Problem 71 - Chemistry: The Central Science in SI Units, Global Edition

Question

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (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

(a) The balanced equation for the overall reaction is: $$ \mathrm{H}_{2}(g)+2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+2\mathrm{HCl}(g) $$ (b) The intermediate in this mechanism is HI. (c) The observed rate law for the overall reaction is: $$ \text{rate} = k_{overall}[\mathrm{H}_{2}][\mathrm{ICl}] $$

Step 1: (a) Balanced equation for the overall reaction:

To find the overall balanced equation for the given mechanism, we should add both reactions and then simplify the equation by canceling out any species that appear on both the reactant and product sides, as these are considered intermediates. Let's add both reactions: $$ \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ +\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) $$ Now combine and cancel out the intermediate (HI): $$ \mathrm{H}_{2}(g)+2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+2\mathrm{HCl}(g) $$ 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: (b) Identification of intermediates:

Intermediates are species that are formed in one step of a mechanism and then consumed in a subsequent step. They are important for understanding a reaction mechanism, but do not appear in the overall balanced equation. In our mechanism, we can see that the species HI appears as a product in the first reaction and as a reactant in the second reaction. Thus, HI is an intermediate in this mechanism.

Step 3: (c) Rate law for the overall reaction:

According to the exercise, the first step is slow and the second step is fast. In mechanisms where one reaction step is much slower compared to others, the overall rate of the reaction is determined by the slowest step, known as the rate-limiting step or rate-determining step. In our case, the slow (rate-determining) step is: $$ \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) $$ The rate law for this elementary reaction is simply determined by the order with respect to each reactant. As it is a bimolecular reaction, the rate law will be: $$ \text{rate} = k[\mathrm{H}_{2}][\mathrm{ICl}] $$ Since this step is determining the overall rate, its rate law is the same for the overall reaction: $$ \text{rate} = k_{overall}[\mathrm{H}_{2}][\mathrm{ICl}] $$ So, the observed rate law for the overall reaction is: $$ \text{rate} = k_{overall}[\mathrm{H}_{2}][\mathrm{ICl}] $$

Key Concepts

Balanced Chemical Equation

In chemistry, a balanced chemical equation is a way to represent a chemical reaction using symbols and formulas. This equation ensures that the same number of each type of atom is present on both sides of the equation, indicating the law of conservation of mass, which states that atoms are not created or destroyed in a chemical reaction.
To achieve a balanced equation for the overall reaction of \(\text{H}_{2}(g) + 2\text{ICl}(g) \rightarrow \text{I}_{2}(g) + 2\text{HCl}(g)\) in this exercise, we add the two given elementary steps and cancel out the intermediates.
Balancing a chemical equation involves several steps:

• Write down the number of atoms of each element for both reactants and products.
• Add coefficients to balance the atoms for each element.
• Simplify the equation by canceling out intermediates, which do not appear in the initial or final form of the equation.
• Check your work to assure the same number of atoms on each side.

This confirms the reaction between hydrogen gas and iodine monochloride involves consuming two molecules of \(\text{ICl}\) to produce iodine and hydrogen chloride.

Intermediates in Reactions

In the mechanism of chemical reactions, intermediates are species that are formed in one step and consumed in another, and they do not appear in the final balanced equation.
Identifying intermediates is crucial for understanding the step-by-step process of how reactants turn into products, although they are only transiently present and can be difficult to detect experimentally.
In our exercise, the intermediate is \(\text{HI}\).

• Produced in the first step: \(\text{H}_{2}(g) + \text{ICl}(g) \rightarrow \text{HI}(g) + \text{HCl}(g)\)
• Consumed in the second step: \(\text{HI}(g) + \text{ICl}(g) \rightarrow \text{I}_{2}(g) + \text{HCl}(g)\)

The presence of \(\text{HI}\) highlights the presence of an intermediate in the mechanism, confirming that the chemical process progresses through multiple steps where \(\text{HI}\) serves as a bridge from reactants to products.

Rate Law

The rate law of a chemical reaction gives a relationship between the concentration of reactants and the rate at which the reaction occurs. Especially in complex reactions, when looking at a mechanism, the rate law is governed by the slowest step, known as the rate-determining step.
This reaction's rate law is derived from the first slow step: \(\text{H}_{2}(g) + \text{ICl}(g) \rightarrow \text{HI}(g) + \text{HCl}(g)\).
The overall rate is dictated by this specific step as it involves direct participation of the reactants \(\text{H}_{2}\) and \(\text{ICl}\) in a bimolecular manner. The rate law can be written as:

• \(\text{rate} = k[\text{H}_{2}][\text{ICl}]\)

This implies that the reaction rate is proportional to the concentration of both hydrogen and iodine monochloride, meaning a change in either concentration will affect how fast the reaction goes."
The rate constant, \(k\), is specific to this reaction and accounts for factors such as temperature and catalyst presence if any.

Elementary Reaction Steps

Elementary reaction steps are simple stages in a series of reactions that describe the mechanisms by which a chemical reaction occurs. Each elementary step shows actual molecular events, revealing how reactants are converted to products.
They are unique because they cannot be broken down further and are characterized by their molecularity—the number of molecules that participate as reactants.
In the exercise:

• The first step (slow): \(\text{H}_{2}(g) + \text{ICl}(g) \rightarrow \text{HI}(g) + \text{HCl}(g)\) is bimolecular, reflecting direct interaction between two molecules.
• The second step (fast): \(\text{HI}(g) + \text{ICl}(g) \rightarrow \text{I}_{2}(g) + \text{HCl}(g)\) is also bimolecular.

Elementary steps are foundational for understanding reactions in catalysis, enzymatic processes, and synthetic chemistry. Recognizing each step's role helps in predicting both qualitative and quantitative aspects of chemical kinetics.