Equilibrium Favoring

Moderators: Chem_Mod, Chem_Admin

KatelinTanjuaquio 1L
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

Equilibrium Favoring

Postby KatelinTanjuaquio 1L » Thu Jan 24, 2019 11:35 pm

I am working on Module #1 and came across this problem:

For the unbalanced reaction, CO (g) + H2 (g) ⇌ CH3OH (g), the equilibrium concentrations are [CO] = 0.0911 M, [H2 ] = 0.0822 M, [CH3OH] = 0.00892 M. What is the value of the equilibrium constant? Does the equilibrium favor reactants or products?

I understand how the equilibrium constant is calculated. However, what I do not understand is how equilibrium favors reactants or products. How can we tell which side is preferred when we are only given equilibrium concentrations, as opposed to when we can calculate the reaction quotient and determine favorability from there?

Ariel Cheng 2I
Posts: 67
Joined: Fri Sep 28, 2018 12:29 am

Re: Equilibrium Favoring

Postby Ariel Cheng 2I » Thu Jan 24, 2019 11:44 pm

I think that the question is asking what the reaction favors when it is at equilibrium (as opposed to when we use Q to see which direction the reaction will go when it is not at equilibrium). I do not remember the exact numbers but I believe in the first few lectures Dr. Lavelle talked about the reaction "sitting" to the right or left. If Kc is really large meaning that there is a greater amount of products (remember that Kc = [products]/[reactants]), then the equilibrium favors products and "sits" to the right. If Kc is really small meaning that there are more reactants, then equilibrium favors reactants and "sits" to the left.

Phan Tran 1K
Posts: 29
Joined: Fri Sep 28, 2018 12:19 am

Re: Equilibrium Favoring

Postby Phan Tran 1K » Fri Jan 25, 2019 1:38 pm

A reaction favoring either side of an equation is really just saying will the reaction proceed in the forwards or the backwards direction. You know this by calculating Q and then comparing it to K. Since in these quotients are a fraction with products on the top and reactants on the bottom, Q being smaller than K means that more products need to be produced to increase Q to match K and the reaction will proceed accordingly, and vice versa.

Ethan Breaux 2F
Posts: 63
Joined: Sat Sep 29, 2018 12:16 am

Re: Equilibrium Favoring

Postby Ethan Breaux 2F » Fri Jan 25, 2019 3:27 pm

The post-module assignment just had it that:

K > 1 means that there are more products at equilibrium (lies to the right)
K < 1 means that there are more reactants at equilibrium (lies to the left)
K = 1 means neither are favored.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 12 guests