When .560 g of Na(s) reacts with excess F2(g) to form NaF(s), 13.8 kJ of heat is evolved at standard-state conditions. What is the standard enthalpy of formation of NaF(s)?

Start off by balancing the equation:
2Na(s) + F2(s) ---> 2NaF(s)

Then make it for one mole

Na(s) + (1/2)F2(s) ---> NaF(s)

.560g/22.99 = .02435 mols Na

kJ/mols = 13.8/.02135 = 570 kJ/mol. Should it be 570 kJ/mol or -570 kJ/mol?

The problem says so much heat is EVOLVED. That means that much heat is released so the sign is negative. However, I would redo the numbers. My calculator reads slightly over 560 ane 570. Also, note that you made a typo on the last line but you plugged in the right number.

The standard enthalpy of formation, represented by ΔH⦵f, is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. Since the given reaction evolves heat (exothermic reaction), the enthalpy change is negative. Therefore, the value -570 kJ/mol should be used as the standard enthalpy of formation of NaF(s).

To determine the sign of the standard enthalpy of formation, we need to consider whether heat is absorbed or evolved during the reaction. In this case, the question states that 13.8 kJ of heat is evolved, which means heat is released from the reaction.

Therefore, the standard enthalpy of formation of NaF(s) is -570 kJ/mol. The negative sign indicates that the reaction is exothermic and heat is released when one mole of NaF(s) is formed.