Help on 11.13

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Joanna Pham - 2D
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Joined: Fri Apr 06, 2018 11:04 am

Help on 11.13

Postby Joanna Pham - 2D » Mon Jan 07, 2019 3:09 pm

For this question, we are asked to find the reaction quotient Q. For part b, the equation is and the answer is .

I am confused why we still need to take aqueous solutions into account. In lecture today, Professor Lavelle said we do not take solids and liquids into account. How do we know the H_2S and H_3PO_4 molecules are not liquids or solids? Should we always assume aqueous solutions are in gas phase?

Neil Hsu 2A
Posts: 61
Joined: Fri Sep 28, 2018 12:16 am

Re: Help on 11.13

Postby Neil Hsu 2A » Mon Jan 07, 2019 3:14 pm

When considering substances in your equilibrium expression, always include aqueous and gas phase molecules while excluding solids and liquids. Aqueous solutions are not pure liquids, rather, they are the substance dissolved in water. Therefore, they have concentrations that change with the reaction which is why we consider them in our equilibrium expression. Note that the water that the substances are dissolved in are not included because it is a pure solvent and the changes to it are negligible. The pure solid is also not included because it does not have concentration.

Miriam Sheetz 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:25 am

Re: Help on 11.13

Postby Miriam Sheetz 2B » Mon Jan 07, 2019 9:33 pm

Since the reactants are a solid and liquid the K value will just be the concentrations of the products multiplied together. You do not include pure solids or liquids are in the equilibrium expression because their concentrations stay constant throughout the reaction. The density of a pure liquid or solid remains the same, regardless of how much is present in the rxn.


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