At what temperature is an equilibrium constant equal to 1.0 given enthalpy..

[UIUCstudent]

So the full question is.

The change in enthalpy for a given reaction is 12.5 kJ/mol. The change in entropy is 25 kJ/mol *K. At what temperature is Keq equal to 1.0?

This is BR passage 4 question 28 in the thermochemistry section.

So BR said if Keq = 1 then change in G is zero. This is due to
change in G=-RTlnKeq. But, I always thought this equation pertained to standard state...The explanation goes on to say since G= 0 and G=H-TS then H=TS and then it's plug and chug.

I'm just confused b/c the question didn't mention this was at standard state nor are there symbols that indicate it in the passage solution.

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[NYR56]

That equation G=-RTlnKeq is actually for G naught. It applies whenever the reaction is at equilibrium, since there is no free energy to do work, hence G=0. I'm not sure if that fully explains what's going on or not to you though.

 

[UIUCstudent]

That equation G=-RTlnKeq is actually for G naught. It applies whenever the reaction is at equilibrium, since there is no free energy to do work, hence G=0. I'm not sure if that fully explains what's going on or not to you though.

Can we use standard state for G= H-TS as well?

 

[NYR56]

I believe so, although both G and G naught seem to be 0 in this case, since they are essentially equal to each other from the EQ G naught = G-RTlnKeq, and -RTlnKeq = 0. To be perfectly honest, I think this question is far too in-depth for a true MCAT question. I'm not an expert in thermo though, so perhaps someone else could chime in if I missed something.

 

[BerkReviewTeach]

So the full question is.

The change in enthalpy for a given reaction is 12.5 kJ/mol. The change in entropy is 25 J/mol*K. At what temperature is Keq equal to 1.0?

This is BR passage 4 question 28 in the thermochemistry section.

So BR said if Keq = 1 then change in G is zero. This is due to
change in G=-RTlnKeq. But, I always thought this equation pertained to standard state...The explanation goes on to say since G= 0 and G=H-TS then H=TS and then it's plug and chug.

I'm just confused b/c the question didn't mention this was at standard state nor are there symbols that indicate it in the passage solution.

They can't (or won't) tell you it's at standard state, because that is what you are solving for (the temperature). If you look at the answer choices, you can see that standard state is not relevant here. They are testing whether you know that if Keq = 1, then deltaG = 0.

Given that when Keq = 1, deltaG = 0, you can simply plug into the fundamental equation: deltaG = deltaH - TdeltaS.

deltaG = deltaH - TdeltaS

0 = deltaH - TdeltaS

TdeltaS = deltaH

T = deltaH/deltaS = 12,500/25 = 500​

It's meant to be a straight-forward calculation question using a fundamental concept coupled with a typical equation. Be careful not to over-complicate things, which is so easy to do.

 

[UIUCstudent]

They can't (or won't) tell you it's at standard state, because that is what you are solving for (the temperature). If you look at the answer choices, you can see that standard state is not relevant here. They are testing whether you know that if Keq = 1, then deltaG = 0.

Given that when Keq = 1, deltaG = 0, you can simply plug into the fundamental equation: deltaG = deltaH - TdeltaS.

deltaG = deltaH - TdeltaS​

0 = deltaH - TdeltaS​

TdeltaS = deltaH​

T = deltaH/deltaS = 12,500/25 = 500​

It's meant to be a straight-forward calculation question using a fundamental concept coupled with a typical equation. Be careful not to over-complicate things, which is so easy to do.

Can you explain why deltaG=0 when Keq = 1 ? The equation when not at standard is deltaG=RTln(Q/K) so even though Keq = 1, we don't know if Q is equal to one?

 

[Renaissance Man]

They can't (or won't) tell you it's at standard state, because that is what you are solving for (the temperature). If you look at the answer choices, you can see that standard state is not relevant here. They are testing whether you know that if Keq = 1, then deltaG = 0.

Given that when Keq = 1, deltaG = 0, you can simply plug into the fundamental equation: deltaG = deltaH - TdeltaS.

deltaG = deltaH - TdeltaS

0 = deltaH - TdeltaS

TdeltaS = deltaH

T = deltaH/deltaS = 12,500/25 = 500​

It's meant to be a straight-forward calculation question using a fundamental concept coupled with a typical equation. Be careful not to over-complicate things, which is so easy to do.

I am also confused at why deltaQ=0 when Keq=1. A Keq of 1 simply means that at equilibrium, there are equal amounts of products and reactants, right? Most reactions don't have a Keq of 1, so I don't understand why knowing "deltaQ=0 when Keq=1" is important? Maybe I am just completely lost about this concept though. Any help would be appreciated.

 

[NYR56]

If a reaction as at equilibrium (K=1), there is no free energy remaining, hence dG=0. This is definitional.

 

[Renaissance Man]

If a reaction as at equilibrium (K=1), there is no free energy remaining, hence dG=0. This is definitional.

Actual no, Keq=1 does not mean that a reaction is at equilibrium. Without knowing anything about the reaction, one can't assume that. Equilibrium happens when the forward reaction equals the reverse reaction, and this can happen whether Keq is very big or very small (or equal to 1).

When Q=K you are at equilibrium though

 

[NYR56]

You're right. I should have said that at standard conditions, if you are at equilibrium, K=1 and thus dG naught = 0. dG=0 for any equilibrium. I think this question isn't giving you enough information, at least I'm missing something.

 

[Renaissance Man]

You're right. I should have said that at standard conditions, if you are at equilibrium, K=1 and thus dG naught = 0. dG=0 for any equilibrium. I think this question isn't giving you enough information, at least I'm missing something.

Yeah, I think we are missing the part about standard conditions, sorry if my last post came off wrongly, wasn't trying to sound the way it came out.

 

[BerkReviewTeach]

You're right. I should have said that at standard conditions, if you are at equilibrium, K=1 and thus dG naught = 0. dG=0 for any equilibrium. I think this question isn't giving you enough information, at least I'm missing something.

You're right. You need at least one other piece of information such as Q or deltaGo. I wonder if it said something in the passage about the reaction considitions?

Great catch UIUCstudent. I believe this question is a typo, because that question for all intents and purposes is a free-standing question that happens to be asked as part of the passage set. I believe they meant to ask if Keq = Qrx, not if it equals 1. My bad for just blindly following the key and trying to simply reword it without thinking it through.

 

[NYR56]

Thumbs up for looking into it and letting us know!