Question

What is the standard enthalpy change for the vaporization of 50g of water at its boiling point (100°C) and 1 atm (pressure of the atmosphere)? Given that the standard enthalpy of vaporization of water is 40.79 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization of 50g of water, we first need to determine the number of moles of water being vaporized. The molecular weight of water  H2O  is 18.015 g/mol.Number of moles = mass / molecular weightNumber of moles = 50g / 18.015 g/mol = 2.775 molesNow, we can use the standard enthalpy of vaporization of water  40.79 kJ/mol  to calculate the total enthalpy change.Standard enthalpy change = number of moles  standard enthalpy of vaporizationStandard enthalpy change = 2.775 moles  40.79 kJ/mol = 113.19 kJTherefore, the standard enthalpy change for the vaporization of 50g of water at its boiling point and 1 atm is 113.19 kJ.