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Chapter 14: Problem 96

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

Short Answer

Expert verified
In summary: 1) The overall reaction obtained from the proposed mechanism is 2NO(g) + H2(g) 鈫� N2O(g) + H2O(g), which is balanced. 2) The rate laws for each elementary reaction are: Rate 1 = k鈧乕NO]虏 and Rate 2 = k鈧俒N2O2][H2]. 3) The intermediate species in this mechanism is N2O2(g). 4) Comparing the observed rate law (rate = k[NO]虏[H2]) with the elementary reactions' rate laws, we can conclude that the second elementary reaction (involving N2O2 and H2) is much faster than the first elementary reaction involving two NO molecules. The first reaction, involving NO molecules, is the rate-determining (slow) step in this reaction mechanism.

Step by step solution

01

Overall Reaction

To find the overall reaction, let's add the two elementary reactions: 1) NO(g) + NO(g) 鈫� N2O2(g) 2) N2O2(g) + H2(g) 鈫� N2O(g) + H2O(g) Adding reactions 1 and 2: NO(g) + NO(g) + N2O2(g) + H2(g) 鈫� N2O2(g) + N2O(g) + H2O(g) Now, we can cancel the intermediate N2O2 from both sides of the equation, resulting in: NO(g) + NO(g) + H2(g) 鈫� N2O(g) + H2O(g) We can simplify it further into: 2NO(g) + H2(g) 鈫� N2O(g) + H2O(g) The overall reaction is a balanced equation, and thus, the mechanism is consistent.
02

Writing Rate Laws

For each elementary step, we need to write its corresponding rate law: For reaction 1: NO(g) + NO(g) 鈫� N2O2(g) Rate 1 = k鈧乕NO]虏 For reaction 2: N2O2(g) + H2(g) 鈫� N2O(g) + H2O(g) Rate 2 = k鈧俒N2O2][H2]
03

Identify Intermediates

Intermediates are species that appear in the mechanism but not in the overall reaction. For this exercise, we have only one intermediate species: Intermediate: N2O2(g)
04

Compare Rate Laws & Determine Reaction Speeds

Based on the given observed rate law: \(rate = k[NO]^2[H2]\) Comparing the observed rate law with the rate laws obtained from the elementary steps, we can observe that Rate 1=k鈧乕NO]虏 contributed to the part [NO]虏 in the observed rate law. Then, we also have Rate 2=k鈧俒N2O2][H2], which contributed the [H2] part in the observed rate law. One key point is that the reaction intermediate N2O2 does not show up in the observed rate law. What this suggests is that the second elementary reaction (involving N2O2 and H2) is much faster than the first elementary reaction involving two NO molecules. It means that the concentration of N2O2 does not significantly build up in the reaction mechanism, which in turn implies that the first reaction (involving NO molecules) is the rate-determining (slow) step in this reaction 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) \(125^{\circ} \mathrm{C}\) ?

The rate of disappearance of \(\mathrm{HCl}\) was measured for the following reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The following data were collected:\begin{tabular}{rl} \hline Time (min) & [HCl] (M) \\ \hline \(0.0\) & \(1.85\) \\ \(54.0\) & \(1.58\) \\ \(107.0\) & \(1.36\) \\ \(215.0\) & \(1.02\) \\ \(430.0\) & \(0.580\) \\ \hline \end{tabular} (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Graph [HCl] versus time, and determine the instantaneous rates in \(M / \mathrm{min}\) and \(\mathrm{M} / \mathrm{s}\) at \(t=75.0 \mathrm{~min}\) and \(t=250 \mathrm{~min}\).

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{gathered} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{gathered} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? How do you know? (c) What is the intermediate in the reaction? How do you distinguish it from the catalyst?

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

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form B according to the hypothetical gas-phase reaction \(\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})\). The following data are collected: \begin{tabular}{lccccc} \hline Time (s) & 0 & 40 & 80 & 120 & 160 \\ \hline Moles of A & \(0.100\) & \(0.067\) & \(0.045\) & \(0.030\) & \(0.020\) \\ \hline \end{tabular} (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 40 -s interval, in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?
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