To calculate the standard enthalpy change for the vaporization reaction of 25.0 g of water at 100C and atmospheric pressure 1 atm , we need to use the following equation:H = m Hvapwhere H is the enthalpy change, m is the mass of water, and Hvap is the molar enthalpy of vaporization for water.The molar enthalpy of vaporization for water at 100C and 1 atm is 40.7 kJ/mol.First, we need to convert the mass of water 25.0 g to moles:moles of water = mass of water / molar mass of watermoles of water = 25.0 g / 18.015 g/mol = 1.387 molesNow, we can calculate the enthalpy change:H = 1.387 moles 40.7 kJ/mol = 56.5 kJThe standard enthalpy change for the vaporization reaction of 25.0 g of water at 100C and atmospheric pressure 1 atm is 56.5 kJ.