/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 The following mechanism has been... [FREE SOLUTION] | 91影视

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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

(a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol}\) ? (b) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\) What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

Indicate whether each statement is true or false. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)
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