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Chapter 14: Problem 107

The following mechanism has been proposed for the reaction of NO with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and $\mathrm{H}_{2} \mathrm{O} :$ $$\mathrm{NO}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)$ $\mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (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
The balanced overall reaction for the proposed mechanism is: NO(g) + NO(g) + H2(g) 鈫� N2O(g) + H2O(g). The rate laws for the elementary reactions are \(rate_1 = k_1[NO]^2\) and \(rate_2 = k_2[N2O2][H2]\). N2O2 is an intermediate in the mechanism. Comparing the observed rate law (\(rate = k[NO]^2[H2]\)) with the elementary reactions, we can conclude that the first reaction is fast and the second reaction is slow, with the second reaction determining the overall reaction rate.

Step by step solution

01

Add the two elementary reactions

Add the given elementary reactions: NO(g) + NO(g) 鈫� N2O2(g) N2O2(g) + H2(g) 鈫� N2O(g) + H2O(g) Now, we can add these two reactions together.
02

Combine and simplify

Combine the equations and cancel any species that appear on both sides: NO(g) + NO(g) + N2O2(g) + H2(g) 鈫� N2O2(g) + N2O(g) + H2O(g) Since N2O2 appears on both sides, it can be cancelled: NO(g) + NO(g) + H2(g) 鈫� N2O(g) + H2O(g) Now we have a balanced overall reaction. #b) Write a rate law for each elementary reaction#
03

Write rate laws for the elementary reactions

The rate law for each reaction depends on the order of the reactants. Since both elementary reactions are bimolecular, their rate laws will be: Reaction 1: \(rate_1 = k_1[NO]^2\) Reaction 2: \(rate_2 = k_2[N2O2][H2]\) Where \(k_1\) and \(k_2\) are the rate constants for the reactions. #c) Identify any intermediates#
04

Determine the intermediates

An intermediate is a species that is produced in one elementary reaction and then consumed in a subsequent elementary reaction. In this case, N2O2 is an intermediate because it's produced in Reaction 1 and consumed in Reaction 2. #d) Compare observed rate law with the mechanism#
05

Analyze the observed rate law

The observed rate law is given as: \(rate = k[NO]^2[H2]\)
06

Compare with the elementary reactions

Compare the observed rate law with the elementary reaction rate laws: Observed: \(rate = k[NO]^2[H2]\) Reaction 1: \(rate_1 = k_1[NO]^2\) Reaction 2: \(rate_2 = k_2[N2O2][H2]\) It's clear that the observed rate law matches closely with the rate laws of both elementary reactions.
07

Conclusion

Based on the comparison of the observed rate law and the elementary rate laws, we can conclude that the first reaction is fast (forming the N2O2 intermediate) and the second reaction is slow (consuming the intermediate and determining the reaction rate). This means that the overall reaction rate depends mostly on the second reaction.

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Most popular questions from this chapter

The following mechanism has been proposed for the gasphase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\quad \mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{-1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

The accompanying graph shows plots of \(\ln k\) versus \(1 / T\) for two different reactions. The plots have been extrapolated to the \(y\)-intercepts. Which reaction (red or blue) has (a) the larger value for \(E_{a}\) and (b) the larger value for the frequency factor, \(A\) ? [Section 14.5]

Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide \((\mathrm{NO})\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is second order with a rate constant of \(0.0796 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(737^{\circ} \mathrm{C}\) and \(0.0815 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(947{ }^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction.

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

In solution, chemical species as simple as \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\)can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right.\)] of a solution containing an acid- catalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\)concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.
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