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[chm] Equilibrium

Discussion in 'DAT Discussions' started by ippie, Jun 5, 2008.

  1. ippie

    ippie ippie

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    #1.
    NO (g) + 1/2 O2 (g) --> NO2 (g)
    what happens if the size of the vessel decreased?
    of course, the equilibrium would shift to the right.
    but how about K(eq)???
    If the equilibrium shifts to the right, it means NO and O2 decrease, and NO2 increases. right? then, K(eq) is increases becuase K(eq) is product over reactant. Product increase and reactant decrease result in K(eq) increase.
    Am I worng?
    but why K(eq) doesn't change???

    #2.
    150ml of a 3M KCl solution is diluted with 500ml of distilled water. the resulting solution is stirred and 300ml are poured out of the beaker. what is the molarity of the remaining solution?
    (150*3)/650 is right?
    it's nothing with the information of 300ml pouring out of the beaker. right?

    #3.
    Al(OH)3, HCl, NaOH
    all are water soluble. right?
     
    #1 ippie, Jun 5, 2008
    Last edited: Jun 5, 2008
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  3. arpitpatel86

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    for number2 you are definatley right

    number 3 i thin your right not sure tho

    NUMBER 1 I WAS WONDERING THE SAME THING WHEN I WENT THROUGH THIS PROBLEM neone can help?
     
  4. creative8401

    creative8401 Im Anush Hayastan

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    Hi. I’m going to address your first question (kinda tired right now), but here it goes. You are right, when we decrease the size of the vessel, the equilibrium would shift to the right. But from what I am getting is that you are confusion two properties: equilibrium position and equilibrium constant. When a reaction conditions changes, such as changes in pressure, concentration, volume, etc, the position of the equilibrium would change. So yes, as you say it, the product would increase and the reaction would decrease. Great! But where are you coming short? What are you are saying is that the equilibrium position has changed: an increase in one and a decrease in the other, but the equilibrium constant is the same. An increase in either the products or reactants would cause the opposite change in the other: inc in products, dec in react, vice versa. As a result, even though we have changed the equilibrium position, the equilibrium constant will stay the same. Put a post-it somewhere you always see: equilibrium constant changes with temperature because we are given new conditions that are specific to a unique equilibrium constant.
     
  5. osimsDDS

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    Let me put it simply,

    Pressure, Volume, concentration, catalyst, and Temperature and whatever else im missing will shift the Keq (favoring products or reactants in regards to Le Chatliers Principle)...

    BUUUUUTTTT.....

    The only thing that changes the equilibrium constant Keq itself is Temperature, nothing else...
     
  6. baedero1

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    #3. Al(OH)3 is insoluble
     

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