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You cannot use values but must determine from looking The equation I have trouble with is as follows:

$$\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2}$$

According to my teacher the entropy is positive. I also got the entropy to be positive, but now that I think of it, the moles is decreasing and one is becoming aqueous. So does turning into a gas trump everything?

What if NaOH was solid? Would it mean still that the entropy would be positive?

Also finally according to the textbook the entropy values is as follows:

$$\ce{s<l<<g}$$

But where would the (aq) be?

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Aqueous salts will always have less order than solid salts. There is much less order when ions are floating around in solution than when they are in their rigid lattice structures. This is what I think you are failing to consider. The reaction is not so much:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2NaOH_{(aq)} + H2_{(g)}}$$

But really:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2Na^{+}_{(aq)} +2OH^{-}_{(aq)} + H2_{(g)}}$$

What results from the reaction of 2 atoms of a very structured metal and 2 molecules of liquid water, is 4 aqueous ions that are free to move within the solution, and one molecule of gas. Not only are there now more species present, but they are also in a relatively more free states than before.

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  • $\begingroup$ Yeah thanks, I was thinking about that dissociation. $\endgroup$ – Asker123 Apr 15 '15 at 1:32

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