Question

Calculating the standard enthalpy change for the vaporization of water is a common example used in chemistry. Suppose that you have 50.0 g of liquid water at 25°C and 1 atm pressure. What is the standard enthalpy change for the vaporization of this amount of water at its boiling point, which is 100°C and 1 atm pressure?

Answer 1

To calculate the standard enthalpy change for the vaporization of water, we need to use the molar heat of vaporization  Hvap  of water, which is the amount of heat required to vaporize one mole of water at its boiling point under standard conditions  100C and 1 atm pressure . The molar heat of vaporization for water is 40.7 kJ/mol.First, we need to convert the mass of water  50.0 g  to moles using the molar mass of water  18.02 g/mol :moles of water =  50.0 g  /  18.02 g/mol  = 2.775 molesNext, we can calculate the standard enthalpy change for the vaporization of this amount of water using the molar heat of vaporization:H = moles of water  HvapH =  2.775 moles    40.7 kJ/mol  = 112.93 kJTherefore, the standard enthalpy change for the vaporization of 50.0 g of water at its boiling point  100C and 1 atm pressure  is approximately 112.93 kJ.