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^$\circ$_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 is: 2Na(s) + F2(g) → 2NaF(s). The molar mass of Na is 22.99 g/mol, so 0.560 g of Na is 0.560/22.99 = 0.02435 mol. Since 13.8 kJ is evolved for 0.02435 mol of Na, for 1 mol of Na, it would be 13.8 kJ / 0.02435 mol = 566.8 kJ/mol. Since the reaction forms 2 moles of NaF, the enthalpy change per mole of NaF is 566.8 kJ/mol / 2 = 283.4 kJ/mol. The negative sign indicates exothermic reaction, so the standard enthalpy of formation is -570 kJ/mol (rounded).

When 0560 G of Na(s) Reacts with Excess F2(g) to Form

When 0560 G of Na(s) Reacts with Excess F₂(g) to Form