Problem 72 - Chemistry: The Central Science in SI Units, Global Edition

Question

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)(\text { slow }) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad(\text { fast }) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

Step-by-Step Solution

Answer

(a) The overall chemical equation for the decomposition of hydrogen peroxide catalyzed by iodide ions is: \(\mathrm{2H}_{2} \mathrm{O}_{2}(a q) + \mathrm{I}^{-}(a q) \longrightarrow \mathrm{2H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g) + \mathrm{I}^{-}(a q)\) (b) The intermediate in the mechanism is \(\mathrm{IO}^{-}(a q)\). (c) The rate law for the overall process, assuming the first step is rate determining, is: Rate \(= k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\).

Step 1: 1. Writing the chemical equation for the overall process

To find the overall chemical equation, we can add the two given equations and then cancel out any species that appear on both sides of the equation. First, let's write down the two equations provided: (i) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) (ii) \(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q)\)

Step 2: 2. Identifying the intermediate

An intermediate is a species that is produced in one step of a mechanism and consumed in another step. In this case, \(\mathrm{IO}^{-}(a q)\) is produced in step (i) and consumed in step (ii), so it is the intermediate.

Step 3: 3. Predicting the rate law for the overall process

Since the first step of the mechanism is rate determining (slow), the rate law for the overall process will be determined by the rate law of the first step. In the first step, we have: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) The rate law for this elementary reaction is given by: Rate \(= k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\) where k is the rate constant, and the concentration of each reactant is raised to the power of its stoichiometric coefficient (which is 1 for both reactants in this case). So, the rate law for the overall process is given by: Rate \(= k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\).

Key Concepts

Reaction Mechanism

In chemical kinetics, a reaction mechanism provides insight into the step-by-step transformations that occur during a reaction. It breaks down the overall chemical reaction into a series of simpler reactions or steps.

These individual steps are known as elementary reactions. The decomposition of hydrogen peroxide catalyzed by iodide ion is an example of such a process that proceeds in a stepwise manner. The reaction involves two steps:

• The first step is the slow formation of water and hypoiodite ion (\(\text{IO}^{-}\)).
• The second step, which is faster, reacts \(\text{IO}^{-}\) with another molecule of hydrogen peroxide, resulting in water, oxygen gas, and regeneration of \(\text{I}^{-}\).

This breakdown helps us understand how bonds are broken and formed, and how energy is transferred throughout the process. Identifying the slowest step as a rate-determining step is crucial for understanding the reaction's kinetics, which we'll explore more in the rate law section.

Rate Law

The rate law is a mathematical expression that predicts how the rate of a chemical reaction is affected by the concentration of its reactants. It is derived from the reaction mechanism. For our hydrogen peroxide decomposition, the rate law focuses on the rate-determining step, which is the pace controlling how fast the overall reaction can occur.

Since the first step of our mechanism is slow, it dictates the rate for the entire process. Thus, the rate law is determined by this step alone.

• The reaction for this step: \[\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\]
• This leads to the rate law: \[\text{Rate} = k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}]\]

Here, \(k\) is the rate constant, which is specific to the reaction at a given temperature, and the concentrations of hydrogen peroxide and iodide ion define how frequently they collide to form products.Understanding rate laws helps chemists control reactions and optimize conditions for desired results.

Intermediates

Intermediates are transient species that appear in the sequence of reaction steps, forming in one step and being used up in another. They are important in connecting various elementary steps of a reaction mechanism.

For the decomposition of hydrogen peroxide, we identify \(\mathrm{IO}^{-}(aq)\) as an intermediate. It is generated during the slow first step:

• \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\)

Then, \(\mathrm{IO}^{-}(aq)\) quickly reacts in the second step to regenerate \(\mathrm{I}^{-}(aq)\) and produce oxygen gas:

• \(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q)\)

Understanding intermediates gives us a deeper comprehension of the reaction process and how molecules transform over a reaction pathway. This insight can guide changes to reaction conditions in order to favor particular pathways and optimize reaction efficiency.