Q and K
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 100
- Joined: Wed Sep 30, 2020 10:08 pm
Q and K
I was wondering if someone could clarify this, we use Q when we know that [P] and [R] are not at equilibrium right?
-
- Posts: 86
- Joined: Wed Sep 30, 2020 10:11 pm
Re: Q and K
Yes, that's correct. Q can be used to see which direction the reaction will go to reach equilibrium. If Q<K it will proceed forward, but if Q>K it will go backwards and form more reactants.
-
- Posts: 95
- Joined: Thu Oct 08, 2020 12:16 am
Re: Q and K
Yes, that is correct. K is just Q when the products are at equilibrium, so you can use Q as a measure of determining which way the reaction will move. When K>Q, the reaction will shift towards the products, and if K< Q, the reaction will shift towards the reactants.
-
- Posts: 101
- Joined: Wed Sep 30, 2020 9:39 pm
- Been upvoted: 3 times
Re: Q and K
Everything above is true: I would also just add on that if we are not sure that the reaction is at equilibrium but we are given K, I believe we can also have Q = K and prove the reaction is at equilibrium, since the process for finding Q is the same as the process for finding K. :)
-
- Posts: 97
- Joined: Wed Sep 30, 2020 10:00 pm
Re: Q and K
The process of finding Q and K is the same, if you calc the concentration ratio and find it doesn't equal K then you know it is the reaction quotient and is not yet at equilibrium (either a forward or reverse reaction will exist). You use Q when you are not at equilibrium and you are then able to determine the direction of the reaction by comparing Q and K.
-
- Posts: 86
- Joined: Wed Sep 30, 2020 9:41 pm
Re: Q and K
When Q<K, the reaction has a tendency to proceed toward products. When Q=K, the mixture has its equilibrium composition and has no tendency to change in either direction. When Q<K, the products tend to decompose into the reactants.
-
- Posts: 91
- Joined: Wed Sep 30, 2020 9:33 pm
Re: Q and K
How would you be able to calculate K if you're only given concentrations for Q? They would typically give K to you right?
-
- Posts: 86
- Joined: Wed Sep 30, 2020 9:31 pm
Re: Q and K
Hey Melissa,
That is correct! We use Q when calculating the reaction quotient of a reaction that isn't at equilibrium. Because Q is calculated the same way as K ([P]/[R]), then if the reaction is at equilibrium the ratio of the concentrations of products to reactants would just be equal to K. If the reaction is not at equilibrium, then you will get a value that is not equal to K and is instead called Q. Hope this helps!
That is correct! We use Q when calculating the reaction quotient of a reaction that isn't at equilibrium. Because Q is calculated the same way as K ([P]/[R]), then if the reaction is at equilibrium the ratio of the concentrations of products to reactants would just be equal to K. If the reaction is not at equilibrium, then you will get a value that is not equal to K and is instead called Q. Hope this helps!
-
- Posts: 91
- Joined: Mon Oct 05, 2020 12:16 am
-
- Posts: 83
- Joined: Wed Sep 30, 2020 9:55 pm
Re: Q and K
Hi! As stated above, K is when the system is at equilibrium and Q is when it is not. So, if you are given the K and asked to find Q and then Q=K, you'd know the system is at equilibrium. Q can also be used to find out whether the reactants or the products are favored.
-
- Posts: 88
- Joined: Wed Sep 30, 2020 9:59 pm
Re: Q and K
This is correct. Q is a value of the ration of product to reactant at some point in the reaction, but not at equilibrium. You can have Q as a result of not waiting long enough for equilibrium to be reached. Furthermore, the inequality between Q and K could be a result of different temperatures that the reactions took place in.
Re: Q and K
Yes. Q is used as a tool to decipher whether the system is at equilibrium, tends towards the product side of the reaction (Q>K), or tends towards the reactant side of the reaction (Q<K).
-
- Posts: 85
- Joined: Wed Sep 30, 2020 9:33 pm
-
- Posts: 87
- Joined: Wed Sep 30, 2020 10:03 pm
Re: Q and K
Yep when P and R do not give you the K value(when you plug the numbers into the equation) then the reaction is not at equilibrium and we refer to it as Q instead of K.
Return to “Equilibrium Constants & Calculating Concentrations”
Who is online
Users browsing this forum: No registered users and 2 guests