Problem 72 - Chemistry The Central Science In Si Units

Question

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{IO}^{-}(a q) \quad(\text { slow }) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \end{aligned} $$ (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) \( 2\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \). (b) \( \mathrm{IO}^{-} \). (c) \( \text{Rate} = k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}] \).

Step 1: Combine the reactions

To find the chemical equation for the overall process, we must sum all the steps of the mechanism. In this case, we add the two given steps. The first step is: \( \mathrm{H}_{2} \mathrm{O}_{2}(aq)+\mathrm{I}^{-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(aq) \), and the second step is: \( \mathrm{IO}^{-}(aq)+\mathrm{H}_{2} \mathrm{O}_{2}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(aq) \). By adding these equations, we cancel \( \mathrm{IO}^{-}(aq) \) and \( \mathrm{I}^{-}(aq) \), resulting in the overall reaction: \( 2\mathrm{H}_{2} \mathrm{O}_{2}(aq) \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \).

Step 2: Identify the intermediate

Intermediates are species that are produced in one step of a reaction mechanism and consumed in a subsequent step. From the given mechanism, \( \mathrm{IO}^{-}(aq) \) is generated in the first step and consumed in the second step. Therefore, \( \mathrm{IO}^{-} \) is the intermediate.

Step 3: Derive the rate law

The rate law is determined by the rate-determining step (slow step) of the mechanism. The first step is slow: \( \mathrm{H}_{2} \mathrm{O}_{2}(aq) + \mathrm{I}^{-}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{IO}^{-}(aq) \). The rate law is formulated from the reactants of this step: \( \text{Rate} = k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}] \), where \( k \) is the rate constant for the reaction.

Key Concepts

Reaction Mechanism

In chemical kinetics, a reaction mechanism is a step-by-step description of the pathway from reactants to products. It breaks down the complex transformation into a series of simpler, elementary steps. Each step shows the molecular interactions leading to the overall chemical change. Understanding a reaction mechanism helps chemists predict the behavior of reactions under different conditions.

• For instance, the decomposition of hydrogen peroxide catalyzed by iodide ion occurs via two steps.
• The first step is a slow combination of hydrogen peroxide and iodide ion, producing water and the intermediate iodide ion (\( \mathrm{IO}^- \)).
• The second step is faster and involves the reaction of \( \mathrm{IO}^- \) with more hydrogen peroxide, yielding water, oxygen gas, and regenerating iodide ion.

By summing the steps, we eliminate intermediates and determine the overall reaction. Understanding each step provides insights into the speed and control of reactions.

Rate Law

The rate law is an expression that relates the rate of a chemical reaction to the concentration of its reactants. It provides important information about the speed of the reaction and its reliance on reactant concentrations. The form of the rate law is determined by the reaction's slowest step, often called the rate-determining step.
For the catalyzed decomposition of hydrogen peroxide, the slow step involves hydrogen peroxide and iodide ion:

• The rate law based on this step is expressed as \( \text{Rate} = k[\mathrm{H}_{2} \mathrm{O}_{2}][\mathrm{I}^{-}] \), where \( k \) is the rate constant.
• This equation indicates the reaction rate depends on the concentration of hydrogen peroxide and iodide ion.
• The order of the reaction with respect to each reactant is determined by their respective exponents. In this case, both are first order.

Understanding the rate law helps control reaction conditions to optimize product formation.

Catalysis

Catalysis involves the acceleration of a chemical reaction by a substance called a catalyst. Catalysts function by providing an alternative pathway with a lower activation energy, increasing the rate of reaction without being consumed. The catalyst is involved in the mechanism but appears unchanged in the overall reaction.

• In our example, iodide ion acts as a catalyst in the decomposition of hydrogen peroxide.
• It facilitates the reaction by forming an intermediate that reacts further, regenerating the catalyst itself.
• This results in a more efficient reaction that proceeds faster than the uncatalyzed version.

By using catalysts, chemists can increase the efficiency of chemical processes and make them more environmentally friendly by reducing necessary energy input.

Intermediates

Intermediates are species that are formed during one step of a reaction mechanism and consumed in a subsequent step. They are crucial to understanding how a reaction proceeds through its mechanism. Intermediates do not appear in the overall balanced equation for the reaction as they are not present in the final products.
In the decomposition of hydrogen peroxide catalyzed by iodide, the intermediate is

• \( \mathrm{IO}^- \), which is formed in the first step from hydrogen peroxide and iodide ion.
• This intermediate then reacts in the second step with more hydrogen peroxide to form the final products.

Identifying intermediates provides insight into the detailed pathways of a reaction. Knowing about intermediates assists in grasping the full picture of how reactants transform into products as the mechanism progresses.