N2(g) + 3 F2(g) → 2 NF3(g) ΔH° = –264 kJ/mol ΔS° = –278 J/(mol∙K)

a. Using the information provided above (only), calculate the maximum amount of non-PΔV work that can be accomplished through this reaction at a temperature of 500°C.
b. Determine the temperature at which the equilibrium constant for this reaction is equal to 1.00.
c. Use the information above and the table of average bond enthalpies below to calculate the average bond enthalpy of the F–F bond. (You may find it helpful to draw Lewis structures…)

Bond Average Bond Energy (kJ mol-1)
N≡N 946
N–F 272
F–F ?

To answer these questions, we will need to use the equations and concepts of thermodynamics. I will guide you through the steps to find the answers to each question.

a. To calculate the maximum amount of non-PΔV work, we can use the equation: non-PΔV work = -ΔH.

Given: ΔH° = -264 kJ/mol

Converting ΔH° from kJ/mol to J/mol:
ΔH° = -264,000 J/mol

Since the reaction has 2 moles of NF3, we need to multiply ΔH° by 2 to account for this:
non-PΔV work = -(ΔH° × 2)

Substituting the values:
non-PΔV work = -(-264,000 J/mol × 2)

Simplifying:
non-PΔV work = 528,000 J/mol
non-PΔV work = 528 kJ/mol

Therefore, the maximum amount of non-PΔV work that can be accomplished through this reaction is 528 kJ/mol.

b. To find the temperature at which the equilibrium constant (K) for this reaction is equal to 1.00, we can use the equation: ΔG° = -RT ln(K).

Given: ΔS° = -278 J/(mol∙K)
Assuming R = 8.314 J/(mol·K) (gas constant)

We need to convert ΔS° from J/(mol·K) to kJ/(mol·K):
ΔS° = -278 J/(mol·K) = -0.278 kJ/(mol·K)

Substituting the values:
ΔG° = -RT ln(K)
-0.278 kJ/(mol·K) = -(8.314 J/(mol·K) × T) ln(1.00)

Simplifying:
0.278 = T ln(1.00)

Since ln(1.00) = 0, we get:
0.278 = 0

This equation is not possible to solve since the left side is non-zero while the right side is zero. Therefore, there is no temperature at which the equilibrium constant for this reaction is exactly equal to 1.00.

c. To find the average bond enthalpy of the F-F bond, we can use the bond enthalpy equation: ΔH° = Σ (bonds broken) - Σ (bonds formed).

Given:
ΔH° = -264 kJ/mol
Average bond enthalpy of N≡N = 946 kJ/mol
Average bond enthalpy of N-F = 272 kJ/mol

Substituting the values and let the average bond enthalpy of F-F be x:
-264 kJ/mol = 2(946 kJ/mol) + 6(272 kJ/mol) - x

Simplifying:
-264 kJ/mol = 1892 kJ/mol + 1632 kJ/mol - x
-264 kJ/mol = 3524 kJ/mol - x

Rearranging the equation to solve for x:
x = 3524 kJ/mol + 264 kJ/mol
x = 3798 kJ/mol

Therefore, the average bond enthalpy of the F-F bond is approximately 3798 kJ/mol.

Note: It may be helpful to draw the Lewis structures of the molecules to visualize the bonds and better understand the calculations.