Thermodynamics

[chiddler]

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If ΔS is positive for a reaction, then as temperature is increased, what is observed?

A. Both Keq-to-Q ratio and ΔG increase
B. The Keq-to-Q ratio decreases, while ΔG increases.
C. The Keq-to-Q ratio increases, while ΔG decreases.
D. Both Keq-to-Q ratio and ΔG decrease.

There are quite a few formulas to use with this question, but they all ultimately lead to ΔG=RT*ln(Q/Keq). I just need some math help with this question.

(otherwise you can use ΔG=ΔG⁰ + RT*ln(Q), and ΔG⁰=-RT*ln(Keq) to get the above equation)

So ΔG is the easy part. With the gibbs free energy equation, the TΔS becomes negative and increasing T makes it more negative. Decreasing.

Keq/Q. I'm not sure how to look at this. On one hand, T is increasing. On the other, ΔG is decreasing. ΔG=RT*ln(Q/Keq). So can somebody help me walk through the logic of concluding that the ratio increases?

Answer is C.

 

[MedPR]

If ΔS is positive for a reaction, then as temperature is increased, what is observed?

A. Both Keq-to-Q ratio and ΔG increase
B. The Keq-to-Q ratio decreases, while ΔG increases.
C. The Keq-to-Q ratio increases, while ΔG decreases.
D. Both Keq-to-Q ratio and ΔG decrease.

There are quite a few formulas to use with this question, but they all ultimately lead to ΔG=RT*ln(Q/Keq). I just need some math help with this question.

(otherwise you can use ΔG=ΔG⁰ + RT*ln(Q), and ΔG⁰=-RT*ln(Keq) to get the above equation)

So ΔG is the easy part. With the gibbs free energy equation, the TΔS becomes negative and increasing T makes it more negative. Decreasing.

Keq/Q. I'm not sure how to look at this. On one hand, T is increasing. On the other, ΔG is decreasing. ΔG=RT*ln(Q/Keq). So can somebody help me walk through the logic of concluding that the ratio increases?

Answer is C.

Well, if T increases then in order for ΔG to decrease, ln(Q/Keq) must decrease.

Try using a simpler expression and other scenarios:
For X=YZ

1. If X remains constant and Y increases, what does Z do? It must decrease.
2. If X decreases and Y remains constant, what does Z do? It must decrease.
3. If X decreases and Y increases, what does Z do? It must decrease again, but even more than in scenario 2 because the Y term is opposing the change in the X term.

So back to the original equation.

If ln(Q/Keq) decreases, that means Keq/Q increases.

As an aside, if they told you that both ΔG and T were decreasing (or increasing) you would not be able to make a qualitative assumption about the Keq/Q ratio. It could remain the same, decrease, or increase depending on the quantitative data.
,

 

[411309]

If ΔS is positive for a reaction, then as temperature is increased, what is observed?

A. Both Keq-to-Q ratio and ΔG increase
B. The Keq-to-Q ratio decreases, while ΔG increases.
C. The Keq-to-Q ratio increases, while ΔG decreases.
D. Both Keq-to-Q ratio and ΔG decrease.

There are quite a few formulas to use with this question, but they all ultimately lead to ΔG=RT*ln(Q/Keq). I just need some math help with this question.

(otherwise you can use ΔG=ΔG⁰ + RT*ln(Q), and ΔG⁰=-RT*ln(Keq) to get the above equation)

So ΔG is the easy part. With the gibbs free energy equation, the TΔS becomes negative and increasing T makes it more negative. Decreasing.

Keq/Q. I'm not sure how to look at this. On one hand, T is increasing. On the other, ΔG is decreasing. ΔG=RT*ln(Q/Keq). So can somebody help me walk through the logic of concluding that the ratio increases?

Answer is C.

Ok since you already figured out the easy part using G=H-TS formula, lets look at the second part...

ΔG=RT*ln(Q/Keq).

Since we know G is decreasing we know that Keq is increasing because they're inversely proportional. We also see that Q is directly proportional to G so as G decreases, Q decreases. Since Keq is getting larger and Q is getting smaller the ratio of Keq to Q is larger because you have more Keq and less Q.

Example: T= S/O. Inreasing the numerator, S, will give us more T, increasing the denominator O will give us less T. Increasing T must mean an increased S.

 

[411309]

Actually this question is kind of confusing it would prob be easier to just go through each narrowed down answer choices (C and D)

D. The Keq to to Q ratio were decreasing and temperature was increasing, then how could G be decreasing? Because now we have more Q which would give a higher G because we also have an increased T.

