Posted by **chamy** on Saturday, April 21, 2012 at 6:38am.

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 ?

## Answer this Question

## Related Questions

- chemistry - N2(g) + 3 F2(g) → 2 NF3(g) ΔH° = –264 kJ/mol ΔS° = –...
- Math - Sorry! I don't get this one either.. limit of (f(x+Δx)-f(x)) / Δ...
- Chemistry - Can you check my calculations, please? It's for a lab we did in ...
- chemistry - Nitroglycerin is a powerful explosive, giving four different gases ...
- Chemistry - Which of the following reactions are spontaneous (favorable)? A. ...
- Chemistry - How do i calculate: What is is ΔG° at 25 °C? 2O3(g) → 3 ...
- Chemistry - Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6...
- Chemistry - How would you exactly start the first problem? My thermodynamic ...
- Chemistry - The standard internal energy change for a reaction can be symbolized...
- Chemistry - If ammonia is manufactured at 356 K, is the reaction spontaneous, ...