Question

Given the following information: N2 bond energy = 941 kJ/mol F2 bond energy = 154 kJ/mol 1/2 N2(g) + 3/2 F2(g) --> NF3 ∆H0 = -103 kJ/mol Calculate the N-F bond energy. Option 1: 397 kJ/mol Option 2: 247 kJ/mol Option 3: 147 kJ/mol Option 4: 94 kJ/mol

Answer 1

To calculate the N-F bond energy, determine the total energy required to break the bonds on the reactant side and the total energy released to form the N-F bond in NF3. Subtract the energy released from the energy required to find the bond energy of N-F.

Step-by-step explanation:

To calculate the N-F bond energy, we need to consider the energy changes involved in the reaction. The bond energy of N2 is 941 kJ/mol and the bond energy of F2 is 154 kJ/mol. Based on the reaction, 1/2 N2(g) + 3/2 F2(g) → NF3, the ∆H0 is -103 kJ/mol. We can use these values to determine the bond energy of N-F.

First, we calculate the total energy required to break the bonds on the reactant side:

Energy for N2 = 1/2 x 941 kJ/mol = 470.5 kJ/mol

Energy for F2 = 3/2 x 154 kJ/mol = 231 kJ/mol

Total energy required = 470.5 kJ/mol + 231 kJ/mol = 701.5 kJ/mol

Next, we calculate the total energy released to form the N-F bond in NF3:

Total energy released = ∆H0 = -103 kJ/mol

Finally, we can subtract the energy released from the energy required to find the bond energy of N-F:

N-F bond energy = Total energy required - Total energy released = 701.5 kJ/mol - (-103 kJ/mol) = 804.5 kJ/mol

Therefore, the N-F bond energy is approximately 804.5 kJ/mol, which is not one of the given options.

Answer 2

To calculate the N-F bond energy, consider the bond energies of the reactants and products in the given reaction. The difference between the total energy needed to break the bonds in the reactants and the total energy released when the bonds in the product are formed will give us the energy associated with the N-F bond in
`NF_(3)`, which is 690 kJ/mol.

To calculate the N-F bond energy, we need to consider the bond energies of the reactants and products in the given reaction. The reactants are 1/2
`N_(2)` (g) and 3/2
`F_(2)` (g), and the product is
`NF_(3)`. We can calculate the total energy needed to break the bonds in the reactants and the total energy released when the bonds in the product are formed. The difference between these two values will give us the energy associated with the N-F bond in
`NF_(3)`.

The bond energy of
`N_(2)` is 941 kJ/mol,

the bond energy of
`F_(2)` is 154 kJ/mol.

To break the bonds in 1/2
`N_(2)`(g), we need to multiply the bond energy by 1/2. Similarly, to break the bonds in 3/2
`F_(2)` (g), we need to multiply the bond energy by 3/2.

The energy needed to break the bonds in the reactants is

(1/2 * 941 kJ/mol) + (3/2 * 154 kJ/mol) = 587 kJ/mol.

The energy released when the N-F bonds are formed in
`NF_(3)` is -103 kJ/mol.

Therefore, the energy associated with the N-F bond is 587 kJ/mol + 103 kJ/mol = 690 kJ/mol.