Problem 9.19 about Standard Entropy

[Sophia Bozone 2G]

For problem 9.19, It asks us to calculate the standard entropy of vaporization of water at 85 degrees. The solution of the problem shows that you have to calculate S of heating the reactants, and then S of vaporization, but then for some reason it says you have to calculate the entropy change as the products cool back down to 85 degrees. Why do you have to calculate is as it cools back down?

[Jenny Cheng 2K]

You should break this problem into 4 steps:
1) heat liquid water (ΔS=C * ln(T2/T1))
2) vaporize water at boiling point (ΔS(vap) is given in the problem)
3) cool water vapor back to 85 degrees C (ΔS=C * ln(T2/T1))
4) sum all the entropies from the above steps to find ΔS(total)

You need to calculate entropy of the water vapor cooling down because the problem asks you to find the standard entropy of water vaporization at a temperature below the boiling temperature of water.

[andrewr2H]

The boiling point of water is 100 degrees C. So to find the entropy of the vaporization of water at 85 degrees you have to account for the cooling down of water. I'm not sure how it is that water is able to be vaporized at 85 degrees though.

[zanekoch1A]

How is it possible for the water vapor to cool back down to 85 degrees without undergoing another phase change back to a liquid?

[Gevork 2E]

Hi, So I was also really confused about this, and I found an article on it.

https://chemistry.stackexchange.com/que ... t-of-water

I found this to be really useful.

Hope this is helpful.

[Evelyn L 1H]

How are we supposed to know to calculate the change in entropy back down to 85 degrees from the way the problem is worded?

[Kayla Ikemiya 1E]

You have to cool it down to find the standard entropy of vaporization for 85 degrees not 100.