Question

Calculate the enthalpy of vaporization of water given the following information: - The boiling point of water is 100°C- The specific heat capacity of liquid water is 4.18 J/g·°C- The specific heat capacity of steam is 1.84 J/g·°C- The heat of fusion of water is 40.7 kJ/mol - The molar mass of water is 18.015 g/mol- The heat capacity of the container is negligible- Assume that the vaporized water behaves ideally and has a constant heat capacity

Answer 1

To calculate the enthalpy of vaporization of water, we can use the Clausius-Clapeyron equation:Hvap =  Tb * vapS  / Rwhere Hvap is the enthalpy of vaporization, Tb is the boiling point in Kelvin, vapS is the entropy of vaporization, and R is the gas constant  8.314 J/molK .First, we need to find the entropy of vaporization  vapS . We can do this using the following equation:vapS = Cp,gas - Cp,liquidwhere Cp,gas is the molar heat capacity of steam and Cp,liquid is the molar heat capacity of liquid water.We are given the specific heat capacities of liquid water  4.18 J/gC  and steam  1.84 J/gC . To convert these to molar heat capacities, we multiply by the molar mass of water  18.015 g/mol :Cp,liquid = 4.18 J/gC * 18.015 g/mol = 75.3 J/molCCp,gas = 1.84 J/gC * 18.015 g/mol = 33.2 J/molCNow we can find the entropy of vaporization:vapS = Cp,gas - Cp,liquid = 33.2 J/molC - 75.3 J/molC = -42.1 J/molCNext, we need to convert the boiling point of water to Kelvin:Tb = 100C + 273.15 = 373.15 KNow we can use the Clausius-Clapeyron equation to find the enthalpy of vaporization:Hvap =  Tb * vapS  / R =  373.15 K * -42.1 J/molC  / 8.314 J/molKHvap = -157300 J/molSince the enthalpy of vaporization is a positive value, we take the absolute value:Hvap = 157300 J/mol = 157.3 kJ/molTherefore, the enthalpy of vaporization of water is 157.3 kJ/mol.