Question

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?

Answer 1

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.