The standard free energy of formation of NaBrs is 350 kJmol

The standard free energy of formation of NaBr(s) is -350 kJ/mol. ?G

Solution

2NaBr(s) --> 2Na(s) + Br2(l)

dGf (NaBr) = -350 KJ/mol

dG(Na) = 0

dG(Br2) = 0

dG(rxn) = dG (products) -dG (reactants)

dG(rxn) = 0+0-(-2*350) = 700 KJ...............ANS