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Chemistry Question
In Chemistry, what does capital Q stand for and how does it relate to spontaneity and concentrations and chemical equilibrium (and conditions)? Thanks.
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Re: Chemistry Question
Q is used to test how close a system is to equilibrium. Set up your K-expression and plug in the concentrations. If Q=K, you are at equilibrium. If Q>K, then your product concentrations are too high, so the reaction will move in reverse to equilibrium. If Q<K, the reaction will move forward to equilibrium.
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Re: Chemistry Question
Yup, Q is just the temporary relation to K. I always like it because the way we learned to use it, to test hypothetical reaction concentrations, it was totally made up and didn't relate to anything real.
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Re: Chemistry Question
Oops, forgot to relate it to spontaneity. If Q determines the reaction should go forward, Gibbs free energy for the forward reaction is negative and therefore is spontaneous. If Q determines the reaction should go in reverse, Gibbs free energy for the forward reaction is positive and the forward reaction is nonspontanious. If Q=K, Gibbs free energy is zero and the system is at equilibrium
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Re: Chemistry Question
How do you find Q though? Does it relate to pressure or temperature or anything? Basically, what makes it different from K?
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Re: Chemistry Question
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Re: Chemistry Question
Alright, I guess I'll make an example.
In the system of acetic acid (HC2H3O2<----> H+ + C2H3O2-), Ka=1.8x10-5. If [HC2H3O2] = [H+] = [C2H3O2-] = 1.0 M a. Find Q b. Predict which direction the reaction will shift in order to reach equilibrium a. First, write the Ka expression, which will be [H+][C2H3O2-] [HC2H3O2] The expression will be the same as the Ka expression. So Q also equals [H+][C2H3O2-] [HC2H3O2] Plug in the values for concentrations given. This gives you (1x1)/1. Q=1 b. Since Q>Ka, the amount of products is too high (you can think of this as having to make Q smaller until it reaches Ka). Therefore, the reaction needs to shift to the reverse reaction until Q=Ka The same thing can be done with Kc and Kp in gaseous systems. Does this help? |
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