Q and K


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Celio_G_Dis2C
Posts: 33
Joined: Fri Jan 11, 2019 12:16 am

Q and K

Postby Celio_G_Dis2C » Mon Jan 28, 2019 1:02 pm

What’s the difference between them seeing as they’re calculated the same way?

Meachelle_Lum_1I
Posts: 92
Joined: Fri Sep 28, 2018 12:24 am

Re: Q and K

Postby Meachelle_Lum_1I » Mon Jan 28, 2019 1:57 pm

Q is when the reaction is not at equilibrium, K is when the reaction is at equilibrium. If Q<K, the reaction shifts right to favor the products. If Q>K the reaction shifts left to favor the reactants. Most of the time it is Q that changes, not K. K is changed by temperature only.

Fanny Lee 2K
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

Re: Q and K

Postby Fanny Lee 2K » Mon Jan 28, 2019 2:36 pm

Q is used to compare to K when at equilibrium. It tells us which way the non-equilbrium solution will shift in order to reach equilibrium. When this state is reached, Q = K.

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

Re: Q and K

Postby Jennifer Su 2L » Mon Jan 28, 2019 9:18 pm

Q is called the "reaction quotient," and can be calculated anytime during the reaction. K is called the "equilibrium constant", and can be calculated only when the reaction has reached equilibrium. For a specific reaction (at a constant temperature and pressure), you can get different Q values depending on when during the reaction you calculate it and the initial concentrations of reactants/products, but you will only get one K value.

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

Re: Q and K

Postby inlovewithchemistry » Tue Jan 29, 2019 4:02 pm

Q is calculated with values that are at any point in the reaction. We can use our answer from Q and compare to to the constant K value to determine whether the reaction will proceed forward or reverse.

madisondesilva1c
Posts: 91
Joined: Fri Sep 28, 2018 12:16 am

Re: Q and K

Postby madisondesilva1c » Tue Jan 29, 2019 9:29 pm

As the Q and K value approach each other, does the rate at which products and reactants are converted ever decline or is the reaction a constant rate?

Mindy Kim 4C
Posts: 65
Joined: Fri Sep 28, 2018 12:25 am

Re: Q and K

Postby Mindy Kim 4C » Tue Jan 29, 2019 9:33 pm

As Q approaches K, the rate of the forward or reverse reaction might change so that the rates of the forward and reverse reactions are equal.

Yiting_Gong_4L
Posts: 69
Joined: Fri Sep 28, 2018 12:25 am

Re: Q and K

Postby Yiting_Gong_4L » Wed Jan 30, 2019 12:06 am

K is the constant at equilibrium while Q is approaching the equilibrium. Solving for Q and comparing it to the K constant can get you the direction in which the reaction is favoring.

Brianna Brockman 1F
Posts: 60
Joined: Fri Sep 28, 2018 12:23 am

Re: Q and K

Postby Brianna Brockman 1F » Wed Jan 30, 2019 6:14 pm

Since K is only for when it's at equilibrium, the value is only K IF it's at equilibrium. When it's not at equilibrium, it is Q.

Ashe Chen 2C
Posts: 31
Joined: Mon Jan 07, 2019 8:23 am

Re: Q and K

Postby Ashe Chen 2C » Sat Feb 16, 2019 9:59 pm

Q doesn't mean anything until it is compared to K. The comparison helps to determine which direction the reaction will proceed given time.

Lauren Huang 1H
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: Q and K

Postby Lauren Huang 1H » Tue Feb 26, 2019 11:17 pm

Q is when the reaction is not at equilibrium. K is when the reaction is at equilibrium. Q is compared to K to determine whether the reaction will shift to the left or right due to the Le Chatalier's Principle.

Mikka Hoffman 1C
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Re: Q and K

Postby Mikka Hoffman 1C » Wed Feb 27, 2019 7:07 pm

Q can be calculated at any point in the reaction, whereas K is only calculated at equilibrium

Pritish Patil 1K
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

Re: Q and K

Postby Pritish Patil 1K » Tue Mar 12, 2019 12:57 pm

Q is the concentrations of the products over the reactants when the equation is not at equilibrium; K is the concentration of the products over the reactants when the equation is at equilibrium.

jocelyntzeng
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: Q and K

Postby jocelyntzeng » Sat Mar 16, 2019 12:47 pm

Q is not always at equilibrium but it can be; it is most often used to show to which side the reaction will shift to


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