Question

At 1 atm pressure, the heat of sublimation of gallium is 277 kJ/mol and the heat of vaporization is 271 kJ/mol. To the correct number of significant figures, how much heat is required to melt 4.50 mol of gallium at 1 atm pressure?

Answer 1

Ga (s) ----------> Ga (g) ,           delta H = 277 kJ/mol

Ga (l) -----------> Ga (g) ,          delta H = 271 kJ/mol

Therefore, to get the required equation reverse the second equation and then add it with first equation.

Ga (s) -----------> Ga (l) ,            delta H = 277 - 271 = 6.00 kJ/mol

So, heat required to melt 4.50 mol of Ga = 4.50 * 6.00 = 27.0 kJ