Question

What is the standard enthalpy change for the vaporization of 25 grams of water at 100°C and 1 atm pressure? Given the standard enthalpy of vaporization for water is 40.7 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization of 25 grams of water, we need to first determine the number of moles of water and then use the given standard enthalpy of vaporization.1. Calculate the number of moles of water:The molar mass of water  H2O  is 18.015 g/mol  1.008 g/mol for hydrogen and 15.999 g/mol for oxygen .Number of moles = mass / molar massNumber of moles = 25 g / 18.015 g/mol = 1.387 moles2. Calculate the standard enthalpy change:Standard enthalpy change = number of moles  standard enthalpy of vaporizationStandard enthalpy change = 1.387 moles  40.7 kJ/mol = 56.47 kJThe standard enthalpy change for the vaporization of 25 grams of water at 100C and 1 atm pressure is approximately 56.47 kJ.