Problem 71
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
Verified Answer
(a) \(\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\).
(b) Intermediate: \(\mathrm{HI}(g)\).
(c) Rate law: \(\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\).
1Step 1: Write the Balanced Equation for the Overall Reaction
First, sum the two given elementary steps to find the overall reaction. The intermediate species produced in the first step and consumed in the second are canceled out. Thus, the intermediate is HI. Sum the two reactions:\[\begin{align*} \text{Step 1:} & \ \mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g) \ \text{Step 2:} & \ \mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g) \end{align*}\]Cancellation of the intermediate HI results in the overall reaction:\[\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\]This is the balanced equation for the overall reaction.
2Step 2: Identify Intermediate
An intermediate is a species that is produced in one step of the mechanism and consumed in another. From the given mechanism, \(\mathrm{HI}(g)\) is formed in the first step and consumed in the second, so \(\mathrm{HI}(g)\) is the intermediate species in the reaction mechanism.
3Step 3: Determine the Rate Law for the Overall Reaction
The overall rate law depends on the rate-determining step, which is the slowest step in the mechanism. Given that the first step is slow, this will determine the overall rate. The rate law for the slow step is based on its reactants. For step 1:\[\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\]The rate law is:\[\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\]This is the expected rate law for the overall reaction.
Key Concepts
Intermediates in ReactionsRate Law DeterminationElementary Steps in Reactions
Intermediates in Reactions
In chemical reaction mechanisms, intermediates play a crucial role. These are species that form in one step of a mechanism and are consumed in another. They are neither reactants nor final products; instead, they exist temporarily as the reaction occurs. Intermediates can sometimes be tricky to identify because they do not appear in the overall balanced chemical equation. For example, in the reaction mechanism given,
- The intermediate identified is \(\mathrm{HI}(g)\), which is produced in the first reaction step and consumed in the second.
- This substance is key to understanding the progress of the reaction at the molecular level, but it does not appear in the final overall reaction equation.
Rate Law Determination
Understanding rate law determination is a key aspect of analyzing reaction mechanisms. The rate law gives insight into which molecules are involved in the slowest, or rate-determining step, of the mechanism.For the reaction \(\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\),
- The overall rate is determined by the slowest step, which in many complex reactions is the first step.
- In the given mechanism, since the first step is slower, it dictates the reaction’s rate.
- Consequently, the rate law can be written using the reactants from this first step.
Elementary Steps in Reactions
Elementary steps in a reaction provide a detailed view of the mechanism by dissecting larger complex reactions into simpler, individual stages. Each step represents a single molecular event, such as the collision between reactant molecules, that must occur for the reaction to proceed.
- In the given problem, there are two elementary steps:
- The first step: \(\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\), which is the slow step.
- The second step: \(\mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g)\), which is faster.
Other exercises in this chapter
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