/*! 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 gas-phase reaction of \(\mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 107

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form NOF and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(A=6.0 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). 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 structure 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.

Short Answer

Expert verified
(a) The rate constant at \(100^{\circ} \mathrm{C}\) is approximately \(1.1\times 10^6 \,\text{M}^{-1}\text{s}^{-1}\). (b) In the NO molecule, nitrogen and oxygen form a double bond, with one lone pair of electrons on both atoms. In the NOF molecule, nitrogen is the central atom, forming a double bond with oxygen and a single bond with fluorine. Oxygen has two lone pairs, and fluorine has three. (c) The structure of the NOF molecule is bent due to the VSEPR electron pair geometry around the nitrogen atom, which has three electron groups. (d) In the transition state for the formation of NOF, the bond between nitrogen and oxygen, and the bond between the fluorine atoms are partially broken, while new nitrogen-fluorine and oxygen-fluorine bonds are partially formed, represented by dashed lines in the Lewis structure. (e) The low activation energy for the reaction between NO and F2 could be due to the high reactivity of the molecules involved and the strong electrostatic attraction between the positively charged nitrogen atom in NO and the negatively charged fluorine atoms in F2.

Step by step solution

01

Arrhenius Equation

\[ k = A \times e^{-\frac{E_a}{RT}} \] where: k: rate constant; A: frequency factor (\(6.0 \times 10^8 M^{-1} s^{-1}\)); Ea: activation energy (6.3 kJ/mol); R: gas constant (8.314 J/mol K); T: temperature in Kelvin (100掳C + 273.15K = 373.15K). Now, we will plug in the values and solve for k.
02

Calculating k

\[ k = (6.0 \times 10^8 M^{-1} s^{-1}) \times e^{-\frac{6300 J / mol}{(8.314 J / mol K)(373.15 K)}} \] Calculating the result, we get:
03

Rate constant

\[ k \approx 1.1\times 10^6 \,\text{M}^{-1}\text{s}^{-1} \] (b) Draw the Lewis structures for the NO and NOF molecules
04

NO Lewis structure

For the NO molecule: - Nitrogen has 5 valence electrons. - Oxygen has 6 valence electrons. These two atoms form a double bond, with one lone pair of electrons on both nitrogen and oxygen.
05

NOF Lewis structure

For the NOF molecule: - Nitrogen has 5 valence electrons. - Oxygen has 6 valence electrons. - Fluorine has 7 valence electrons. In the NOF molecule, nitrogen is the central atom. Nitrogen forms a double bond with oxygen and a single bond with fluorine. Oxygen has two lone pairs, and fluorine has three lone pairs of electrons. (c) Predict the structure for the NOF molecule
06

NOF molecule structure

To predict the structure of the NOF molecule, we can use the VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the electron pair geometry. The nitrogen atom has three electron groups surrounding it: a double bond with oxygen, a single bond with fluorine, and a lone pair. This gives it a trigonal planar electron pair geometry. However, because the lone pair occupies more space in the electron cloud, it pushes the bonded atoms closer together, resulting in a bent molecular geometry. (d) Draw a possible transition state for the formation of NOF
07

Transition state

In the transition state for the formation of the NOF molecule, the bond between the nitrogen and oxygen in NO and the bond between the fluorine atoms in F2 are partially broken, while new bonds (nitrogen-fluorine and oxygen-fluorine) are partially formed. These weak bonds can be represented by dashed lines in the Lewis structure. (e) Suggest a reason for the low activation energy for the reaction
08

Low activation energy reason

A possible reason for the low activation energy for the reaction between NO and F2 is that the molecules involved are highly reactive. As a result, the reactants can quickly form the transition state, allowing the reaction to proceed with relatively low energy input. Additionally, the presence of a strong electrostatic attraction between the positively charged nitrogen atom in NO and the negatively charged fluorine atoms in F2 may further facilitate the reaction, lowering the activation energy barrier.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M}\) \(\mathrm{HCl}\) occurs according to the reaction \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+\underset{2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)}\) The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the value for the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?(\mathrm{c})\) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\) (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 63 torr \(/ \mathrm{h}\). What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

(a) What part of the energy profile of a reaction is affected by a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst?

In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a dif- ferent gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\). The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}:\) Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} \quad \text { (fast) }}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{2}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} \quad\) (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 steps? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?\)

Americium-241 is used in smoke detectors. It has a rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- 125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00-\mathrm{mg}\) sample of either isotope remains after three half-lives?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.