91Ó°ÊÓ

To calculate the rate constant at 100 °C, we will use the Arrhenius equation, which relates the activation energy (Ea), the frequency factor (A), the rate constant (k), and the temperature (T). The equation is: $$ k = Ae^{\frac{-Ea}{RT}} $$ Where R is the gas constant, which has a value of \(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\), T is the temperature in Kelvin and Ea is the activation energy in J/mol. First, we need to convert 100 °C to Kelvin (T = 273.15 + 100 = 373.15 K), and Ea to J/mol (6300 J/mol). Now we can plug in the values and find the rate constant: $$ k = (6.0 \times 10^{8} M^{-1} s^{-1}) \times e^{\frac{-6300 \mathrm{~J} / \mathrm{mol}}{(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})(373.15 \mathrm{~K})}} $$ After calculating this expression, we get \(k = 1.88 \times 10^{10} M^{-1}s^{-1}\)

Now we'll draw the Lewis structures for NO and NOF molecules. NO: The total number of valence electrons in the NO molecule is 5(electrons from N) + 6(electrons from O) = 11. We place the atoms N and O next to each other and create a single bond between them. This bond accounts for 2 of the valence electrons, leaving 9 electrons. We can distribute these electrons as lone pairs: 6 electrons to the oxygen atom (as 3 lone pairs) and the remaining 3 electrons to the nitrogen atom (as 1 lone pair and 1 unpaired electron). So, we end up with a single bond and one unpaired electron on each atom. NOF: In the case of the NOF molecule, we are given that nitrogen is the central atom, and the total number of valence electrons is 5(electrons from N) + 6(electrons from O) + 7(electrons from F) = 18. We place N at the center and create single bonds with both the O and F atoms. This accounts for 4 of the valence electrons, leaving 14 electrons. We can distribute these electrons as lone pairs: 6 electrons to the oxygen atom (as 3 lone pairs), 6 electrons to the fluorine atom (as 3 lone pairs), and the remaining 2 electrons to the nitrogen atom (as 1 lone pair).

To predict the shape of the NOF molecule, we can use the Valence Shell Electron Pair Repulsion (VSEPR) theory. In the case of NOF, the nitrogen atom has 3 electron regions: a single bond to O, a single bond to F, and a lone pair. According to VSEPR theory, these electron regions will arrange themselves to minimize repulsion, leading to a "bent" molecular shape with an angle of approximately \(120^\circ\) between the O-N-F atoms.

In the transition state for the formation of NOF, the reactants (NO and F2) are in the process of breaking and forming bonds. In this case, the N-O bond is already present and relatively strong. We can depict the forming N-F bond and the breaking F-F bond as dashed lines in the transition state structure. The F atoms from F2 will have an angling between the N and the other F atom, and the O atom will be repelling the forming F atom. This will create a structure where all atoms appear to have partial bonds in the transition state (indicated with dashed lines) compared to solid lines for the reactant molecule.

A possible reason for the low activation energy for this reaction is that the N-O bond remains largely unaffected during the process, only changing its bonding partners as the F2 molecule is replaced by the F atom. The bond formation helps stabilize the structure overall, which could contribute to the lower energy required to initiate the reaction. Additionally, the reaction may proceed through a mechanism that requires fewer steps, further reducing the activation energy.