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 ?

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