Problem 71 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The balanced equation for the overall reaction is: \(\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\). The intermediate in this reaction mechanism is HI. The rate law for the overall reaction, as determined by the slow step (the first step), is: Rate = k[\(\mathrm{H}_{2}\)][\(\mathrm{ICl}\)].

Step 1: Write the balanced equation for the overall reaction.

To find the overall reaction, we need to add both elementary steps given in the exercise: (1) \(\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)\) (2) \(\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g)\) Add both reactions: \(\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\) The balanced equation for the overall reaction is: \(\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\)

Step 2: Identify any intermediates in the mechanism.

Now let's find intermediates in the reaction mechanism. Intermediates are species that are formed and consumed during the reaction process. In the two elementary steps, we can see that HI is produced in the first step and then consumed in the second step. Therefore, HI is an intermediate. Intermediates: HI

Step 3: Determine the rate law for the overall reaction

We are told that the first step is slow and the second step is fast. The slow step determines the rate of the overall reaction, so we need to consider only the first elementary step for the rate law. The rate law is based on the concentrations of the reactants in the slow step (the first step). The reaction has the following form: \(\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)\) The rate law is given by the equation: Rate = k[\(\mathrm{H}_{2}\)][\(\mathrm{ICl}\)] Where k is the rate constant and [\(\mathrm{H}_{2}\)] and [\(\mathrm{ICl}\)] are the molar concentrations of \(\mathrm{H}_{2}\) and ICl, respectively. So, the expected rate law for the overall reaction is: Rate = k[\(\mathrm{H}_{2}\)][\(\mathrm{ICl}\)]

Key Concepts

Elementary Reaction Steps

Elementary reaction steps are the building blocks of a reaction mechanism. Each step represents a simple chemical reaction that occurs as part of a larger process.
These steps showcase how reactants are converted into products, usually involving a single collision event between molecules or a single bond formation or breaking. In the given problem, the reaction between hydrogen (\(\mathrm{H}_2\)) and iodine monochloride (\(\mathrm{ICl}\)) to form hydrogen iodide (\(\mathrm{HI}\)) and hydrogen chloride (\(\mathrm{HCl}\)) is such an elementary reaction step.

• The first step involves the initial reactants (\(\mathrm{H}_{2}(g)+\mathrm{ICl}(g)\)).
• The products, \(\mathrm{HI}(g)\) and \(\mathrm{HCl}(g)\), are transient and may only exist momentarily.
• The second step uses these transient products to form the final solution (\(\mathrm{I}_{2}(g)+\mathrm{HCl}(g)\)).

Understanding these individual steps is crucial as they set up the stage for determining the reaction rate and identifying intermediates.

Intermediates

Intermediates in a chemical reaction are species that are formed during one of the reaction steps and then consumed in a subsequent step of the mechanism.
They do not appear in the overall balanced equation because their existence is fleeting. In the mechanism provided in the problem, \(\mathrm{HI}\) is produced in the first elementary step and consumed in the second. Thus, \(\mathrm{HI}\) is an intermediate.

• Intermediates bridge the two elementary steps.
• They are crucial for understanding the complexity of reaction pathways.
• Although temporary, they play a vital role in advancing the reaction from reactants to products.

Analyzing intermediates helps chemists deduce how reactions proceed at the molecular level.

Rate Law Determination

Determining the rate law for a reaction involves understanding which step controls the reaction rate. This is often referred to as the rate-determining step, frequently the slowest step in the sequence. In this exercise, the first step of the reaction mechanism is given as slow, making it the rate-determining step.
The general idea is that the overall reaction rate is controlled by this slow step because subsequent steps proceed much faster without significant delay. From the first elementary step: \(\mathrm{H}_{2}(g)+\mathrm{ICl}(g)\) to \(\mathrm{HI}(g)+\mathrm{HCl}(g)\), we derive the rate law:

• The rate law is expressed in terms of concentration of the reactants in the slow step.
• It follows the form: \(\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\).
• Here, \(k\) is the rate constant specific to this reaction.

Understanding the rate law is important because it allows chemists to predict how a change in concentration of reactants affects the overall rate of reaction.