Question

Calculate the standard enthalpy change (∆H) for the vaporization of 25.0 g of water at its boiling point (100°C) and 1 atm pressure assuming the heat capacity of water to be constant. Given: The heat of vaporization for water is 40.7 kJ/mol and the molar mass of water is 18.015 g/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization 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, we can calculate the enthalpy change using the heat of vaporization:H = moles of water  heat of vaporizationH = 1.387 moles  40.7 kJ/molH  56.5 kJTherefore, the standard enthalpy change for the vaporization of 25.0 g of water at its boiling point and 1 atm pressure is approximately 56.5 kJ.