K vs Q

Moderators: Chem_Mod, Chem_Admin

Melody Haratian 2J
Posts: 108
Joined: Wed Sep 30, 2020 9:48 pm
Been upvoted: 1 time

K vs Q

Postby Melody Haratian 2J » Sun Jan 10, 2021 9:31 am

Hi guys!
Do you use Q when you are unsure if the reactants/products are at equilibrium? Do you also use Q if the problem doesn’t explicitly state if the reactants and products have reached equilibrium?
Thanks!

Shalyn Kelly 3H
Posts: 83
Joined: Wed Sep 30, 2020 9:55 pm

Re: K vs Q

Postby Shalyn Kelly 3H » Sun Jan 10, 2021 9:41 am

Hi! Coming from another post about this, I think K is when the system is at equilibrium and Q is when the system is still reacting (you can also use it to find which side is favored). So, if Q=K then the reactants and products should have reached equilibrium. Also, I believe the questions will ask for which one they want you to find/use? Otherwise, my thought process would be to do as you say and use Q first.

Margia Adriano 2A
Posts: 84
Joined: Wed Sep 30, 2020 10:00 pm

Re: K vs Q

Postby Margia Adriano 2A » Sun Jan 10, 2021 10:35 am

Hi there! In general, yes you would use Q when you aren't sure if the system has reached equilibrium yet. Most of the time, I think the problem would state what's going on with the system, but since you find Q and K the same way (products over reactants) then I think you should be all right. You can also think of Q as what is going on in the system right now, and K as what the system's end goal looks like. Hope this helps!

Isabel_Eslabon_2G
Posts: 89
Joined: Wed Sep 30, 2020 9:50 pm

Re: K vs Q

Postby Isabel_Eslabon_2G » Sun Jan 10, 2021 10:39 am

Hello.

I'm pretty sure both K & Q are found by dividing product concentrations (to the power of their coefficients) by the reactant concentrations (to the power of their coefficients).

The difference is that K uses equilibrium concentrations. When you are given concentrations of reactants & products, but the reaction is not at equilibrium, that gives you Q.

Lung Sheng Liang 3J
Posts: 85
Joined: Wed Sep 30, 2020 9:33 pm

Re: K vs Q

Postby Lung Sheng Liang 3J » Sun Jan 10, 2021 11:48 am

If the problem doesn't explicitly say it's at equilibrium, we should generally use Q, the reaction quotient. However, we use K when the problem states that it is at equilibrium. If they equal each other, this means that the reaction is a equilibrium.

Giselle Granda 3F
Posts: 104
Joined: Wed Sep 30, 2020 10:00 pm

Re: K vs Q

Postby Giselle Granda 3F » Sun Jan 10, 2021 12:08 pm

K and Q are calculated in the same way (products over reactants with the exclusion of liquids and solids). The difference is that K is the equilibrium constant, meaning that it is only used with concentrations that are at equilibrium, while Q is used with concentrations in general. Therefore, Q can be used to determine if a reaction is at equilibrium. If you are given K and the concentrations of the reactants and products in a reaction, you solve for Q. Then comparing Q and K, if Q=K the reaction is at equilibrium; if Q<K the reaction is not at equilibrium, and the reaction proceeds towards the products; if Q>K, the reaction is also not at equilibrium, and the reaction proceeds towards the reactants. Hope this is helpful!

Joel Meza 3I
Posts: 86
Joined: Wed Sep 30, 2020 9:56 pm

Re: K vs Q

Postby Joel Meza 3I » Sun Jan 10, 2021 12:09 pm

Q is the ratio of products to reactants during any time in a reaction, so I would say to use Q when you are unsure. But I would think that a question in this course would give you the K value so you wouldn't have to be unsure. If you're in a lab environment and don't have a known K value, then you could just take the Q value at a later time and keep doing that until you notice that it doesn't change, thus you can conclude that is the K value for the reaction at equilibrium.

Kat Stahl 2K
Posts: 87
Joined: Wed Sep 30, 2020 10:03 pm

Re: K vs Q

Postby Kat Stahl 2K » Sun Jan 10, 2021 12:24 pm

Yes you would use Q if you are unsure that the reaction is at equilibrium.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest