To calculate the standard enthalpy change for the vaporization reaction of 25.0 g of water, we first need to determine the number of moles of water present. We can do this using the molar mass of water:moles of water = mass of water / molar mass of watermoles of water = 25.0 g / 18.015 g/molmoles of water 1.387 molesNow that we have the number of moles of water, we can use the molar heat of vaporization to calculate the standard enthalpy change:H = moles of water molar heat of vaporizationH = 1.387 moles 40.7 kJ/molH 56.4 kJThe standard enthalpy change for the vaporization reaction of 25.0 g of water at its boiling point of 100C is approximately 56.4 kJ.