# 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.

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.

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.

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?

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.

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?