text: chemistry help please! standard enthalpy of formation? when 0.56 g 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)? first balance the formula and make a formula:  f2(g) + 2na(s)------------------- 2naf(s)  atomic mass na(s): 22.99g/mol  0.56gna * 1mol/22.99gna= 0.0244 mol na  since 13.8 kj evolved from 0.0244mol na divide them together to obtain kj/mol  therefore 13.8kj/0.0244mol= 566 kj/mol na   now using the chemical coefficent from the balanced equation turn it into the stadard enthalpy of naf.  566 kj/mol na * 2 mol naf(s)/2mol na(s)= 566 kj/mol naf(s)  since they used the term evolved in the question it means it is exothermic and threfore the stadard heat of formation of naf(s) is negative ( -566 kj/mol naf(s))