## K vs. Q in experiments

Rachael Cohen 3G
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Joined: Sun Dec 13, 2020 12:17 am

### K vs. Q in experiments

When he introduced Q in the lecture, and the comparisons you can make between Q and K, Dr. Lavelle gave an example of performing an experiment on a lake, where you do a reaction in the lake and measure the concentrations. He said you might find, based on Q and K, that what you thought was the reactant is actually the product. Wouldn't this make K and Q, and the products and reactants since the equilibrium reaction is going both ways, arbitrary? Does it really matter which is which?

Chem_Mod
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### Re: K vs. Q in experiments

The expression for K is defined to be for a specific direction of a reaction- if the reaction is shown as $A+B\rightleftharpoons C$, then the expression for K is $K=[C]/([A][B])$
If the reaction is shown in the opposite direction, as $C\rightleftharpoons A+B$, then the expression is K-1, $K^{-1}=[A][B]/[C]$
So yes, the direction the equation is written is arbitrary, but the expression for K is dependent on what direction the equation is that you're looking at.

Samudrala_Vaishnavi 3A
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### Re: K vs. Q in experiments

So to put it simply, K is when the reaction or system is at equilibrium and the concentrations of the reactants and products can be anything but they don't change further (for instance product concentrations can be greater than reactants or vice versa). What is important to note is that the concentrations of the reactants and products, in this case, are what they are at equilibrium (concentration composition at equilibrium). Q is the concentration composition of the reactants and products when the system hasn't reached equilibrium or K yet. Maybe something was added or the system was perturbed and the system hasn't adjusted yet. K is the long term, Q is the short term. We can therefore use Q to predict what actions the system will take to return to equilibrium. ex.) If Q is less than K, then more products will be eventually formed so Q=K.