Welcome Guest Search | Active Topics |

Tag as favorite
Equilibrium constant
Shahd_6465
#1 Posted : Friday, May 21, 2021 8:04:10 PM
Rank: Member

Groups: Registered
Joined: 5/3/2021
Posts: 20

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
If we add more reactants to a reaction at equilibrium, I know that the concentration of reactants immediately increases and then decreases to form products according to Le Chatelier's Principle, but I am not sure if in the Keq equation, the concentrations of reactants and products are different (I know their ratio is constant because Keq is constant, but I want to know if the concentrations change).

Thank you
INSTR_Katrina_128
#2 Posted : Friday, May 21, 2021 10:05:01 PM
Rank: Advanced Member

Groups: Registered
Joined: 5/18/2021
Posts: 73

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Great question, Shahd.

We can "technically" only use the Keq expression to relate the concentrations of reactants and products when the system is AT equilibrium.

When we know that a system is NOT at equilibrium, or if we want to test and see if whether or not a system is at equilibrium, we use a Q expression. I remember it as Q for "question: are you at equilibrium?"

Here's the annoying thing: the expression looks IDENTICAL to the Keq expression. But its meaning is very different!!!

For example:

For the equilibrium reaction 2NO(g) + Cl2(g) <---> 2NOCl(g), we get the equilibrium expression

Keq = [NOCl]²/([NO]²[Cl2]) = 2.2 x 10³ (this number is from literature and would be given to you)

Now, if you change the concentrations of any of the species, you are no longer at equilibrium. You cannot use the Keq expression anymore. You need to use a Q expression, which is

Q = [NOCl]²/([NO]²[Cl2])

Hey, isn't that the same formula? ALMOST, but not quite. Notice that the Q expression ISN'T EQUAL TO 2.2 x 10³. Instead, the Q expression will be equal to another value.

Then, you will COMPARE that value to the value of Keq, and make a conclusion.

If Q > Keq, then the system is "overbaked" (like a cake), and you have added too many products. The system will now shift towards the reactants side.

If Q = Keq, you are at equilibrium!

If Q < Keq, then the system is "underbaked" (again, like a cake), and you have added too many reactants. The system will now shift towards the products side.

I hope this explanation helps. 😄
Users browsing this topic
Guest (2)
Tag as favorite
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.

Clean Slate theme by Jaben Cargman (Tiny Gecko)
Powered by YAF | YAF © 2003-2009, Yet Another Forum.NET
This page was generated in 0.068 seconds.