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### Equilibrium in relation to K and Q

Posted: 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.

### Re: Equilibrium in relation to K and Q

Posted: 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.

### Re: Equilibrium in relation to K and Q

Posted: 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."

### Re: Equilibrium in relation to K and Q

Posted: 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.

### Re: Equilibrium in relation to K and Q

Posted: 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.

### Re: Equilibrium in relation to K and Q

Posted: 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.