Question

The gas-phase reaction of 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 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.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

(a) The rate constant at $100^{\circ} \mathrm{C}$ can be calculated using the Arrhenius equation: $k = (6.0 \times 10^{8}) e^{\frac{-6300}{(8.314)(373.15)}}$. The rate constant is approximately $1.86 \times 10^{7} M^{-1} \mathrm{s}^{-1}$. (b) Lewis structures: NO has a double bond between N and O and one unpaired electron on N; NOF has a double bond between N and O and a single bond between N and F. (c) The NOF molecule has a bent or V-shaped molecular geometry due to three electron regions around the central nitrogen atom (one double bond to oxygen, one single bond to fluorine, and one lone pair). (d) In the transition state, the N atom from NO is connected to O with a partial double bond and to F from F2 with a partial single bond, both indicated by dashed lines. The F2 molecule is being broken to form a single F atom. (e) The low activation energy for the reaction may be due to the gas-phase nature of the reactants, which allows them to collide readily with low energy requirements and the effective overlap of atomic orbitals in the transition state structure, making the formation of weak bonds more energetically favorable.

Step-by-step solution

Part (a): Calculate the rate constant at 100°C

To calculate the rate constant, we use the Arrhenius equation: \[ k = Ae^{\frac{-E_{a}}{RT}} \] where: - k: rate constant - A: frequency factor, \(6.0 \times 10^{8} M^{-1} \mathrm{s}^{-1}\) - \(E_{a}\): activation energy, \(6.3 \mathrm{kJ/mol}\) or \(6300 \mathrm{J/mol}\) - R: gas constant, \(8.314 \mathrm{J/mol\cdot K}\) - T: temperature in Kelvin, which can be found by adding 273.15 to the given temperature in Celsius (100°C), so T = 373.15 K. Now, plug the values into the equation: \[ k = (6.0 \times 10^{8}) e^{\frac{-6300}{(8.314)(373.15)}} \]

Calculate the result

Solve the equation to get the rate constant: \[ k \approx 1.86 \times 10^{7} M^{-1} \mathrm{s}^{-1} \]

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

To draw the Lewis structures, first, find the number of valence electrons for each atom involved. NO: - N has 5 valence electrons - O has 6 valence electrons NOF: - N has 5 valence electrons - O has 6 valence electrons - F has 7 valence electrons Now, draw the Lewis structures keeping in mind that the nitrogen atom is the central atom in the NOF molecule. NO: N is connected to O with a double bond and has one unpaired electron NOF: N is connected to O with a double bond and connected to F with a single bond

Part (c): Predict the shape for the NOF molecule

The NOF molecule has three electron regions around the central nitrogen atom: one double bond to oxygen, one single bond to fluorine, and one lone pair. According to the VSEPR (valence shell electron pair repulsion) theory, three electron regions around a central atom result in a bent or V-shaped molecular geometry.

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

To draw a possible transition state for the formation of NOF, we must show a state between the reactants and the products, with dashed lines representing the weak bonds that are beginning to form. In the transition state: - N atom from NO is connected to O with a partial double bond (indicated by dashed lines) - N atom from NO is connected to F from F2 with a partial single bond (indicated by dashed lines) - F2 molecule is being broken to form a single F atom

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

The low activation energy for the reaction may be due to the fact that the reacting molecules are in the gas phase, allowing them to readily collide with each other with relatively low energy required to initiate the reaction. Additionally, the transition state structure may have low energy requirements due to effective overlap of atomic orbitals. This could make the formation of weak bonds in the transition state to be more energetically favorable, thus lowering the activation energy needed for the reaction to occur.

Key concepts

Arrhenius Equation

The Arrhenius equation is a fundamental formula that offers insight into the rate at which chemical reactions occur. It expresses the rate constant (\( k \)) of a reaction as a function of the temperature and the activation energy (\( E_a \)), as well as a pre-exponential factor (\( A \)), which represents the frequency of collisions with the correct orientation for a reaction to proceed.

In the context of the exercise, we have used the Arrhenius equation to calculate the rate constant. By substituting the given values into the equation, like the activation energy and the temperature converted into Kelvin, students can see the direct effect temperature and activation energy have on the rate at which a reaction will proceed. A lower activation energy or a higher temperature will both serve to increase the rate constant, allowing the reaction to occur more quickly.

Lewis Structures

Lewis structures are visual representations that depict the bonding between atoms of a molecule as well as the lone pairs of electrons that may exist. This sketching exercise provides a blueprint of the molecule's structure, highlighting the distribution of electrons among the atoms.

When drawing Lewis structures, the goal is to use dots to represent valence electrons and lines for bonds, ensuring each atom achieves a stable electron configuration. The structures for NO and NOF were constructed by allocating electrons to form a double bond between the nitrogen and oxygen for NO, and for NOF, a double bond between nitrogen and oxygen with an additional single bond between nitrogen and fluorine - revealing the central role of nitrogen in this molecule.

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules based on the repulsion between the electron-containing regions around a central atom. The shape of a molecule is significant as it influences the molecule's polarity, reactivity, phase of matter, color, magnetism, and biological activity.

Following VSEPR theory, the lone pair on the NOF molecule contributes to a bent molecular geometry due to the repulsion it exerts on the bonding electron pairs. This illustrates the three-dimensional nature of molecules and is crucial in understanding how the molecules interact with each other and their environment.

Rate Constant

The rate constant (\( k \)) is a proportionality constant in the rate law of a chemical reaction and is critically dependent on the reaction conditions, notably the temperature. It is not a universal constant but is specific to each chemical reaction and reflects how quickly a reaction proceeds under certain conditions.

By understanding and calculating the rate constant, as demonstrated in the solution, students can better comprehend the kinetics of a reaction. This allows for the prediction of reaction rates, the planning of synthesis, and the optimization of reaction conditions in both laboratory and industrial settings.

Transition State

The transition state refers to the highest-energy state during a chemical reaction, which represents a particular configuration along the reaction coordinate. It is the point at which the reactant molecules have just enough energy to form the products, often visualized as the peak on a reaction energy diagram.

Understanding transition states helps chemists to understand the mechanism of reactions. In the provided exercise, drawing the transition state with partial bonds allows students to visualize the 'in-between' state of reactants turning into products, which is essential in understanding the detailed process of chemical transformations.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of the atoms within a molecule. Determining molecular geometry is crucial for predicting the physical and chemical properties of a molecule, such as reactivity, polarity, and intermolecular forces.

By applying VSEPR theory, we can determine that the NOF molecule has a bent shape, which is essential to predicting the molecule's behavior. The bent shape affects the molecule's dipole moment and its interaction with other molecules, thus playing a significant role in the chemical and physical properties it exhibits.