Chapter 14

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\) (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?\) (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

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 \(=k[A]^{2} \cdot(\mathbf{b})\) 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.

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(30^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.5 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(2.0 \times 10^{-2} \mathrm{M},\) what is the rate of formation of \(\mathrm{H}^{+} ?\)

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \((\mathrm{NO}]=0.040 \mathrm{M}\), and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of NO is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8 ?\)

You perform a series of experiments for the reaction \(\mathrm{A} \rightarrow 2 \mathrm{~B}\) and find that the rate law has the form, rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) The rate increases by a factor of \(6.25,\) when \([A]_{0}\) is increased by a factor of \(2.5 .\) (b) There is no rate change when \([A]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2\), when \([A]\) is cut in half.

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{\underline{\phantom{xx}}}^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}:\) \begin{tabular}{llll} \hline Experiment & {\(\left[\mathrm{HgCl}_{2}\right](M)\)} & {\(\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline 1 & 0.164 & 0.15 & \(3.2 \times 10^{-5}\) \\ 2 & 0.164 & 0.45 & \(2.9 \times 10^{-4}\) \\ 3 & 0.082 & 0.45 & \(1.4 \times 10^{-4}\) \\ 4 & 0.246 & 0.15 & \(4.8 \times 10^{-5}\) \\ \hline \end{tabular} (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}{\underline{\phantom{xx}}}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: \begin{tabular}{llll} \hline Experiment & {\([\mathrm{X}]_{0}(M)\)} & {\([\mathrm{Z}]_{0}(\mathrm{M})\)} & Rate \((\mathrm{M} / \mathrm{s})\) \\ \hline 1 & 0.25 & 0.25 & \(4.0 \times 10^{1}\) \\ 2 & 0.50 & 0.50 & \(3.2 \times 10^{2}\) \\ 3 & 0.50 & 0.75 & \(7.2 \times 10^{2}\) \\ \hline \end{tabular} (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{X}\) is \(0.75 \mathrm{M}\) and that of \(\mathrm{Z}\) is \(1.25 \mathrm{M} ?\)

The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to \(\mathrm{C}_{4} \mathrm{~F}_{\mathrm{s}}\) has a rate constant \(k=0.045 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(450 \mathrm{~K}\). (a) Based on the unit of \(k\), what is the reaction order in \(\mathrm{C}_{2} \mathrm{~F}_{4} ?(\mathbf{b})\) If the initial concentration of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) is \(0.100 \mathrm{M}\), how long would it take for the concentration to decrease to \(0.020 \mathrm{M}\) at \(450 \mathrm{~K} ?\)

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is \(3.05 \times 10^{4} s,\) what is the rate constant \(k\) at this temperature? What percentage of A will not have reacted after one day?

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O} .\) At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1}\). Calculate the half- life at this temperature. (b) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

Urea (NH_2CONH \(_{2}\) ) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{MHCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ 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 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 \(0.500 \mathrm{M} ?\) (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

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 \(1.00-\mathrm{cm}\) 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 \(0.100 ?\)

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a \(1-\mathrm{cm}\) cuvette in a spectrometer and obtain the following data: \begin{tabular}{cc} \hline Time (min) & Absorbance at \(608 \mathrm{nm}\) \\ \hline 0 & 1.254 \\ 30 & 0.941 \\ 60 & 0.752 \\ 90 & 0.672 \\ 120 & 0.545 \\ \hline \end{tabular} From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene \(\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\). A \(0.0400 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{6}\) was monitored as a function of time as the reaction \(2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}\) proceeded. The following data were collected: \begin{tabular}{cc} \hline Time (s) & {\(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)\)} \\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{tabular} Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, \(\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, and \(1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature \((\mathrm{K})\) & Rate Constant \(\left(\mathrm{s}^{-1}\right)\) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

At \(25^{\circ} \mathrm{C}\), raw milk sours in \(6.0 \mathrm{~h}\) but takes \(60 \mathrm{~h}\) to sour in a refrigerator at \(5{ }^{\circ} \mathrm{C}\). Estimate the activation energy in \(\mathrm{kJ} / \mathrm{mol}\) for the reaction that leads to the souring of milk.

The following mechanism has been proposed for the reaction of NO with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (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 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?

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step \(1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (fast) Step 2: \(\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so,

The following mechanism has been proposed for the gasphase reaction of chloroform (CHCl \(_{3}\) ) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \stackrel{\mathrm{k}_{1}}{\mathrm{k}_{-1}} 2 \mathrm{Cl}(g)\) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CCl}_{4}\) (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 rate-determining step? (e) What is the rate law predicted by this mechanism?

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: A+B \longrightarrow C+X \\ \text { Step } 2: A+X \longrightarrow C+D \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

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 different 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: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \stackrel{\mathrm{k}_{\mathrm{L}}}{2}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{h}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{\mathrm{k}_{\mathrm{j}}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) (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 rate-determining 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 \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?\)

Platinum nanoparticles of diameter \(-2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face- centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .\) (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .\) (b) Esti- mate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one \(\mathrm{Pt}\) atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and \((b),\) calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0-\mathrm{nm}\) platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

The human body ischaracterized 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.

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{l} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \text { (slow) } \end{array} $$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, \(\mathrm{ES}=\) enzyme- substrate complex, and \(\mathrm{P}=\) product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of \(\mathrm{E}\) with \(\mathrm{I}\), an inhibitor.

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\) has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.1 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(30^{\circ} \mathrm{C}\). (b) A solution of KOH in ethanol is made up by dissolving 0.500 g KOH in ethanol to form \(500 \mathrm{~mL}\) of solution. Similarly, \(1.500 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{l}\) is dissolved in ethanol to form \(500 \mathrm{~mL}\). of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(30^{\circ} \mathrm{C} ?\) (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? ((d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(40^{\circ} \mathrm{C}\)

The gas-phase reaction of 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 \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). 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 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.