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Topic: Why does water affect equilibrium?  (Read 4631 times)

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Offline purplecinnamon

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Why does water affect equilibrium?
« on: February 18, 2016, 12:35:26 AM »
Hello! So I want to know why adding water to an eqalibrim reaction does change our Q value. This is what my book says:

2 H2O(l)  2 H2(g) + O2(g) .. Kc = [H2]2 [O2]
"Adding H2O(l) will have no effect on the equilibrium because the reactant is a liquid. Notice that the concentration of H2O(l) is not a part of the equilibrium constant expression. Since the concentration of product gases do not change when H2O(l) is added, Q is still equal to K and there is no driving force for the system to change. As long as some liquid is present to establish the equilibrium, adding more will have no effect on the concentration of products."

Any ideas?

Thank you for your time.

Offline clinz63

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Re: Why does water affect equilibrium?
« Reply #1 on: February 18, 2016, 01:10:02 AM »
How do you calculate the concentration of water? Moles of water divided by volume of water? Does this change as the reaction progresses? I think you can safely say it remains constant since the volume changes as the water reacts. So it doesn't really matter how much water you add since it always has the same concentration.

Offline Borek

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Re: Why does water affect equilibrium?
« Reply #2 on: February 18, 2016, 03:51:23 AM »
This is a poorly thought of explanation. Liquid water doesn't occupy whole volume (whole as in volume occupied by products and reactants, reaction mixture), so its concentration is different in different places. Then, the water evaporates, so the reaction equation is in fact

2H2O(g) :lequil: 2H2(g) + O2(g)

In this case it is is general the same situation as with solids - we don't care about their amount, as long as they are present.
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