Equilibrium in relation to K and Q

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Rogelio Bazan 1D
Posts: 64
Joined: Tue Nov 14, 2017 3:01 am

Equilibrium in relation to K and Q

Postby Rogelio Bazan 1D » Sun Jan 13, 2019 10:57 pm

I am reviewing the modules for equilibrium under Dr. Lavelle's website and I was hoping that I could get some help understanding what is the difference when solving for both Q and K while working on an equilibrium equation. I know that to solve for K you put products over reactants.

Example) aA+bB⇋cC+dD K=[C]c[D]d/[A]a[B]b

To clarify is solving for Q the same as solving for K?

Thank you.

Rebecca Altshuler 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

Re: Equilibrium in relation to K and Q

Postby Rebecca Altshuler 1D » Sun Jan 13, 2019 11:07 pm

Solving for Q and solving for K are the same, K just indicates that the reaction is at equilibrium.

AditiSax
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: Equilibrium in relation to K and Q

Postby AditiSax » Sun Jan 13, 2019 11:16 pm

K is solved using concentrations of reactants and products at equilibrium. Q is solved using concentrations of reactants and products when the system is not necessarily in equilibrium, and can be compared to K to determine in which direction the equilibrium will "sit."

Lia Inadomi 1I
Posts: 62
Joined: Fri Sep 28, 2018 12:15 am

Re: Equilibrium in relation to K and Q

Postby Lia Inadomi 1I » Tue Jan 15, 2019 11:54 am

Solving for Q is the same as solving for K in regards to setting up the problem. The only difference is that K is when the reaction is at equilibrium whereas Q is at any arbitrary stage in the reaction.

Casandra
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

Re: Equilibrium in relation to K and Q

Postby Casandra » Tue Jan 15, 2019 8:59 pm

Calculating the values of K and Q requires the same process of solving. The only difference is that K tells you the equilibrium of the reaction, whereas Q tells you where you are on the path to equilibrium. This is why Q is useful when comparing it to K.

Samantha Chang 2K
Posts: 69
Joined: Fri Sep 28, 2018 12:17 am

Re: Equilibrium in relation to K and Q

Postby Samantha Chang 2K » Wed Jan 16, 2019 12:46 pm

Solving for Q is the same as solving for K. When Q is less than K, forward reaction is favored. When Q is greater than K, reverse reaction is favored.


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