What is the entropy change for the vaporization of 1.8 mol H2O at 100 degrees Celsius and 1 ATM?

I think you meant to use the Word enthalpy. Enthalpy refers to the flow of heat, while entropy means the "randomness" of a system.

Because The water is already at its boiling point (100o C), it doesn't need to be heated any more to start boiling. To find out how much energy it takes, you'll need to look up the molar heat of vaporization. Simply put, this is the amount of energy it takes to boil one mol of a substance. Water's molar heat of vaporization is 40.2 kJ/mol.

Now, all we need to do is multiply:

1.8mol * 40.2kJ/mol = 72kJ

(Note: The 1 atm is important because that is the pressure at which water's boiling point is 100oC. If there was any less pressure, it would have already boiled. If there was more pressure, it would have to heat up to the new boiling point, then use 72kJ to boil.)