When 0.560 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 ($\Delta$H°_f)of NaF(s)?

Question

Quizplus · Accepted Answer

The reaction is Na(s) + 1/2 F2(g) → NaF(s). The heat evolved is 13.8 kJ for 0.560 g of Na, which is 0.02435 mol (0.560 g / 22.99 g/mol). The enthalpy change per mole of NaF formed is 13.8 kJ / 0.02435 mol = -570 kJ/mol.

When 0560 G of Na(s)reacts with Excess F2(g)to Form NaF(s), 13

When 0560 G of Na(s)reacts with Excess F₂(g)to Form NaF(s), 13