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Chapter 14: Problem 125

As shown in Figure 14.23 , the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the "sticking" of ethylene to a metal surface is the interaction of the electrons in the \(\mathrm{C}-\mathrm{C} \pi\) bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?

Short Answer

Expert verified
(a) Ethane is not expected to adsorb to a metal surface, as it does not have \(\pi\) electrons available for interaction with the vacant orbitals on the surface. Hence, it would not bind to the metal surface as ethylene does. (b) Yes, ammonia can adsorb to a metal surface due to the lone pair of electrons on the nitrogen atom, which can interact with the vacant orbitals on the surface. This interaction is similar to the ethylene case, although the specific details and strength of the interaction may differ.

Step by step solution

01

Understand the adsorption mechanism of ethylene on the metal surface

According to the given information, the adsorption of ethylene on a metal surface is due to the interaction of its \(\mathrm{C}-\mathrm{C}\) \(\pi\) bond electrons with the vacant orbitals on the metal surface. Ethylene has a C=C double bond, which has a \(\sigma\) bond that holds the atoms together and a \(\pi\) bond made up of the overlapping p orbitals.
02

Determine the binding of ethane on the metal surface

Ethane has a C-C single bond, so it does not have \(\pi\) electrons. Instead, the single bond between the carbons is a \(\sigma\) bond. To determine if ethane would bind to a metal surface, and how strongly, we need to compare the availability of electrons for interaction with vacant orbitals on the surface. Since ethane does not have \(\pi\) electrons, it would not be expected to adsorb to a metal surface as ethylene does.
03

Analyzing the Lewis structure of ammonia

Ammonia has a Lewis structure consisting of one nitrogen atom bonded to three hydrogen atoms, with a lone pair of electrons on the nitrogen atom: NH3. The lone pair of electrons on the nitrogen atom can engage in interactions with vacant orbitals on the metal surface, similar to the \(\pi\) electrons in ethylene.
04

(a) Answer for if ethane would adsorb to a metal surface and how strongly it would bind compared to ethylene

Ethane is not expected to adsorb to a metal surface, as it does not have \(\pi\) electrons available for interaction with the vacant orbitals on the surface. Hence, it would not bind to the metal surface as ethylene does.
05

(b) Answer for if ammonia would adsorb to a metal surface using a similar explanation as for ethylene

Yes, ammonia can adsorb to a metal surface due to the lone pair of electrons on the nitrogen atom, which can interact with the vacant orbitals on the surface. This interaction is similar to the ethylene case, although the specific details and strength of the interaction may differ.

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Most popular questions from this chapter

(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

The human body is characterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

The addition of NO accelerates the decomposition of $\mathrm{N}_{2} \mathrm{O}$, possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (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 \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.
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