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sanddrag 08-06-2005 21:34

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.

Daniel Brim 08-06-2005 21:50

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.

Greg Marra 08-06-2005 22:01

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.

Daniel Brim 08-06-2005 22:25

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

Does this help?

sanddrag 08-06-2005 22:32

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?

Daniel Brim 08-06-2005 22:36

Re: Chemistry Question
 
Quote:

Originally Posted by sanddrag
How do you find Q though? Does it relate to pressure or temperature or anything? Basically, what makes it different from K?

You can use pressure or concentration. It is different from K because it can be found at any point, given pressure or concentration. The expression for Q is the same, but the pressures or concentrations given do not have to be at equilibrium. Q determines how the system will get to equilibrium.

Daniel Brim 09-06-2005 00:09

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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