Quick GChem Destroyer Question 21

[Andrew324]

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So for question 21, I get them all but C

C) Changing the pressure will not change K

Since there are 4 moles of gas on the right and 2 on the left. I was thinking a pressure change would change the Concentrations of B and C and since K has to do with concentrations it would change.

In the answers all it says is
A catalyst will have no effect on equilibrium. It does not mention pressure in the answer.

A little help?

 

[Danny289]

IMO, you are right, pressure will change the "K".

 

[DRHOYA]

I just took a look at the question. The question asks which statement is false. C is a true statement, thus a wrong answer. The reaction is reversible, and the beginning of the question even says "equilibrium". Based on that, the "K" that choice C is referring to is the Keq, equilibrium constant, and the only thing that changes that is temperature. Hope that helps man.

Btw, it took me a while to figure out choice B, I couldn't get the math down.

 

[Danny289]

I just took a look at the question. The question asks which statement is false. C is a true statement, thus a wrong answer. The reaction is reversible, and the beginning of the question even says "equilibrium". Based on that, the "K" that choice C is referring to is the Keq, equilibrium constant, and the only thing that changes that is temperature. Hope that helps man.

Btw, it took me a while to figure out choice B, I couldn't get the math down.

👍 very good point

 

[osimsDDS]

I just took a look at the question. The question asks which statement is false. C is a true statement, thus a wrong answer. The reaction is reversible, and the beginning of the question even says "equilibrium". Based on that, the "K" that choice C is referring to is the Keq, equilibrium constant, and the only thing that changes that is temperature. Hope that helps man.

Btw, it took me a while to figure out choice B, I couldn't get the math down.

Isnt choice B because when you reverse a reaction you are taking the inverse and when you divide the reaction by 2 then you will take the square root of it. This makes sense because think about this:

4A + 2B ----> 4C

K= [C]^4/^2 =50

Now reverse it:

4C ----> 2B + 4A

K= ^2/[C]^4 = 1/50

Reversing it will cause the inverse of the beginning....And then simply dividing all the concentrations by 2 will give you the square root of 50...if you multiply the concentrations by 2 then you will square it to find the K...

 

[DRHOYA]

Isnt choice B because when you reverse a reaction you are taking the inverse and when you divide the reaction by 2 then you will take the square root of it. This makes sense because think about this:

4A + 2B ----> 4C

K= [C]^4/^2 =50

Now reverse it:

4C ----> 2B + 4A

K= ^2/[C]^4 = 1/50

Reversing it will cause the inverse of the beginning....And then simply dividing all the concentrations by 2 will give you the square root of 50...if you multiply the concentrations by 2 then you will square it to find the K...

Ok, I understand the reversing part to get 1/50, but still a little unclear on how it goes to be radical 1/50. I still don't understand how the radical comes into play from dividing the concentrations by 2?

 

[osimsDDS]

Alright consider this:

K = C^4/B^2

Plug in a concentration of 2 for both C and B...

K= 2^4/2^2 = 16/4 = 4

Now double the moles of each and it will now be:

K = C^8/B^4

Now plug back in 2 for both C and B...

K = 2^8/2^4 = 256/16 = 16

So when we double the moles for each you can see that the K is squared...4^2 = 16

If you decrease the moles by a factor of 2 then the K will be the square root of the initial K...

Hope that helps...

 

[DRHOYA]

Alright consider this:

K = C^4/B^2

Plug in a concentration of 2 for both C and B...

K= 2^4/2^2 = 16/4 = 4

Now double the moles of each and it will now be:

K = C^8/B^4

Now plug back in 2 for both C and B...

K = 2^8/2^4 = 256/16 = 16

So when we double the moles for each you can see that the K is squared...4^2 = 16

If you decrease the moles by a factor of 2 then the K will be the square root of the initial K...

Hope that helps...

Awesome! Ok, yeah I see it now. I made an error in the exponents when I originally double the moles of each. Thanks man.