Mini-Thermo Passage

[SaintJude]

View attachment Temp jpeg.jpg

Which of the following statements are true?

I. The equilibrium constant of a reaction is directly proportional to the temperature at which the reaction is run.
II. The equilibrium constant of a reaction is inversely proportional to the temperature at which the reaction is run.

A. I only
B. II only
C. Neither

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[MedPR]

B, inversely proportional.

G=-RTlnKeq

 

[SaintJude]

They actually define the equation you cite MedPr in the passage in an earlier paragraph, but then say nothing about it and then move on to the passage I included here

Kaplan: Both choices are wrong since the passage states that the relationship between temperature and equilibrium constant depends on the thermodynamics of the reaction (i.e. endothermic versus exothermic), so no clear proportionality can be used.

I still don't get it.

 

[MedPR]

They actually define the equation you cite MedPr in the passage in an earlier paragraph, but then say nothing about it and then move on to the passage I included here

Kaplan: Both choices are wrong since the passage states that the relationship between temperature and equilibrium constant depends on the thermodynamics of the reaction (i.e. endothermic versus exothermic), so no clear proportionality can be used.

C was what I was thinking initially, but the relationship is still inversely proportional regardless of which direction the equation is shifted.

By G=-RTlnK, if you increase temperature, Keq decreases. So for an exothermic reaction, if you increase T, the reaction gets shifted to the left and Keq goes down. For an endothermic reaction, if you increase T, the reaction gets shifted to the right and Keq goes up.

Crap, I should've wrote that out before I picked my answer. C seems right.

 

[chiddler]

Having read the answer (), I think medpr is right. I'm going to explain it as well both for myself and in case you need another perspective.

G=-RTlnK

the lnK means that it can be either positive or negative so it depends on the specific reaction like the passage says.

So if it's positive (Keq = products/reactants. So products > reactants), then more temp means a more spontaneous reaction as G dips deeper in the negative.

If it's negative (Keq = products/react, so reactants > products), then more temp means a less spontaneous reaction as G becomes more positive rather than negative.

It can swing both ways.

 

[SaintJude]

So for an exothermic reaction, if you increase T, the reaction gets shifted to the left and Keq goes down. For an endothermic reaction, if you increase T, the reaction gets shifted to the right and Keq goes up.

Son of a gun, you're right! This passage was so demoralizing, but I guess Kaplan is just trying to show me some tough love and make me realize a concept that goes unnoticed. Keq is a ratio that will be affected depending on the thermodynamic profile of the reaction.

So quite interestingly, if we don't know the thermodynamics of a reaction we can't determine whether temperature is proportional to equilibrium constant.

That's what that last sentence was trying to tell us by saying : "For example, an endothermic reaction will respond to an increase in temperature differently than will an exothermic reaction; this follows directly from Le Chatelier's principle." Although that was supposed to be a clue, I'm not sure I would have deciphered that. Thank you. I rarely ever heeded the G= -RTlnK equation.

 

[MedPR]

Son of a gun, you're right! This passage was so demoralizing, but I guess Kaplan is just trying to show me some tough love and make me realize a concept that goes unnoticed. Keq is a ratio that will be affected depending on the thermodynamic profile of the reaction.

So quite interestingly, if we don't know the thermodynamics of a reaction we can't determine whether temperature is proportional to equilibrium constant.

That's what that last sentence was trying to tell us by saying : "For example, an endothermic reaction will respond to an increase in temperature differently than will an exothermic reaction; this follows directly from Le Chatelier's principle." Although that was supposed to be a clue, I'm not sure I would have deciphered that. Thank you. I rarely ever heeded the G= -RTlnK equation.

Yea, that's why I was leaning towards C initially. Then I talked myself out of it with a half-ass explanation of why. What I typed (about endo/exo) was what I began to explain to myself in my head, but I stopped halfway through and didn't get to the part about how T up for endo = Keq up and T up for exp = Keq down.

For what it's worth, I'd say about 20-30% of the gen chem I've done on AAMC FLs thus far could be answered with le chatelier's alone. It's very, very powerful!

 

[kasho11]

Keq problems like this on the MCAT can be misleading because Keq is only viable at an exact temperature. If temp changes, Keq changes. So Q should be used. Anyways, not that helpful of a question to learning - just a little trick question