Question

Calculate the standard enthalpy change for the vaporization of 1 mole of water at 100°C and 1 atm pressure. The enthalpy of formation of liquid water is -285.8 kJ/mol and the enthalpy of formation of water vapor is -241.8 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization of 1 mole of water, we can use the following equation:H_vaporization = H_f  water vapor  - H_f  liquid water where H_vaporization is the standard enthalpy change for vaporization, and H_f is the standard enthalpy of formation for water vapor and liquid water.Given values:H_f  liquid water  = -285.8 kJ/molH_f  water vapor  = -241.8 kJ/molNow, we can plug these values into the equation:H_vaporization =  -241.8 kJ/mol  -  -285.8 kJ/mol H_vaporization = 44.0 kJ/molThe standard enthalpy change for the vaporization of 1 mole of water at 100C and 1 atm pressure is 44.0 kJ/mol.