5G.11


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Andrea_Torres
Posts: 98
Joined: Sun Sep 15, 2019 12:15 am

5G.11

Postby Andrea_Torres » Thu Jan 09, 2020 5:53 pm

Is reaction Quotient (Q) the same thing as K? If so then how is my answer not the same as what I would get for K? How do you solve if you don't have any values?

Alexis Webb 2B
Posts: 124
Joined: Thu Jul 11, 2019 12:15 am

Re: 5G.11

Postby Alexis Webb 2B » Thu Jan 09, 2020 5:57 pm

It’s the same equation, but the concentrations used are during a different time in the reaction, which may or may not be at equilibrium depending on if Q is less than, equal to, or greater than K.

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

Re: 5G.11

Postby Uisa_Manumaleuna_3E » Thu Jan 09, 2020 7:03 pm

Just think of the Reaction Quotient like you proofing the equation. You do just calculate it the same way you find K (with the concentrations raised to the power of their coefficients), but you're trying to see if the value of Q, with the given concentrations, will give you K. Because if it gives you K, then at that given Q is at equilibrium. If it Q and K don't match up, then you've proved that the system hasn't reached equilibrium yet. I noticed that a lot of times our work shows us that Q and K aren't the same and then we have to predict which way the equation will "lean".

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: 5G.11

Postby DHavo_1E » Thu Jan 09, 2020 8:21 pm

Hi,

I agree with the statements above that Q is a reaction quotient at a time that may or may not be at equilibrium and is solved the same way as if solving for K. It helps determine where the reaction is at. For example, if Q < K, then the equation will lie to the right (because a smaller Q means that there are more reactants, and to reach equilibrium, products will be made at the expense of the reactants) , if Q > K, then the equation will lie to the left (by similar logic to that above), and if Q=K, then the reaction is at equilibrium.

Amy Xiao 1I
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am

Re: 5G.11

Postby Amy Xiao 1I » Fri Jan 10, 2020 12:18 am

I think if you are not given any values, you would simply write out the Q expression with the molecular formula placeholders in the square brackets. Q and K would be the same if no values are given, considering Q is calculated during a specific point in the reaction when there must be actual values you can plug in.

Matt Sanruk 2H
Posts: 131
Joined: Wed Sep 18, 2019 12:21 am

Re: 5G.11

Postby Matt Sanruk 2H » Fri Jan 10, 2020 12:42 am

So the Q shows whether the reaction is forward or reverse when compared to K, right?

andrewcj 2C
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

Re: 5G.11

Postby andrewcj 2C » Fri Jan 10, 2020 1:14 am

Yes, the reaction quotient Q can indicate the direction of the reaction by comparing it to K. If Q < K, then the products side is favored and the forward reaction will proceed, and vice versa if Q > K.


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