Question

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

Short answer

The proposed mechanism is consistent with the rate law: \(\text{rate} = k [\text{H}_2] [\text{I}_2]\).

Step-by-step solution

- Identify the Rate-Determining Step

The rate-determining step (RDS) is the slowest step and determines the rate law of the overall reaction. Here, Step 3: \(\text{H}_2\text{I}(g) + \text{I}(g) \rightarrow 2 \text{HI}(g)\) is given as the slow step.

- Write the Rate Law for the RDS

Write the rate law based on the rate-determining step: \( \text{rate} = k \text{[H}_2\text{I] [I]} \).

- Express Intermediates in Steady-State Approximation

Identify intermediates that are not in the overall reaction, like \(\text{H}_2\text{I}\) and \(\text{I}\). Use the steady-state approximation for them to link their concentrations to the reactants.

- Apply Steady-State to Intermediate I

For the intermediate \(\text{I}\), set up a steady-state condition: \(0 = k_1 [\text{I}_2] - k_{-1} [\text{I}]^2 - k_2 [\text{H}_2][\text{I}] + k_{-2} [\text{H}_2 \text{I}]\).

- Simplify Steady-State Expression for I

Solve the steady-state condition to express \([\text{I}]\) in terms of \([ \text{I}_2 ]\) and \([ \text{H}_2]\), approximating that reactions involving iodine atoms are fast.

- Apply Steady-State to Intermediate H2I

For intermediate \(\text{H}_2\text{I}\), set up steady-state condition: \(0 = k_2 [\text{H}_2][\text{I}] - k_{-2} [\text{H}_2 \text{I}] - k_3 [\text{H}_2\text{I}] [\text{I}]\).

- Solve for Intermediate H2I Concentration

Solve for \([\text{H}_2 \text{I}]\) in terms of \( [\text{H}_2] \) and \( [\text{I}_2]\). Use steady-state results for \([\text{I}]\) from step 5.

- Substitute Intermediate Concentrations into Rate Law

Substitute expressions for \([\text{I}]\) and \([\text{H}_2\text{I}]\) from steady-state approximations into the rate law from Step 2.

- Simplify the Final Rate Law

Simplify the resulting expression to show that the rate law is in the form \( k' [\text{H}_2] [\text{I}_2]\), where \(k'\) is a combined constant.

Key concepts

Rate-Determining Step

The rate-determining step (RDS) is the slowest step in a reaction mechanism and it defines the overall reaction rate. Think of it as the bottleneck in a process. In our example, Step 3: \(\text{H}_2\text{I}(g) + \text{I}(g) \rightarrow 2 \text{HI}(g)\) is identified as the RDS because it is the slowest. This step limits the overall speed of the reaction.

When we focus on the RDS, we can write a rate law that shows how the concentrations of different reactants affect the overall reaction rate. For the given step, the rate law is: \(\text{rate} = k \text{[H}_2\text{I] [I]}\). This tells us that the rate depends on the concentrations of both \(\text{H}_2\text{I}\) and \(\text{I}\).

Steady-State Approximation

In chemistry, the steady-state approximation simplifies the analysis of complex reactions involving intermediate species. Here, we assume that the concentration of intermediates remains almost constant over the course of the reaction. This means the rate of formation of intermediates equals their rate of consumption.

For the intermediate \(\text{I}\) atoms in our reaction, we establish: \0 = k_1 [\text{I}_2] - k_{-1} [\text{I}]^2 - k_2 [\text{H}_2][\text{I}] + k_{-2} [\text{H}_2 \text{I}]\. Similarly, for \(\text{H}_2\text{I}\), the condition is: \0 = k_2 [\text{H}_2][\text{I}] - k_{-2} [\text{H}_2 \text{I}] - k_3 [\text{H}_2\text{I}] [\text{I}]\. Using these approximations helps us express intermediate concentrations in terms of reactants like \(\text{H}_2\) and \(\text{I}_2\).

Intermediate Species

Intermediates are molecules or atoms formed temporarily in a multi-step reaction but are not present in the final product. In the provided reaction, \(\text{I}\) and \(\text{H}_2\text{I}\) are intermediates.

These intermediates play crucial roles in the reaction pathway. They are created and consumed within the reaction steps, following certain kinetics. For the step involving \(\text{I}\), its formation and consumption are linked to other reactions involving \(\text{I}_2\) and \(\text{H}_2\). Expressing these intermediates correctly through steady-state approximations helps us understand the detailed reaction mechanism.

Rate Law

The rate law expresses how the reaction rate depends on the concentration of reactants. For the given mechanism, the overall rate law is determined by the RDS (Step 3): \(\text{rate} = k \text{[H}_2\text{I] [I]}\).

To simplify this, we substitute the concentrations of intermediates using steady-state approximations. By solving for \([\text{I}]\) and \([\text{H}_2\text{I}]\), we can rewrite the rate law in terms of \([\text{H}_2]\) and \([\text{I}_2]\). This yields a final form like \(\text{rate} = k' [\text{H}_2] [\text{I}_2]\), consistent with the original rate law.

Bimolecular Reaction

A bimolecular reaction involves two reactant molecules colliding and reacting with each other. It is the most common type of elementary reaction. In the provided mechanism, each step involves such interactions.

For example, in Step 1: \(\text{I}_2(g) \rightleftharpoons 2 \text{I}(g)\), two iodine atoms form from one iodine molecule. Another bimolecular step, Step 2: \(\text{H}_2(g) + \text{I}(g) \rightleftharpoons \text{H}_2\text{I}(g)\), involves a collision between \(\text{H}_2\) and \(\text{I}\). Understanding these bimolecular steps helps in grasping how rate laws are formed based on collision theory.