Chemistry Question

Other Math and Science
1

In Chemistry, what does capital Q stand for and how does it relate to spontaneity and concentrations and chemical equilibrium (and conditions)? Thanks.

2

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.

3

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.

4

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?

5

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

6

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.

7

Alright, I guess I’ll make an example.

In the system of acetic acid (HC[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]<----> H+ + C[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]-), K[sub]a[/sub]=1.8x10-5. If [HC[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]] = [H+] = [C[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]-] = 1.0 M

a. Find Q
b. Predict which direction the reaction will shift in order to reach equilibrium

a. First, write the K[sub]a[/sub] expression, which will be
[H+][C[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]-]
[HC[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]]

The expression will be the same as the K[sub]a[/sub] expression. So Q also equals
[H+][C[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]-]
[HC[sub]2[/sub]H[sub]3[/sub]O[sub]2[/sub]]

Plug in the values for concentrations given. This gives you (1x1)/1.

Q=1

b. Since Q>K[sub]a[/sub], the amount of products is too high (you can think of this as having to make Q smaller until it reaches K[sub]a[/sub]). Therefore, the reaction needs to shift to the reverse reaction until Q=K[sub]a[/sub]

The same thing can be done with K[sub]c[/sub] and K[sub]p[/sub] in gaseous systems.

Does this help?

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