[katrina007]

Hi again,

Need help with one more question if anyone don't mind. Any help is appreciated. Thanks in advance.

Arrange these three processes in order of their decreasing tendencies toward spontaneity.

I.   H2O (l) --> H2O (g) at 105 C degrees, 1.00 atm
II.  H2O (l) --> H2O (s) at 25 C degrees, 1.00 atm
III. H2O (l) --> H2O (s) at 0 C degrees, 1.00 atm

Select one answer
a) I > III > II
b) II > III > I
c) I > II > III
d) III > II > I
e) II > I > III

EDIT:
I think the answer is (a) because gas is more spontaneous than solids. And then we look at degrees, and since III is  zero degrees that comes second and then the solid.

[Borek]

Sorry for directly postingi the questions without any attempt at it first. It wont happen again.

[katrina007]

Sorry. I forgot to give in my input of what I think of the question.
I editted. Please let me know if im wrong or correct.

[enahs]

Could you look at your question and check II.
Maybe I am misinterpreting what the question says, but you can not have solid water (Ice) at 25^oC at 1.00 atm. You can have Ice at 25^oC if you change the pressure; but that is not what this is saying.


So then yes the answer would be A, as II is impossible.

[Yggdrasil]

(A) is the correct answer, but I'm not sure I understand your explanation.  Could you explain your reasoning a little bit more?

[ARGOS++]

Dear Katrina007,

(Sorry, dear Yggdrasil, but I can understood, if Katrina007 has made their decision out, more of her stomach, then of real knowledge of the required “Phase-Diagram” of Water.)

To answer the question, you must first know the so called “Phase-Diagram” of Water, and realise at the same time, that you have to keep the pressure all times constant at 1.0 atm.
That means that you have to draw a line into the graph at this const. pressure and make the required proof for your answers.

Now you can easily see, that at 105°C only H₂ O_(g) is existing on your Line, and that means, if you are able to bring H₂ O_(l) to 105°C, it will dramatically (mini explosion like) change to the gaseous state, at least as long as enough energy is usable.
On the other hand, you will see too, that it is impossible to generate ice under this condition (25°C), because the first time ice can exist on your Line is just at 0°C (or below), where your line is crossed from the phase-line of water-ice, which tells you that at that temperature both aggregate states exist at the same time.

Now it’s only left to you to correctly classify/determine the resulting Spontaneity of the postulated “reactions” of your question.
I hope it’s not too difficult anymore.

Good Luck!
                    ARGOS⁺⁺