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Chapter 14: Problem 107

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 anyintermediatesin 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: (a) The elementary reactions add up to the balanced equation: \(NO(g) + NO(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\). (b) Rate laws for elementary reactions: \(rate_1 = k_1[NO]^2\) and \(rate_2 = k_2[N_{2}O_{2}][H_{2}]\). (c) The intermediate in the mechanism is \(N_{2}O_{2}\). (d) Based on the observed rate law \(rate = k[NO]^2[H_{2}]\) and the derived rate laws, we conclude that the first reaction is fast and the second reaction is slow.

Step by step solution

01

(a) Verify the elementary reactions

First, we need to add up the elementary reactions in the given mechanism to check if they yield a balanced equation. Given reactions: 1st reaction: \(NO(g) + NO(g) \rightarrow N_{2}O_{2}(g)\) 2nd reaction: \(N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) Let's add them: \(NO(g) + NO(g) + N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O_{2}(g) + N_{2}O(g) + H_{2}O(g)\) Now, we can remove the species found on both sides of the equation: \(NO(g) + NO(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) This is the balanced equation for the reaction.
02

(b) Rate Law for Each Elementary Reaction

For the given elementary reactions, we need to write the rate laws. 1st reaction: \(NO(g) + NO(g) \rightarrow N_{2}O_{2}(g)\) Rate law: \(rate_1 = k_1[NO]^2\), where \(k_1\) is the rate constant for the first reaction. 2nd reaction: \(N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) Rate law: \(rate_2 = k_2[N_{2}O_{2}][H_{2}]\), where \(k_2\) is the rate constant for the second reaction.
03

(c) Identify Intermediates

In a reaction mechanism, intermediates are those substances that are produced and then consumed during the sequence of elementary reactions. In our case, N2O2 is an intermediate, as it is formed in the first reaction and consumed in the second reaction. Intermediate: \(N_{2}O_{2}\)
04

(d) Conclusions About Relative Speeds of Reactions

The observed rate law is given as \(rate = k[NO]^2[H_{2}]\). Comparing this with the rate laws derived for the elementary reactions, we can infer that the second reaction is the rate-determining step since it involves the concentrations of \(NO\) and \(H_{2}\) while the intermediate (\(N_{2}O_{2}\)) is not present. In order for the rate-determining step to match the overall rate law, we can deduce that the first reaction must be faster than the second reaction. As such, we can conclude that the first reaction is fast and the second reaction is slow, in accordance with the proposed mechanism.

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Most popular questions from this chapter

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of $5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\( at \)520 \mathrm{nm} .$ (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at $30.0 \mathrm{~min} .\( Calculate the rate constant in units of \)\mathrm{s}^{-1}$. (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Indicate whether each statement is true or false. (a) If you measure the rate constant for a reaction al 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.

(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?

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of $E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\(. and a frequency factor of \)A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}$. The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)(\text { slow }) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad(\text { fast }) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.
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