Qc vs Kc

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Bryce Ramirez 1J
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Joined: Sat Aug 24, 2019 12:16 am

Qc vs Kc

Postby Bryce Ramirez 1J » Sat Mar 14, 2020 8:30 pm

What are the rules for when Qc is greater than Kc and when Qc is less than Kc? And what happens when the two are exactly equal?

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

Re: Qc vs Kc

Postby Indy Bui 1l » Sat Mar 14, 2020 8:33 pm

If they are equal then the reaction is at equilibrium. If Q>K then the forward reaction or products are favored. The opposite is true for Q<K the reactants are favored.

Jared_Yuge
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

Re: Qc vs Kc

Postby Jared_Yuge » Sat Mar 14, 2020 8:33 pm

Q will tend to go towards Kc, so when Q >Kc it will shift towards the reactants and when Q<Kc it will shift towards the products.

chimerila
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Re: Qc vs Kc

Postby chimerila » Sat Mar 14, 2020 8:34 pm

Q > K, system shifts left
Q < K, system shifts right
Q = K, system is at equilibrium, no shift in either direction

Jared_Yuge
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

Re: Qc vs Kc

Postby Jared_Yuge » Sat Mar 14, 2020 8:35 pm

When Q and Kc are equal then the reaction has reached equilibrium and Q will stay the same. This doesn't mean that the reaction is no longer occurring rather that both the forward reaction and the reverse reaction are happening at the same rate

Jacey Yang 1F
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Re: Qc vs Kc

Postby Jacey Yang 1F » Sat Mar 14, 2020 8:36 pm

If Q>K, then the products are greater and the reverse reaction is favored. If Q<K, then reactants are greater and the forward reaction is favored.

Frank He 4G
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Joined: Tue Nov 12, 2019 12:19 am

Re: Qc vs Kc

Postby Frank He 4G » Sat Mar 14, 2020 8:50 pm

Since Q is calculated similarly, when Q is higher than K, we know that there are more products than reactants at Q. If there are more products than at equilibrium, then the reaction has to be proceeding in reverse to go to equilibrium. And the reverse is also true.

Adriana_4F
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Re: Qc vs Kc

Postby Adriana_4F » Sat Mar 14, 2020 8:56 pm

Q > K ----> Reactants are favored
Q < K -----> Products favored
Q = K -----> Equilibrium

Think about the concentration of reactants and products: When Q > K, the concentration (or activities) of products increase (numerator increases) causing the rxn to make more reactants

Gurmukhi Bevli 4G
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Re: Qc vs Kc

Postby Gurmukhi Bevli 4G » Wed Mar 18, 2020 1:20 am

For Q=K, the reaction is at equilibrium
For Q>K, the forward reaction is favored (towards the products)
For Q<K, the reverse reaction is favored (towards the reactants)

austin-3b
Posts: 54
Joined: Wed Nov 11, 2020 12:18 am

Re: Qc vs Kc

Postby austin-3b » Mon Jan 04, 2021 11:50 pm

When Qc is more than Kc, reactants would be formed to get that ratio back to equilibrium.
When Qc is less than Kc, products would be formed to get the ratio up to equilbrium.
When they are equal, it's at equilibrium.

Uyenvy Nguyen 1D
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Joined: Wed Sep 30, 2020 9:36 pm

Re: Qc vs Kc

Postby Uyenvy Nguyen 1D » Tue Jan 05, 2021 2:24 am

If Q < K, the forward reaction is favored. If Q > K, the reverse reaction is favored.

John Calonia 1D
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Joined: Wed Sep 30, 2020 9:49 pm

Re: Qc vs Kc

Postby John Calonia 1D » Tue Jan 05, 2021 4:20 pm

Q is just the same calculation as K, just at a certain point in time and does not necessarily have to be at equilibrium, correct?

Tanya Bearson 2K
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Joined: Thu Dec 17, 2020 12:18 am

Re: Qc vs Kc

Postby Tanya Bearson 2K » Tue Jan 05, 2021 5:21 pm

John Calonia 1D wrote:Q is just the same calculation as K, just at a certain point in time and does not necessarily have to be at equilibrium, correct?


Yes, that is correct. This makes it so the Q and K values are useful to compare like people explained in previous comments.

Kelly Tran 1J
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Re: Qc vs Kc

Postby Kelly Tran 1J » Tue Jan 05, 2021 6:03 pm

When Qc > Kc, the reverse reaction is favored (reactants are favored).
When Qc < Kc, the forward reaction is favored (products are favored).
When Qc = Kc, neither reactants nor products are favored.

Olivia Monroy 1A
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Joined: Wed Sep 30, 2020 10:00 pm

Re: Qc vs Kc

Postby Olivia Monroy 1A » Tue Jan 05, 2021 6:07 pm

When Q>K the system shifts left (the amount of products is greater (P/R) so to reach equilibrium it moves towards reactant)
When Q<K the system shifts right (amount of reactant is greater than product so to reach eq moves toward products)
When Q=K the neither are favored, Q is considered K (the equilibrium constant)
this applies for both Qc/Qp and Kc/Kp

DMaya_2G
Posts: 97
Joined: Wed Sep 30, 2020 9:58 pm

Re: Qc vs Kc

Postby DMaya_2G » Tue Jan 05, 2021 6:27 pm

Q<K = Reaction is going to the product side (right)- forward direction.
Q>K = Reaction is going in the reverse direction, when making too many products.


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