Posted by **chamy** on Sunday, April 22, 2012 at 6:08am.

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

a. 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° = –...
- Chemistry - Can you check my calculations, please? It's for a lab we did in ...
- chemistry - The standard internal energy change for a reaction can be symbolized...
- Chemistry - The standard internal energy change for a reaction can be symbolized...
- Chemistry - If ammonia is manufactured at 356 K, is the reaction spontaneous, ...
- chemistry - Nitroglycerin is a powerful explosive, giving four different gases ...
- Enthalpy Change Calculations - Find the heat of reaction (ΔH) for each of ...
- Chemistry - Acetylene, C2H2, can be converted to ethane, C2H6, by a process ...
- Math - Sorry! I don't get this one either.. limit of (f(x+Δx)-f(x)) / Δ...
- Chemistry - How would you exactly start the first problem? My thermodynamic ...

More Related Questions