Kaplan MCAT General Chem - Concept Check 6.1

[paintbucketgreen]

Hey guys, I had a question about a problem in Chapter 6 of the General Chem Kaplan book, concept check 6.1, question #1.

Given:
Q = 5.0 x 10^-2
Keq = 5.0 x 10^-3

The question then asks what the sign of the free energy change (delta G) would be.



My logic was that since the Q > Keq, the reaction will proceed to the left to reform more reactants in order to reach equilibrium. The answer key says the delta G of the reaction would be positive. But I am confused, because I thought that if a reaction has exceeded equilibrium, the reverse reaction to try and get back to equilibrium would be spontaneous (and thus delta G = negative) since equilibrium is essentially the "ideal" state the reaction wants to be in. Why would this reverse reaction be nonspontaneous? What's even more confusing is that there is a chart in the book that shows delta G to be negative when going from a [ Q > Keq ] state back to a [ Q = Keq ] state.


Any clarification would be appreciated

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[NextStepTutor_1]

Hi @paintbucketgreen -

It sounds like what's going here is fairly simple -- when they give ΔG values, they are always talking about the forward reaction unless otherwise specified. So if the reverse reaction is spontaneous, the forward reaction is nonspontaneous, and that's what they're looking for in the answer.

Hope this helps & best of luck!

 

[Dr. Meliodas]

Another way to put it is that your logic is off because you keep changing your point of reference.

 

[paintbucketgreen]

Another way to put it is that your logic is off because you keep changing your point of reference.

What?

 

[paintbucketgreen]

Hi @paintbucketgreen -

It sounds like what's going here is fairly simple -- when they give ΔG values, they are always talking about the forward reaction unless otherwise specified. So if the reverse reaction is spontaneous, the forward reaction is nonspontaneous, and that's what they're looking for in the answer.

Hope this helps & best of luck!

Okay that sort of makes sense. So if one reaction is spontaneous then the other one must be the opposite? and vice versa? Why is that exactly?

 

[JustinM88]

[ignore]

 

[aldol16]

The question then asks what the sign of the free energy change (delta G) would be.

Q is given in terms of [product]/[reactant] for a given process reactant ---> product. So if Q > K, that means that for the reaction reactant ---> product, the reaction will proceed to the left and therefore the delta G for the process reactant ---> product will be positive.

 

[aldol16]

They are assuming the forward reaction is spontaneous.

You understand already that the reaction will go in the reverse direction (“proceed to the left”). So, answer is reaction is nonspontaneous.

Your statements here directly contradict each other. If the reverse reaction is spontaneous, then the forward is non-spontaneous. You don't need to assume anything about the forward reaction here - they give you adequate information.

 

[JustinM88]

Q is given in terms of [product]/[reactant] for a given process reactant ---> product. So if Q > K, that means that for the reaction reactant ---> product, the reaction will proceed to the left and therefore the delta G for the process reactant ---> product will be positive.

I think I see it now, although I could be wrong lol. So for this particular Q and Keq, shifting the reaction to the left is the spontaneous direction (Le Chatelier's principle), and so the actual Reactants --> Products (how we'd write this questions actual reaction) forward direction is nonspontaneous (+deltaG)...

 

[aldol16]

I think I see it now, although I could be wrong lol. So for this particular Q and Keq, shifting the reaction to the left is the spontaneous direction (Le Chatelier's principle), and so the actual Reactants --> Products (how we'd write this questions actual reaction) forward direction is nonspontaneous (+deltaG)...

Correct. Always identify what process you're trying to characterize first. Here, we know that Q is expressed in terms of /[A], so the process is A ---> B for the given Q and K values. And whatever delta G we come up with must characterize the process A -> B.

 

[paintbucketgreen]

Correct. Always identify what process you're trying to characterize first. Here, we know that Q is expressed in terms of /[A], so the process is A ---> B for the given Q and K values. And whatever delta G we come up with must characterize the process A -> B.

This makes sense. Thank you! So would it then be safe to assume that if we come across this type of question on the exam, they will always be expecting us to characterize the forward reaction A-->B?

 

[JustinM88]

This makes sense. Thank you! So would it then be safe to assume that if we come across this type of question on the exam, they will always be expecting us to characterize the forward reaction A-->B?

Will have to wait on @aldol16 for the definitive answer just to be sure, but I think in chemistry this is just a convention that’s always followed for a reaction

 

[aldol16]

This makes sense. Thank you! So would it then be safe to assume that if we come across this type of question on the exam, they will always be expecting us to characterize the forward reaction A-->B?

Forward or reverse doesn't matter here. They give you a relationship that is relative. To see this, let's say Q > K above was referring to process B -> A. This is the reverse of the process I used above. Then Q = [A]/[ B ]. The system will want to reduce A and make B. In other words, it would want to go backwards from the direction I have written. That means that the process B -> A is non-spontaneous.

"Forward" and "reverse" reaction are just relative here. What's forward and reverse doesn't matter. We just know that the process that Q refers to, whichever direction it is, must be non-spontaneous.