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Chapter 14: Problem 80

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$\begin{array}{c}{\mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g)} \\ {2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)}\end{array}$$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(N_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

Short Answer

Expert verified
a) The overall chemical equation for the reaction can be obtained by adding the two elementary steps and canceling out any species that appear on both sides: N2O(g) -> N2(g) + O2(g). b) NO is a catalyst in this reaction, as it is consumed in the first elementary step and regenerated in the second elementary step, remaining unchanged. c) The absence of accumulating NO2 during the decomposition of N2O in experiments does not rule out the proposed mechanism. It could mean that NO2 is formed and then quickly converted back to NO and O2 in the second elementary step.

Step by step solution

01

a) Overall chemical equation

To find the overall chemical equation, we first need to add the two given elementary steps: 1) NO(g) + N2O(g) -> N2(g) + NO2(g) 2) 2 NO2(g) -> 2 NO(g) + O2(g) Next, we'll cancel any species that appear on both sides of the reaction equation. We notice that 2 NO2(g) is formed in step 1 and completely consumed in step 2, so it will be canceled out. Similarly, NO is consumed in step 1 and regenerated in step 2, so it will also cancel out: Overall reaction: N2O(g) -> N2(g) + O2(g)
02

b) NO as a catalyst or an intermediate

To determine whether NO is a catalyst or an intermediate, we need to check if it is consumed and regenerated during the course of the reaction. From our previous step, we can see that NO is consumed in the first elementary step and regenerated in the second elementary step. Since NO is regenerated during the reaction and is not consumed, it remains unchanged. Therefore, NO is serving as a catalyst in this reaction.
03

c) Accumulation of NO2 and the validity of the proposed mechanism

The absence of accumulating NO2 during the decomposition of N2O in experiments does not rule out the proposed mechanism. The proposed mechanism suggests that NO2 is formed and then quickly converted back to NO and O2 in the second elementary step. The rapid conversion of NO2 could explain why it does not accumulate in measurable quantities during the experiment.

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Most popular questions from this chapter

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } 2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)} \\ {\mathrm{CH}_{2}} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)} \\ {\text { (c) } \mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)}\end{array}\)

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?

A flask is charged with 0.100 mol of A and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(A(g) \longrightarrow \mathrm{B}(g) .\) The following data are collected:(a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 s interval in units of mol/s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?
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