Question

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.

Short answer

The rate constant at 100°C is approximately \(6.0 \times 10^5 M^{-1} s^{-1}\). The Lewis structures for NO and NOF are :N = O: and :F - N = O:, respectively. The NOF molecule has a bent or angular shape based on the VSEPR theory. The transition state for the formation of NOF can be represented as :NO···F-F:, with dashed lines indicating weak bonds. The low activation energy for the reaction could be due to weak interactions between reactants, high energy levels of initial reactants, or the presence of a suitable catalyst.

Step-by-step solution

(a) Calculate the rate constant at 100°C

: To calculate the rate constant, you need to use the Arrhenius equation: \[k = A \cdot e^{-\frac{E_a}{RT}}\] where \(k\) represents the rate constant, \(A\) is the frequency factor (in this case = \(6.0 \times 10^{8} M^{-1} s^{-1}\)), \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol⋅K), and \(T\) is the temperature in Kelvin. First, convert the temperature to Kelvin: \(T_{Kelvin} = 100 + 273.15 = 373.15 K\) Next, convert the activation energy to J/mol: \(E_a = 6.3 kJ/mol \times \frac{1000 J}{1 kJ} = 6300 J/mol\) Now, plug all variables into the Arrhenius equation and solve for \(k\): \[k = (6.0 \times 10^{8} M^{-1} s^{-1}) \cdot e^{-\frac{6300 J/mol}{(8.314 J/mol \cdot K)(373.15 K)}}\] After calculating, you should find that the rate constant, \(k\), is approximately \(6.0 \times 10^5 M^{-1} s^{-1}\).

(b) Draw the Lewis structures for NO and NOF molecules

: To draw the Lewis structures, you have to: 1. Count the total number of valence electrons in each molecule. 2. Place the least electronegative atom in the center. 3. Distribute the valence electrons. 4. Check if each atom has a stable octet. For NO: - N has 5 valence electrons - O has 6 valence electrons Total valence electrons = 11 N is the central atom and forms a double bond with O, leaving one unpaired electron on N. Therefore, the Lewis structure for NO is: :N = O: For NOF: - N has 5 valence electrons - O has 6 valence electrons - F has 7 valence electrons Total valence electrons = 18 N is the central atom. It forms a double bond with O and a single bond with F. The Lewis structure for NOF is: :F - N = O:

(c) Predict the shape for the NOF molecule

: To predict the shape of the NOF molecule, use the VSEPR (Valence Shell Electron Pair Repulsion) theory: 1. Determine the central atom's electron geometry. 2. Identify the molecular geometry based on the electron geometry. In the case of NOF: - Electron geometry: Trigonal planar (Three electron groups - one double bond, one single bond, and one lone pair) - Molecular shape: Based on the electron geometry, the shape of the NOF molecule is bent or angular.

(d) Draw a possible transition state for the formation of NOF

: A transition state represents an intermediate stage in a reaction where bonds are breaking and new bonds are forming. In the formation of NOF, a bond between NO and F2 is breaking, and new bonds between N-O and N-F are forming. To draw the transition state, show the weak bonds forming using dashed lines. The transition state can be represented as: :NO···F-F: where the dashed lines indicate weak bonds.

(e) Suggest a reason for the low activation energy for the reaction

: The low activation energy for this reaction could be due to the following reasons: - The formation of NOF might involve weak interactions between reactants, like van der Waals forces or dipole-dipole interactions. - The initial reactants, NO and F2, might have high energy levels, which lower the energy required for the reaction to proceed. - The reaction might be occurring on a catalytic surface or in the presence of a suitable catalyst, which lowers the energy barrier. Overall, the low activation energy means that the reaction can easily proceed, even at relatively low temperatures.

Key concepts

Activation Energy

In chemical kinetics, activation energy is the minimum energy required for a reaction to proceed. This is a crucial concept because it indicates the energy threshold that must be overcome for reactants to transform into products. For instance, consider the reaction between nitrogen monoxide (\(\text{NO}\)) and fluorine (\(\text{F}_2\)) to form nitrosyl fluoride (\(\text{NOF}\)). Here, the activation energy is given as \(6.3\,\text{kJ/mol}\).

Activation energy can be influenced by several factors such as temperature, catalysts, and the intrinsic nature of the molecules involved. A lower activation energy often implies that a reaction can proceed more readily at a given temperature, facilitating faster reaction rates. In our example, the relatively low activation energy suggests that the reaction between \(\text{NO}\) and \(\text{F}_2\) proceeds easily, requiring less energy to reach the transition state.

• This helps reactants to overcome energetic barriers.
• Typically affected by catalysts which lower the energy needed.
• Gives insight into the speed and feasibility of chemical reactions.

Rate Constant Calculation

The rate constant (\(k\)) is fundamental for calculating the speed of a chemical reaction. It's determined using the Arrhenius equation, which relates the rate constant to the activation energy and temperature. For the reaction involving \(\text{NO}\) and \(\text{F}_2\), the rate constant at \(100^{\circ} \text{C}\) can be found using:

\[k = A \cdot e^{-\frac{E_a}{RT}}\]

where \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

• Convert temperatures to Kelvin: \(T = 373.15\, \text{K}\).
• Convert activation energy to \(\text{J/mol}\): \(E_a = 6300\, \text{J/mol}\).
• Calculate using the given \(A\) value: \(6.0 \times 10^{8} \text{M}^{-1}\text{s}^{-1}\).

The result, \(k = 6.0 \times 10^5 \text{M}^{-1}\text{s}^{-1}\), shows how rapidly the reaction occurs at the specified temperature.

Molecular Geometry Prediction

Predicting molecular geometry involves examining the layout of a molecule's atoms and bonds using the VSEPR (Valence Shell Electron Pair Repulsion) theory. This approach helps in understanding the spatial arrangement based on electron repulsion around the central atom.

For \(\text{NOF}\), the central atom is nitrogen because it can form more bonds compared to fluorine or oxygen. With VSEPR, determine the shape:

• Calculate total valence electrons: Nitrogen (5), Oxygen (6), Fluorine (7), which equals 18 electrons.
• Structure: \(\text{N}\)-F (single bond) and \(\text{N}\)=O (double bond), leaving a lone pair on nitrogen.
• Three regions of electron density, leading to a trigonal planar electron geometry.
• Molecular shape is bent, as lone pairs influence the geometry drawing the angles closer to around 120 degrees.

The bent shape explains the non-linear alignment due to lone pairs, which affects molecular polarity and interactions.

Transition State Theory

Transition State Theory helps to visualize and analyze reactions at the molecular level, particularly focusing on the transition state - an ephemeral state high in energy, where old bonds are breaking and new bonds are forming.

During the formation of \(\text{NOF}\) from \(\text{NO}\) and \(\text{F}_2\), a transition state might depict the simultaneous weakening and forming of bonds. Imagine this process as:

• Bonds of \(\text{NO}\) interacting with \(\text{F}_2\).
• The nitrogen atom forms partial bonds, shown with dashed lines, to indicate bond formation with fluorine while simultaneously maintaining a bond with oxygen.
• Quickly traversing this high-energy state can depend on external factors including the presence of catalysts.

Understanding this theory aids in visualizing kinetic stability and the transformative stages reactants undergo, thus giving insights into the reaction's energy requirements.