C. If we have an increasing Keq to Q ratio then we have a higher denominator which in comparison to the numerator which would certainly help counteract the raising T and make G more negative.

Choice D would actually make G more positive. Make sense?

 

[chiddler]

very well put, thanks for the help.

 

[chiddler]

Actually this question is kind of confusing it would prob be easier to just go through each narrowed down answer choices (C and D)

D. The Keq to to Q ratio were decreasing and temperature was increasing, then how could G be decreasing? Because now we have more Q which would give a higher G because we also have an increased T.

C. If we have an increasing Keq to Q ratio then we have a higher denominator which in comparison to the numerator which would certainly help counteract the raising T and make G more negative.

Choice D would actually make G more positive. Make sense?

No not understanding well....hmm

Keq/Q. Keq is generally static, but Q is dynamic. So if the ratio is increasing, Q is decreasing. An increasing Q means a...MORE NEGATIVE DELTA G! oh damn!

thanks xD

 

[411309]

No not understanding well....hmm

Keq/Q. Keq is generally static, but Q is dynamic. So if the ratio is increasing, Q is decreasing. An increasing Q means a...MORE NEGATIVE DELTA G! oh damn!

thanks xD

yeah pretty much. A higher Keq to Q ratio implies Q is decreasing or Keq is increasing which would contribute to more NEGATIVE GIBBS FREE ENERGY WHICH MAKES US MORE SPONTANEOUS YAY.

 

[MedPR]

No not understanding well....hmm

Keq/Q. Keq is generally static, but Q is dynamic. So if the ratio is increasing, Q is decreasing. An increasing Q means a...MORE NEGATIVE DELTA G! oh damn!

thanks xD

Wait, what? If you increase Q, you decrease the Keq/Q ratio. Answer C says the Keq/C ratio is increasing.

 

[411309]

Wait, what? If you increase Q, you decrease the Keq/Q ratio. Answer C says the Keq/C ratio is increasing.

I think he made a typo since he got the first part right in that sentence.

 

[chiddler]

typo

 

[MedPR]

I think he made a typo since he got the first part right in that sentence.

Oh I see. So what you're getting at is this:

logQ/K (using log instead of ln because its easier 🙂 and the concept is the same).

K is a constant, say 10.

When Q is 100, log(Q/K)=1
When Q is 10, log(Q/K)=0
When Q is 1, log(Q/K)=-1

So as Q decreases, ΔG becomes more negative. Cool. This isn't the way I figured it out the first time, but as I typed out my reasoning I realized it didn't make sense.

However, I do want to know why (if it is wrong) you couldn't do the following.

I know that ΔG and ΔG⁰ aren't the same, but would it lead you in the wrong direction if you set ΔG=-RTln(Keq)? Then, as you make ΔG more negative, you can see that Keq is increasing, so you could say that the Keq/Q ratio is increasing?

 

[chiddler]

Oh I see. So what you're getting at is this:

logQ/K (using log instead of ln because its easier 🙂 and the concept is the same).

K is a constant, say 10.

When Q is 100, log(Q/K)=1
When Q is 10, log(Q/K)=0
When Q is 1, log(Q/K)=-1

So as Q decreases, ΔG becomes more negative. Cool. This isn't the way I figured it out the first time, but as I typed out my reasoning I realized it didn't make sense.

However, I do want to know why (if it is wrong) you couldn't do the following.

I know that ΔG and ΔG⁰ aren't the same, but would it lead you in the wrong direction if you set ΔG=-RTln(Keq)? Then, as you make ΔG more negative, you can see that Keq is increasing, so you could say that the Keq/Q ratio is increasing?

I'm not certain but with that reasoning, you're only looking at half the picture. How is Q behaving while Keq is increasing?

 

[MedPR]

I'm not certain but with that reasoning, you're only looking at half the picture. How is Q behaving while Keq is increasing?

By using ΔG=-RTlnKeq you are only talking about standard conditions so Q is constant/irrelevent for now. But since ΔG isn't the right ΔG for standard conditions, I wasn't sure if would work.

 

[chiddler]

By using ΔG=-RTlnKeq you are only talking about standard conditions so Q is constant/irrelevent for now. But since ΔG isn't the right ΔG for standard conditions, I wasn't sure if would work.

oh i see what you're saying.

i would think that such reasoning is inaccurate because they are fundamentally different. but i can't think of any math to back it up.

that's a good idea though. i'm curious.