Question

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. \(\begin{array}{ll}{\text { Step } 1 :} & {\mathrm{O}_{3}(g) \Longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g) \text { (fast) }} \\ {\text { Step } 2 :} & {\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(g) \quad(\text { slow })}\end{array}\) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is O a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Short answer

The overall balanced equation for the gas-phase decomposition of ozone is \(2 O_3(g) \rightarrow 3 O_2(g)\). The rate law consistent with the given mechanism is \(\text{rate} = k_1[O_3]^2\). O is an intermediate in this reaction mechanism. If the reaction occurred in a single step, the rate law would still be \(\text{rate} = k[O_3]^2\).

Step-by-step solution

Add the two steps together

To obtain the balanced equation for the overall reaction, we first add the two steps together and then simplify: Step 1: \(O_3(g) \rightarrow O_2(g) + O(g) \) (fast) Step 2: \(O(g) + O_3(g) \rightarrow 2 O_2(g) \) (slow) #a) Balanced Equation for the Overall Reaction#

Simplify the combined reactions

Adding both steps, we get: \[O_3(g) + O(g) + O_3(g) \rightarrow O_2(g) + O(g) + 2 O_2(g)\] On simplifying, the \(O(g)\) on both sides cancel out, and we are left with the following balanced equation for the overall reaction: \[2 O_3(g) \rightarrow 3 O_2(g)\] #b) Derive the Rate Law#

Determine the reaction rate of the slow step

The reaction rate of the slow step (Step 2) determines the overall reaction rate. So, now we can write the rate law for the reaction based on the slow step: \[\text{rate} = k_2[O][O_3]\] #b) Derive the Rate Law#

Use the steady-state approximation for O:

To derive the complete rate law, we need to take into account the concentration of O, which is an intermediate. To do that, we use the steady-state approximation for O. This means that the rate of change of the concentration of O is zero, and we can set the formation and consumption rate of O equal: For the first step of the reaction: \[\mathrm{\frac{d[O]}{dt}} = k_1[O_3] - k_2[O][O_3] = 0\] #b) Derive the Rate Law#

Solve for O concentration

Now we can solve for the concentration of O: \[[O] = \frac{k_1}{k_2} [O_3]\] #b) Derive the Rate Law#

Substitute for O in the overall rate law

Now, replace the concentration of O in the rate equation from the slow step with the result from step 3: \[\text{rate} = k_2 \left(\frac{k_1}{k_2} [O_3]\right)[O_3] = k_1[O_3]^2\] #b) Derive the Rate Law#

Write the final rate law

The final rate law consistent with the given mechanism is: \[\text{rate} = k_1[O_3]^2\] #c) O as a Catalyst or Intermediate#

Identify the role of O in the reaction

O is an intermediate, as it is produced in one step (Step 1) and consumed in the subsequent step (Step 2) of the reaction. Catalysts, on the other hand, are not consumed or produced in the reaction mechanism and remain unchanged. #d) Rate Law for a Single-Step Reaction#

Write the overall balanced equation

As we previously found, the overall balanced equation for the reaction is: \[2 O_3(g) \rightarrow 3 O_2(g)\] #d) Rate Law for a Single-Step Reaction#

Write the rate law for a single-step reaction

Assuming the reaction occurs in a single step, with an unknown rate constant k, we can write the rate law as follows: \[\text{rate} = k[O_3]^2\] In this case, the rate law would be the same as in the previous (two-step) mechanism.

Key concepts

Reaction Mechanism

When understanding the decomposition of ozone ( O_3 ), the reaction mechanism is essential. It breaks down a complex reaction into simpler steps that occur. For ozone, the process happens in two steps:

• Step 1 is fast: \( O_3(g) \longrightarrow O_2(g) + O(g) \)
• Step 2 is slow: \( O(g) + O_3(g) \longrightarrow 2 O_2(g) \)

These steps tell us how ozone decomposes, starting with the breaking of an ozone molecule to form an oxygen molecule and an oxygen atom. The slow step is crucial as it determines the rate of the entire reaction. This detailed breakdown helps scientists understand what's happening at a molecular level, and it's key to deriving important equations, like the rate law.

Rate Law

The rate law gives us a formula to calculate how fast a reaction occurs. It's based on the mechanism's slowest step, which is often the bottleneck. For ozone decomposition, it's step 2: \( O(g) + O_3(g) \longrightarrow 2 O_2(g) \). Here, we can write the rate law as:\[\text{rate} = k_2[O][O_3]\]However, since oxygen atoms are intermediates, we use the steady-state approximation to express its concentration in terms of the stable reactants. Solving the equations from the mechanism, we get:\[[O] = \frac{k_1}{k_2} [O_3]\]Substituting this back, we find the final rate law for the reaction: \[\text{rate} = k_1[O_3]^2\]This equation shows how the concentration of ozone affects the rate, with k_1 being a constant that includes multiple contributing factors. By understanding the mechanism, deriving the rate law becomes straightforward.

Intermediate in Reactions

Intermediates are species formed in one step of a mechanism and consumed in another. In the decomposition of ozone, O (an oxygen atom) acts as an intermediate. It appears between the two steps: it's a product in step 1 and a reactant in step 2. Unlike catalysts, intermediates do not reappear in the final product.
Understanding intermediates is crucial because:

• They don't show up in the overall balanced equation.
• Their presence heavily influences the derived rate laws.

By identifying intermediates, chemists can better understand reaction pathways and optimize conditions for desired outcomes. This is why the step-by-step analysis of mechanisms is so valuable in fields like environmental chemistry, where reactions like ozone decomposition play a vital role.