is the process of making sure that the number of atoms of each element is the same on both sides of a chemical equation. This is important because it reflects the conservation of mass, a fundamental principle of physics. In the context of our reaction mechanism involving the reaction of \( \mathrm{H}_{2} \) with \( \mathrm{ICl} \), balancing ensures that the total matter is consistent, aligning with the law of conservation of mass.
To balance the overall equation derived from the reaction mechanism, you need to combine both given steps and ensure that all atoms are accounted for. From Step 1: \( \mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g) \) and from Step 2: \( \mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g) \), the overall balanced equation can be written as:
\[ \mathrm{H}_{2}(g) + 2 \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + 2 \mathrm{HCl}(g) \]
Each element's count is equal on both sides, confirming the equation is now balanced.

Intermediates in Reactions

Intermediates are species that form in one step of a reaction mechanism and are consumed in another. They do not appear in the overall equation because they are not final products. Understanding intermediates is essential because they help us understand how reactions progress at the molecular level.
In our reaction mechanism example, \( \mathrm{HI} \) is the intermediate. It is produced in the first step and consumed in the second step:

• Step 1: Production - \( \mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g) \)
• Step 2: Consumption - \( \mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g) \)

Recognizing intermediates is crucial for understanding the detailed path that transformations occur in multistep reactions.

Rate Laws

Rate laws express the relationship between the rate of a chemical reaction and the concentration of its reactants. They provide insights into the reaction's speed and how altering concentrations can affect that speed. For a given elementary reaction, the rate law can be determined from its stoichiometry.
In our mechanism:

• Step 1 has the rate law \( rate_1 = k_1 [\mathrm{H}_{2}][\mathrm{ICl}] \), showing dependence on the concentrations of \( \mathrm{H}_{2} \) and \( \mathrm{ICl} \).
• Step 2 has the rate law \( rate_2 = k_2 [\mathrm{HI}][\mathrm{ICl}] \), highlighting that \( \mathrm{HI} \) is involved as a reactant at this step.

Understanding these rate laws allows chemists to predict how fast the reactions will occur under different conditions and to identify which reactants have the greatest influence on reaction velocity.

Rate-Determining Step

A rate-determining step in a reaction mechanism is the slowest step, which ultimately controls the overall reaction rate. It acts as a bottleneck, much like the slowest point in a traffic jam determines the flow of cars.
For the gas-phase reaction involving \( \mathrm{H}_{2} \) and \( \mathrm{ICl} \), the first step is noted as the slow step. As such, it is the rate-determining step. This affects the overall reaction rate law, which primarily depends on this step's reactants. Thus, the overall rate law is the same as that of the first step:
\[ rate = k_1 [\mathrm{H}_{2}][\mathrm{ICl}] \]
This concept is important for understanding which parts of a reaction mechanism are most influential on the overall rate, guiding strategies to optimize and control reaction conditions.