The heat vaporization of water is 40.66 kj/mol . ? the heat vaporization of water is 40.66 kj/mol. how much heat is absorbed when 3.45g of water boils at atmospheric pressure?

To answer the question above, first convert the given amount of water from grams to mol by dividing it with its molar mass which is 18 g/mol. 
                           amount in mol = 3.45 g x (mol/ 18 g) = 23/120 mol
Then, multiply this amount in mol with the heat of vaporization given.
                        (23/120) mol x (40.66 kj/mol) = 7.793 kJ
Thus, the heat absorbed is approximately equal to 7.793 kJ. 

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.