Question

Calculate the standard enthalpy change for the vaporization reaction of 2.00 moles of water at 100°C and 1 atm pressure, given that the molar enthalpy of vaporization of water is 40.7 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization reaction of 2.00 moles of water, we can use the following equation:H = n * Hvapwhere H is the standard enthalpy change, n is the number of moles of water, and Hvap is the molar enthalpy of vaporization.Given:n = 2.00 molesHvap = 40.7 kJ/molNow, we can plug in the values into the equation:H = 2.00 moles * 40.7 kJ/molH = 81.4 kJThe standard enthalpy change for the vaporization reaction of 2.00 moles of water at 100C and 1 atm pressure is 81.4 kJ.