posted by Samantha on .
What is the entropy change for the vaporization of 2.6 mol H2O(§¤) at 100◦C and 1 atm? dH= 40700 J/mol
Answer in units of J/K
Since it is a constant-temerature process, the entropy change is the heat of vaporization (2.6 * 40,700 J), divided by the temperature in K (373 in this case).