K=1

Moderators: Chem_Mod, Chem_Admin

Olivia Young 1A
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

K=1

Postby Olivia Young 1A » Thu Jan 10, 2019 10:55 am

Is there a certain significance to when K is equal to 1? In the lecture it was discussed that it is rare when K = 1, but I was not sure if it meant anything specific concerning reactants and products.

Ashley Bouma 1F
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

Re: K=1

Postby Ashley Bouma 1F » Thu Jan 10, 2019 11:08 am

When K equals one the concentration of products and reactants are equal, which is very rare. The only significance of K equalling one is that we can use it as a marker and say a K above one favors products and a K below one favors reactants

Cynthia Ulloa
Posts: 52
Joined: Fri Feb 23, 2018 3:02 am

Re: K=1

Postby Cynthia Ulloa » Thu Jan 10, 2019 11:08 am

When K = 1, this means that the reaction has reached equilibrium. So, the amount of reactants and products would be the same.

Courtney Quan 1C
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am
Been upvoted: 2 times

Re: K=1

Postby Courtney Quan 1C » Thu Jan 10, 2019 11:11 am

As mentioned above, when K=1, we can derive that the amounts of concentration of the products and reactants are equal at equilibrium. Such an occurrence indicates that there is equal stability in both the reactants and products and the reaction favors neither reactants nor products at equilibrium.

charlotte_jacobs_4I
Posts: 63
Joined: Fri Sep 28, 2018 12:29 am

Re: K=1

Postby charlotte_jacobs_4I » Thu Jan 10, 2019 11:12 am

I think about like a proportion. So if there are 3 particles of reactant there has to be 3 particles of products. Dividing those numbers would equal 1.

Cynthia Aragon 1B
Posts: 47
Joined: Mon Apr 09, 2018 1:38 pm

Re: K=1

Postby Cynthia Aragon 1B » Thu Jan 10, 2019 11:45 am

If K is about equal to 1, the reaction will reach equilibrium as reaction intermediate meaning that the amounts of products and reactants will be equal in amount.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 17 guests