Q and K

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Samantha Lee 1A
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Q and K

Postby Samantha Lee 1A » Wed Jan 06, 2021 11:39 am

I have an application question regarding Q and K. If Q and K are the same (at equilibrium) and we add more of either reactants or products, causing the reaction to not be at equilibrium anymore, will the K value still remain the same and we would just need to find Q, or would both Q and K change since more of either reactants or products were added?

Libby Dillon - 1A
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Re: Q and K

Postby Libby Dillon - 1A » Wed Jan 06, 2021 11:44 am

K does not change, so in this situation you would have to calculate Q again to see how adding more products or reactants affected the reaction quotient. If Q comes out to be lower than K, then the forward reaction is favored toward the products. If Q is greater than K, the reverse reaction is favored toward the reactants.

Stuti Pradhan 2J
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Re: Q and K

Postby Stuti Pradhan 2J » Wed Jan 06, 2021 11:46 am

K only changes due to a change in temperature. In this situation, adding more reactants or products would change Q because the concentrations are no longer at equilibrium. You could then make an ICE table and find the new equilibrium concentrations, which would still have the same ratio and therefore, the same K value.

Hope this helps!

Juwon Lim 2A
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Re: Q and K

Postby Juwon Lim 2A » Wed Jan 06, 2021 11:48 am

To calculate K, you would have to wait until the reaction reaches equilibrium, and K does not change in this situation. However, Q will change because we are inputting new concentrations for products and/or reactants.

Becca Nelson 3F
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Re: Q and K

Postby Becca Nelson 3F » Wed Jan 06, 2021 12:05 pm

In cases of adding more reactants or products, K does not change, but Q will change as you are no longer in equilibrium. If you add more products Q will now be greater than K and if you add more reactants Q will now be less than K

Karl Yost 1L
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Re: Q and K

Postby Karl Yost 1L » Wed Jan 06, 2021 12:09 pm

Q tells us about the relative concentrations of products and reactants at a given time. When Q = K, the system is at equilibrium. So if we add more reactant or product to a system at equilibrium, the value of Q will temporarily change until equilibrium is restored (i.e. Q = K). The value of K will only change with temperature.

Emmeline Phu 1G
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Re: Q and K

Postby Emmeline Phu 1G » Wed Jan 06, 2021 12:14 pm

Hi! The equilibrium constant K will always remain the same for a particular reaction. However, in the case that we add either more reactants and products when the reaction is already at equilibrium, then we would have to recalculate Q because concentrations at any time during the reaction can be used to calculate Q. After recalculating Q, compare it to K to determine which direction the reaction will proceed. Hope this helps! :)

Nadya Higgins 3F
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Re: Q and K

Postby Nadya Higgins 3F » Wed Jan 06, 2021 12:28 pm

K will not change because no matter the initial conditions, the ratio should be the same. Q will be different just because you changed the amount of product/reactant at the time, throwing the reaction out of equilibrium but it will eventually return to the value of K.

Eileen Quach Dis 2A
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Re: Q and K

Postby Eileen Quach Dis 2A » Wed Jan 06, 2021 1:38 pm

Only Q would change since the equilibrium constant,K, will always be the same amount.

Namita Shyam 3G
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Re: Q and K

Postby Namita Shyam 3G » Wed Jan 06, 2021 1:53 pm

The reaction quotient, Q, is different from K, even though we are still dividing products by reactants. This is because the P and R values in Q are not concentrations at equilibrium. The only way K will ever change is if we change the temperature. We use Q mainly to find what the ratio is when something is not in equilibrium. So for example, if we add more products or reactants to one side, K will not change---but we will be able to find a Q value. After adding products or reactants to one side, we won't have a K value at this point in time because P and R aren't at their equilibrium concentrations. If we wait some time, however, we can find K after the reaction has come to equilibrium. Yes, our concentrations will now be a bit higher, but K remains the same.
So basically: changing volume and adding products/reactants changes concentration, and changing temperature alters K value itself.

Joshua Swift
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Re: Q and K

Postby Joshua Swift » Wed Jan 06, 2021 9:53 pm

K will always be the same value because it is the equilibrium constant and will only be relevant at equilibrium. At any point the reaction is not at equilibrium you want to solve for Q, the reaction quotient.

Alexandra Salata 2L
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Re: Q and K

Postby Alexandra Salata 2L » Thu Jan 07, 2021 8:45 pm

K will always be the same.
Q would change if you added more products are reactants.

Lily Anne Garcia 1C
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Re: Q and K

Postby Lily Anne Garcia 1C » Thu Jan 07, 2021 8:50 pm

K stays the same, but Q is what you calculate after more reactants are added

John_Tran_3J
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Re: Q and K

Postby John_Tran_3J » Thu Jan 07, 2021 8:52 pm

Q would alter because Q represents the concentration (reaction quotient) at any point in the reaction, K represents the concentration at equilibrium.

Brianna Chen 3F
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Re: Q and K

Postby Brianna Chen 3F » Sat Jan 09, 2021 12:48 am

K only changes when the temperature changes. In this case, you would have different values for Q in which after time, these values should equal K.

tholz11
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Re: Q and K

Postby tholz11 » Sat Jan 09, 2021 1:04 pm

You would need to calculate Q again to figure out whether or not the equation is at equilibrium

Liam Bertrand 3
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Re: Q and K

Postby Liam Bertrand 3 » Sat Jan 09, 2021 1:22 pm

K will stay the same, but Q will need to be calculated again and then you can compare the K and Q to determine reaction direction.

Dominic Benna 2E
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Re: Q and K

Postby Dominic Benna 2E » Sat Jan 09, 2021 2:58 pm

Only Q would change, so you would only need to find the value of Q in that situation. The only time that K changes is if the temperature changes, not if the amount of reactants or products changes.


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