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?

Short answer

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

Step-by-step solution

Writing the balanced equation for the overall reaction

To obtain the overall reaction, we need to add the two given elementary reactions that constitute the reaction mechanism: \(1)~\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)\) \(2)~\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g)\) To find the overall reaction, we simply add these two reactions and eliminate any species that appear on both sides of the arrows: \(\mathrm{H}_{2}(g)+2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+2\mathrm{HCl}(g)\)

Identifying the intermediates in the mechanism

Intermediates are species that are produced in one step of the reaction mechanism and consumed in a subsequent step. In this mechanism, we can see that HI (hydrogen iodide) is formed in the first step and consumed in the second step. Therefore, HI is an intermediate in this mechanism.

Determining the rate law of the overall reaction

The rate law for a reaction can be determined based on the slowest (rate-determining) step in the reaction mechanism. Since we are told that the first step is slow and the second step is fast, the rate law for the overall reaction will be determined by the first step. The first step is: \(\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)\) For an elementary reaction, the rate law can be directly inferred from the reaction equation. In this case, the rate law for the first step will be: \(rate = k[\mathrm{H}_{2}][\mathrm{ICl}]\) Since the rate law for the overall reaction is determined by the slowest step (the first step), the rate law for the overall reaction will be the same as the rate law for the first step: \(rate = k[\mathrm{H}_{2}][\mathrm{ICl}]\)

Key concepts

Rate Law

Understanding the rate law is crucial when studying chemical kinetics, as it expresses the relationship between the rate of a chemical reaction and the concentration of the reactants. It can be represented by an equation of the form:
\(rate = k[\text{A}]^m[\text{B}]^n\), where:

• \(k\) is the rate constant,
• \([\text{A}]\) and \([\text{B}]\) are the molar concentrations of reactants A and B,
• \(m\) and \(n\) are the orders of the reaction with respect to A and B.

In the exercise's proposed mechanism, since the first step is the slowest, it determines the overall rate law of the entire reaction. This step-wise approach implies that the concentration of the intermediate, HI, doesn't appear in the rate law because it is formed and consumed over the course of the reaction and the rate-determining step doesn't include it. For reactions involving several steps, only the rate-determining (slowest) step affects the overall rate law, as exemplified by the provided exercise.

Reaction Intermediates

Reaction intermediates are species that appear in the middle of a chemical reaction mechanism: they are formed in one step and consumed in another. Intermediates are not the same as transition states; they have a real, albeit transient, existence, whereas transition states are theoretical constructs representing the point of highest energy along the reaction coordinate.
In the exercise solution, hydrogen iodide (HI) is identified as an intermediate. It's created in the first step of the mechanism and used up in the second step. These intermediates are vital for understanding mechanisms because they reveal the step-by-step process through which reactants are converted into products. Notably, intermediates do not appear in the balanced chemical equation for the overall reaction, since they are not present at the start or end of the reaction.

Balanced Chemical Equations

Balanced chemical equations are the bedrock of stoichiometry, illustrating the principle of conservation of mass: matter cannot be created or destroyed in a chemical reaction. To achieve a balanced equation, there must be the same number of atoms for each element on both sides.
In our exercise, we balance the overall equation by ensuring that the number of atoms of each element is equal before and after the reaction. This provides a clear and quantitative description of the reaction, which is essential for calculating reactant and product quantities.
It is important to note that in multi-step reactions, like the one in the exercise, balancing is done for the overall reaction, not for individual steps. However, understanding each step is vital for grasping the overall reaction mechanism, which reveals not only what reacts and what is produced but also how the reaction proceeds.