# Gibbs Energy QuestionWatch

#1
Hi all,

Looking for some help with the later parts of this question:

I have answers for a, b and c.
For D, I thought about using the equation Delta G = Delta Go + RTln(B/A) ? But I'm not sure. And then % yield will be B/A * 100?

For E, I figured, if the method for D is correct, that increasing T, the value for B will increase. (By looking at how you calculated B in the first place)

For F, is seting up two van't hoff equations for 25 degrees and 50 degrees, subtract one from the other and find K2. (ln(k1/k2)=-deltHO/R(1/t1-1/t2).

Pretty stumped by the latter half of this question, so any help will be appreciated!
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3 years ago
#2
(Original post by Sam_Chem)
Hi all,

Looking for some help with the later parts of this question:

I have answers for a, b and c.
For D, I thought about using the equation Delta G = Delta Go + RTln(B/A) ? But I'm not sure. And then % yield will be B/A * 100?

For E, I figured, if the method for D is correct, that increasing T, the value for B will increase. (By looking at how you calculated B in the first place)

For F, is seting up two van't hoff equations for 25 degrees and 50 degrees, subtract one from the other and find K2. (ln(k1/k2)=-deltHO/R(1/t1-1/t2).

Pretty stumped by the latter half of this question, so any help will be appreciated!
Too complicated.

You are given the equilibrium constant in the question...

A <==> B

You have the initial moles.

The equilibrium moles of A = initial - x

you see where to go?
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#3
(Original post by charco)
Too complicated.

You are given the equilibrium constant in the question...

A <==> B

You have the initial moles.

The equilibrium moles of A = initial - x

you see where to go?
Not entirely could you explain a bit more please?
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3 years ago
#4
(Original post by Sam_Chem)
Not entirely could you explain a bit more please?
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#5
(Original post by KombatWombat)
(This where I realise its all gone wrong haha!)

A) K=[B]/[A]

B) Delta G = 0, Delta Go = 1426.5Jmol-1
C)102.6Jmol-1K-1- +ve dirsorder is increasing, therefore favouring the forward reaction
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3 years ago
#6
(Original post by Sam_Chem)
(This where I realise its all gone wrong haha!)

A) K=[B]/[A]

B) Delta G = 0, Delta Go = 1426.5Jmol-1
C)102.6Jmol-1K-1- +ve dirsorder is increasing, therefore favouring the forward reaction
No, they're all good!

You know that in total there'll be 0.5 M of A and B, i.e. [A]+[B]=0.5 M

K=0.56 = [B]/[A] =[B]/(0.5 M - [B])

Solve that to give you [B], and get your percentage yield

For e and f, you know the entropy, assume enthalpy is constant with temperature, and you can work out a new value of âˆ†G and so K. I'd rather do f first because that tells you the answer to e, maybe they want you to e with some handwaving.
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#7
(Original post by KombatWombat)
No, they're all good!

You know that in total there'll be 0.5 M of A and B, i.e. [A]+[B]=0.5 M

K=0.56 = [B]/[A] =[B]/(0.5 M - [B])

Solve that to give you [B], and get your percentage yield

For e and f, you know the entropy, assume enthalpy is constant with temperature, and you can work out a new value of âˆ†G and so K. I'd rather do f first because that tells you the answer to e, maybe they want you to e with some handwaving.
So [B] = 0.179M?

%Yield will be 35.8%

F) Came out as 1.52 (forgot units) What about the second part though? D:

E) I imagine part of the answer for E will involve the previous answers, but not sure how to explain it? It does seem intuitive that temp will increase % yield.]
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3 years ago
#8
(Original post by Sam_Chem)
So [B] = 0.179M?

%Yield will be 35.8%

F) Came out as 1.52 (forgot units) What about the second part though? D:

E) I imagine part of the answer for E will involve the previous answers, but not sure how to explain it? It does seem intuitive that temp will increase % yield.]
What are the units of [A] and [B]? This tells you the units of K. Your second value of K will also have the same units as given in the question!

You can calculate [B] at 50Â°C in the same way as you did before, and then get a new percentage yield to get the second part of f, and this also tells you e.

(Actually, I should also point out that using Van 't Hoff is fine too, to calculate your second value of K. It's essentially doing the same thing. I prefer not to since it's another formula to remember).

And to explain what's happening in e, how does gibbs energy change with temperature?
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#9
(Original post by KombatWombat)
What are the units of [A] and [B]? This tells you the units of K. Your second value of K will also have the same units as given in the question!

You can calculate [B] at 50Â°C in the same way as you did before, and then get a new percentage yield to get the second part of f, and this also tells you e.

(Actually, I should also point out that using Van 't Hoff is fine too, to calculate your second value of K. It's essentially doing the same thing. I prefer not to since it's another formula to remember).

And to explain what's happening in e, how does gibbs energy change with temperature?
Ah I see.

I'm more familar with the Van't Hoff method so I'll probably use that. Is the answers I put in the previous post the same as what you got?

Well, entropy is positive, so looking at delta G = delta H - T delta S, so as temperature increases gibbs free energy is decreased, so max product yield is decreased?
Reaction is endothermic - standard enthalpy change is positive - so increases to temperature increases the equilibrium constant K, thus max yield will increase.

It definitely makes sense im just trying to make sense of getting the first part of D. Also, part E is throwing me off; struggling to understand this sufficiently to apply this to different examples, can anyone summarise this idea?
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3 years ago
#10
(Original post by KombatWombat)
What are the units of [A] and [B]? This tells you the units of K. Your second value of K will also have the same units as given in the question!
(Original post by Sam_Chem)
F) Came out as 1.52 (forgot units) What about the second part though? D:
Not sure how many times i've had to say this on TSR now but here we go...

K IS DIMENSIONLESS. ALWAYS.
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#11
(Original post by langlitz)
Not sure how many times i've had to say this on TSR now but here we go...

K IS DIMENSIONLESS. ALWAYS.
Yeah... my lecturer put so much emphasis on units for everything I was just on auto pilot looking for units that are not there, I guess if in doubt put the units of stuff in the equation. My mistake
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3 years ago
#12
(Original post by langlitz)
Not sure how many times i've had to say this on TSR now but here we go...

K IS DIMENSIONLESS. ALWAYS.
Yes, but ...

.. the exam boards expect the students to work out the 'dimensions' assuming that the concentrations are used for the law of mass action and not activities.

A' level (and IB) students have to be able to jump through the required hoops even if they are incorrect.
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3 years ago
#13
(Original post by charco)
Yes, but ...

.. the exam boards expect the students to work out the 'dimensions' assuming that the concentrations are used for the law of mass action and not activities.

A' level (and IB) students have to be able to jump through the required hoops even if they are incorrect.
The label says undergraduate and he refers to his lecturer...?
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3 years ago
#14
(Original post by langlitz)
Not sure how many times i've had to say this on TSR now but here we go...

K IS DIMENSIONLESS. ALWAYS.
Yep, quite right, my apologies!
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