## Self-test 4I.4A

Philip Lee 1L
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

### Self-test 4I.4A

"Confirm that liquid water and water vapor are in equilibrium when the temperature is 100. °C and the pressure is 1 atm. Data are available in Tables 4C.1 and 4G.1."

I'm not sure how to show that ΔStot = 0 at 100. °C

Relevant data from the two tables (maybe... again, I'm not really sure how to approach this question):
Sm° of liquid water at 100 °C = 86.8 J/(K x mol)
Sm° of water vapor at 100 °C = 196.9 J/(K x mol)
ΔHvap° of water at 100 °C = 40.7 kJ / mol

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

### Re: Self-test 4I.4A

So if you're going from gas to liquid, the reaction is H2O (g) -----> H2O (s) and if you wanted to write out the entropy of the reaction you could say it is the standard entropy of products - standard entropy of reactants. We're given those values:

196.9 J/molK - 86.8 J/molK = 110.1 J/molK and let's work with only one mole so we have 110 J/K

If we convert the ΔHvap° of water at 100 °C to ΔSvap°, we take the ΔHvap° and divide by temperature in Kelvin, which is 273+100=373K (we learned the equation ΔSvap°= (ΔHvap°)/ (boiling point in K)

40.7 kJ/mol to J/mol is 40700 J/mol and for one mole (we only worked with one mole above so we do the same here) we get 40700 J. Divide this by our boiling point in Kelvin (373) and we get 109.11 J/K which is very similar to the value of 110 J/K we got above. This shows that they are at equilibrium because their entropies are the same at this temperature.

Philip Lee 1L
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

### Re: Self-test 4I.4A

Why is it that when the two values are similar we can say that ΔStot = 0?
I thought that ΔStot = ΔSsys + ΔSsurr.
What are the system and surrounding in this problem?

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