BR Thermodynamics problem error, unless someone can solve~

[hellocubed]

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The ΔH for a given reaction is 12.5 kJ/mole. The ΔS for the reaction is 25J/molexK. At what temperature does the equilibrium constant equal 1.0?

A.) 0 C
B.) 227 C
C.) 327 C
D.) 500 C

Answer: (Highlight) B

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BR's reasoning was that
ΔG=-RTlnKeq

So if Keq = 1, then ΔG=0
Hence:

0=ΔH-TΔS

and then you solve it as

ΔH=TΔS




It would have been a good question, but they missed one thing:
ΔG does NOT =-RTlnKeq
This is the equation for standard ΔG, and you CANNOT use this equation for this problem because standard ΔG does not equal =ΔH-TΔS.



The correct formula is.
ΔG=RTln(Qrx/Keq)=ΔH-TΔS

And when Keq=1,
RTln(Qrx)=ΔH-TΔS

Which renders the problem unsolvable.

 

[milski]

Never mind, that does not help at all.

I'm not sure how you got ln(Qrx/Keq).

In general ΔG=ΔG0+RTln(Q). At equilibrium Q=K and ΔG=0, from there ΔG0=-RTln(K)

Regardless of that, ΔG0=ΔΗ0-TΔS0, which you have given in this case.

That assumes that ΔH0 and ΔS0 are independent of temperature, which is relatively correct for small temperature ranges.

 

[hellocubed]

http://imgur.com/PH0Ol

All equations I bring to the forum are sourced, I almost never derive anything myself (within my notes).



So, is it assumed that this problem is at equilibrium and Qrx=Keq?
That would solve the question, but frustrating because it never says so.

 

[MedPR]

I went through TBR pretty thoroughly (except organic) and I can pretty confidently say that there are no errors in the gen chem thermodynamics chapter.

Can't prove it to you though since I don't remember any of these equations anymore 😀

 

[milski]

http://imgur.com/PH0Ol

All equations I bring to the forum are sourced, I almost never derive anything myself (within my notes).

Right, you can actually get that from the ΔG/ΔG0 formula too.

So, is it assumed that this problem is at equilibrium and Qrx=Keq?
That would solve the question, but frustrating because it never says so.

It does not have to be at equilibrium. But since K is constant for a given T, you can solve for the equilibrium condition (for which you know ΔG). T and K will stay the same. Solving it for arbitrary condition will require extra information (ΔG) but will still lead to the same T.