Problem 102 - Chemistry The Science in Context

Question

A proposed mechanism for the decomposition of hydrogen peroxide consists of three elementary steps: \(\begin{array}{ll}\text { (1) } & \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{OH}(g)\end{array}\) (2) \(\quad \mathrm{H}_{2} \mathrm{O}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HO}_{2}(g)\) (3) \(\quad \mathrm{HO}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)\) If the rate law for the reaction is first order in \(\mathrm{H}_{2} \mathrm{O}_{2},\) which step in the mechanism is the rate-determining step?

Step-by-Step Solution

Answer

The rate-determining step in this mechanism is Step 1 (\(\mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{OH}(g)\)), as its rate equation (Rate\(_1 = k_1 [\mathrm{H}_2\mathrm{O}_2]\)) matches the given rate law (Rate\(_{reaction} = k[\mathrm{H}_2\mathrm{O}_2]\)).

Step 1:

For the first step, the reaction is: \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{OH}(g)\). The rate equation for this step can be written as: Rate\(_1 = k_1 [\mathrm{H}_2\mathrm{O}_2]\).

Step 2:

For the second step, the reaction is: \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HO}_{2}(g)\). The rate equation for this step can be written as: Rate\(_2 = k_2 [\mathrm{H}_2\mathrm{O}_2][\mathrm{OH}]\).

Step 3:

For the third step, the reaction is: \(\mathrm{HO}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)\). The rate equation for this step can be written as: Rate\(_3 = k_3 [\mathrm{HO}_2][\mathrm{OH}]\). 2. Identifying the rate-determining step

Step 4: Rate law comparison:

We are given that the rate law for the reaction is first order in \(\mathrm{H}_{2}\mathrm{O}_{2}\): Rate\(_{reaction} = k[\mathrm{H}_2\mathrm{O}_2]\). Comparing this with each of our derived rate equations, we can see that the rate equation for Step 1 matches the given rate law: Rate\(_1 = k_1 [\mathrm{H}_2\mathrm{O}_2]\).

Step 5: Rate-determining step:

As the rate equation for Step 1 matches the given rate law, we can conclude that the rate-determining step in this mechanism is Step 1.

Key Concepts

Rate-Determining Step

In any reaction mechanism, the rate-determining step is crucial because it acts as a bottleneck for the reaction. Imagine a narrow bridge on a busy highway; no matter how fast the cars move before or after, they all must slow down at the bridge. Similarly, the rate-determining step is the slowest step in a reaction mechanism, controlling the overall reaction rate. This step determines the speed at which the entire sequence of reactions proceeds.
The rate-determining step often has the highest energy barrier to overcome compared to other elementary steps. Therefore, it limits the rate at which products form. If you want to understand a reaction's overall speed, pay careful attention to this step. Specifically, in the decomposition of hydrogen peroxide, the first elementary step, where \({H}_2 ext{O}_2(g) ightarrow 2 ext{OH}(g)\) occurs at a pace that defines the reaction's speed.
It matches the rate law given for the entire reaction, indicating that this first step is indeed rate-determining. By identifying this, chemists can focus on finding catalysts or conditions that help overcome this bottleneck, thereby increasing the reaction rate.

Elementary Steps

Elementary steps are the individual stages of a reaction mechanism. Each step represents a single molecular event and provides insight into the specifics of how reactants turn into products.
Unlike an overall chemical equation, which offers a broad overview, elementary steps break down each detailed interaction, such as bond formations and breaks, at a molecular level. Imagine them as the scenes in a film, each contributing to the full story.
In the decomposition mechanism of hydrogen peroxide, we see three elementary steps:

• First, the breakdown of \(\text{H}_2\text{O}_2\) into hydroxyl radicals \(\text{OH}\).
• Then, \(\text{H}_2\text{O}_2\) reacts with \(\text{OH}\) to produce \(\text{H}_2\text{O}\) and hydroperoxyl \(\text{HO}_2\)
• Finally, \(\text{HO}_2\) and \(\text{OH}\) react to form water and oxygen \(\text{O}_2\).

Understanding each elementary step is crucial for predicting reaction mechanisms and tailoring conditions for desired outcomes. It also helps explain the overall stoichiometry and energy changes occurring during the reaction.

Rate Law

The rate law is an expression that links the reaction rate with the concentrations of reactants. It tells you how fast a reaction proceeds given the amounts of reactants present.
Mathematically, for a simple reaction, the rate law can be expressed as \(\text{Rate} = k [\text{Reactant}]^n\), where \(k\) is the rate constant, \(\text{Reactant}\) is the concentration of the reactant, and \(n\) is the reaction order. The reaction order is an experimentally determined value, often indicating the number of molecules that come together to react in the rate-determining step.
For the hydrogen peroxide decomposition, we’re told it's first-order in \(\text{H}_2\text{O}_2\), expressed as \(\text{Rate} = k[\text{H}_2\text{O}_2]\). This implies that the reaction rate is directly proportional to the concentration of hydrogen peroxide. Understanding and determining the rate law helps chemists develop models to predict reaction behavior, optimize conditions for industrial processes, and design inhibitors or catalysts effectively.
Moreover, by comparing the experimental rate law with derived rate laws for each elementary step, you can pinpoint which step controls the overall reaction rate, aligning perfectly with identifying the rate-determining step.