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)? A)-570 kJ/mol B)-24.8 kJ/mol C)-7.8 kJ/mol D)24.8 kJ/mol E)570 kJ/mol

Question

Quizplus · Accepted Answer

The reaction involves 0.560 g of Na, which is 0.0243 mol (0.560 g / 22.99 g/mol). Since 13.8 kJ of heat is evolved for this amount, the enthalpy change per mole of NaF formed is 13.8 kJ / 0.0243 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