Question

What is the enthalpy of vaporization of water? Given that the heat of vaporization of water at its boiling point (100°C) is 40.7 kJ/mol, calculate the enthalpy of vaporization of water at 25°C.

Answer 1

To calculate the enthalpy of vaporization of water at 25C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization. The equation is:ln P2/P1  = -Hvap/R *  1/T2 - 1/T1 where P1 and P2 are the vapor pressures at temperatures T1 and T2  in Kelvin , Hvap is the enthalpy of vaporization, and R is the gas constant  8.314 J/molK .First, we need to find the vapor pressures of water at 25C and 100C. We can use the Antoine equation to estimate these values:log10 P  = A - B /  T + C For water, the constants A, B, and C are 8.07131, 1730.63, and 233.426, respectively. Let's calculate the vapor pressures at 25C  T1 = 298.15 K  and 100C  T2 = 373.15 K :log10 P1  = 8.07131 - 1730.63 /  298.15 + 233.426 P1 = 10^3.169  = 1469.5 Palog10 P2  = 8.07131 - 1730.63 /  373.15 + 233.426 P2 = 10^4.757  = 56800.6 PaNow we can use the Clausius-Clapeyron equation to find the enthalpy of vaporization at 25C:ln 56800.6/1469.5  = -Hvap/8.314 *  1/298.15 - 1/373.15 3.93 = Hvap *  0.00336 - 0.00268  / 8.3143.93 = Hvap * 0.00068 / 8.314Solving for Hvap:Hvap = 3.93 * 8.314 / 0.00068 = 47800 J/mol = 47.8 kJ/molSo, the enthalpy of vaporization of water at 25C is approximately 47.8 kJ/mol.