Chemistry Question

[Daniel Brim]

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]

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]

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]

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

[Daniel Brim]

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.