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

Expert verified
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

01

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#
02

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#
03

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#
04

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#
05

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#
06

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#
07

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#
08

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#
09

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#
10

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.

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Most popular questions from this chapter

The activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)

Consider the following reaction: $$\mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q)$$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-} .\) When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at 298 \(\mathrm{K}\) is 0.0432 \(\mathrm{M} / \mathrm{s}\) . (a) What is the value of the rate constant? (\mathbf{b} )What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH \(^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{B}]^{2}\) . (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B} ?\) What is the overall reaction order? (c) What are the units of the rate constant?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2} .\) The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{s}^{-1} .\) Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from 1.00 \(\mathrm{L}\) of 0.600 \(\mathrm{MN}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of 20.0 \(\mathrm{h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(C l_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

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