Question

Calculate the standard enthalpy change for the vaporization of water at its normal boiling point of 100 degrees Celsius and a pressure of 1 atm. The molar enthalpy of vaporization for water is 40.7 kJ/mol. What is the amount of heat required to vaporize 100 g of water?

Answer 1

To calculate the amount of heat required to vaporize 100 g of water, we can use the molar enthalpy of vaporization and the molar mass of water.First, let's convert the mass of water to moles:Molar mass of water  H2O  = 18.015 g/mol100 g of water / 18.015 g/mol = 5.55 moles of waterNow, we can use the molar enthalpy of vaporization to calculate the heat required:Molar enthalpy of vaporization = 40.7 kJ/molHeat required = moles of water * molar enthalpy of vaporizationHeat required = 5.55 moles * 40.7 kJ/mol = 225.885 kJTherefore, the amount of heat required to vaporize 100 g of water is approximately 225.9 kJ.