N2(g) + 3F2(g) > 2NF3(g) H298 = -264kj mol^-1. S298 = -278 JK^-1 mol^-1 The following questions

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Question:

N2(g) + 3F2(g) —> 2NF3(g) ΔH°298 = -264kj mol^-1. ΔS°298 = -278 JK^-1 mol^-1

The following questions relate to the synthesis reaction represented by the chemical equation and data above.

1) Calculate the value of the standard free energy change for the reaction.

2) Determine the temperature at which the equilibrium constant, Keq, for the reaction is equal to 1.00.

3) Calculate the standard enthalpy change ΔH°, that occurs when a 0.256 mole sample of NF3(g) is formed from N2(g) and F2(g) at 1.00 atm and 298 K.

The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and energy released by bond formation in the products.

4) How many bonds are formed when two molecules of NF3 are produced according to the equation above?

5) Use both the information from the formula above and the table of average bond enthalpies below to calculate the average enthalpy of the F - F bond.