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K vs Q

Posted: Thu Jan 10, 2019 7:23 pm
by Karyn Nguyen 1K
What's the difference between the formula for the equilibrium constant K and the formula for the reaction quotient Q?

Re: K vs Q

Posted: Thu Jan 10, 2019 7:48 pm
by ariana_apopei1K
Q changes as a reaction occurs, and K is when said reaction reaches equilibrium. If the reaction does't reach equilibrium, you can't use K, only Q

Re: K vs Q  [ENDORSED]

Posted: Thu Jan 10, 2019 7:49 pm
by Chloe Likwong 2K
The formula is the same. What's different between the two is the concept behind it. The equilibrium constant K is the ratio at equilibrium; it describes a reaction that is at equilibrium. On the other hand, the reaction quotient pertains to the ratio at any given time of the reaction.

This might elucidate the topic further: viewtopic.php?f=50&t=24065

Re: K vs Q

Posted: Thu Jan 10, 2019 7:50 pm
by Matthew Tran 1H
The formula for K and Q is the same; the difference is that the numbers you plug in for K must be the equilibrium partial pressures or concentrations while for Q you can plug in the partial pressures or concentrations at any point in the reaction, not just at equilibrium.

Re: K vs Q

Posted: Thu Jan 10, 2019 8:47 pm
by Aleeque Marselian 1A
The formula to find K and Q are the same. However, K is the ratio of at equilibrium and therefore includes the values of the reaction at equilibrium. Meanwhile, Q is the reaction quotient and is the ratio at any given time in the reaction.

Re: K vs Q

Posted: Sun Jan 13, 2019 11:39 am
by vaishali 1D
The formula is the same, but K is the ratio at equilibrium, whereas Q is the reaction quotient. The numbers you plug in decide whether you're using K or Q.

Re: K vs Q

Posted: Sun Jan 13, 2019 8:43 pm
by Jchellis 1I
So can there be only one value for K but multiple values for Q as the reaction progresses?

Re: K vs Q

Posted: Sun Jan 13, 2019 8:49 pm
by allisoncarr1i
The partial pressures and concentrations do not have to be at equilibrium for Q but they do for K. Other than that, it's the same equation.