College

The normal boiling point of water is 100.0 °C, and its molar enthalpy of vaporization is 40.67 kJ/mol. What is the change in entropy in the system in J/K when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point?

Answer :

ΔS= nΔHvap/T,

Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K

Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K

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