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### Topic: Gibb's Free Energy and Equilibrium Constant Relation  (Read 460 times)

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#### person1234

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• Mole Snacks: +0/-0 ##### Gibb's Free Energy and Equilibrium Constant Relation
« on: June 09, 2019, 02:43:55 AM »
I'm confused about the following equation:

ΔG=-RTln(K)

The spontaneity of the reaction should depend on the reaction quotient, but it doesn't appear in the right hand side. The only possible solution I can see is that it is assumed the initial reaction quotient is 1, so that when Q=K, neither the forward or reverse reaction is favored. Is this correct?

#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #1 on: June 10, 2019, 03:58:16 AM »
Your equation is wrong. The correct equations are
ΔG = ΔG° + RTlnQ and ΔG° = -RTlnK
At equilibrium ΔG = 0 and Q = K
Do you appreciate the difference between ΔG and ΔG°?
« Last Edit: June 10, 2019, 06:41:34 AM by mjc123 »

#### person1234

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• Mole Snacks: +0/-0 ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #2 on: June 10, 2019, 01:34:04 PM »
ΔG° is Gibb's Free Energy at STP for unmixed reactants of equal concentrations, which clarifies things completely. Thank you!

#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #3 on: June 11, 2019, 04:27:40 AM »
ΔG° is the Gibbs free energy change (not Gibb's, the guy wasn't a Bee Gee) at the temperature T for reagents in their standard states.

#### David Tan

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• Gender: • I am a Chemistry teacher + author. ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #4 on: June 14, 2019, 04:32:08 PM »
ΔG=-RTln(K)

ΔG=Gibbs free energy and its unit is kJ/mol. If it is a negative value, it means the reaction is thermodynamically spontaneous. If it is zero, the reaction is at equilibrium. If it is positive, then the reaction is thermodynamically non-spontaneous. R is the molar gas constant or universal gas constant and is equal to 8.31 J per Kelvin per mole. In = natural logarithm. K is not a reaction quotient, it is the equilibrium constant.

Let's look at the math. When K equals one, the system is at equilibrium and In(1) = zero. Hence, ΔG equals zero and the system is at equilibrium. Example: ΔG = zero for water at its melting point and boiling point. Hope it helps.
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#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #5 on: June 17, 2019, 04:45:18 AM »
This is not correct; see my post above. The system is not "at equilibrium when K = 1"; it is at equilibrium when Q = K, whatever the value of K is. In this situation ΔG = 0.

#### David Tan

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• Gender: • I am a Chemistry teacher + author. ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #6 on: June 17, 2019, 04:52:50 AM »
This is not correct; see my post above. The system is not "at equilibrium when K = 1"; it is at equilibrium when Q = K, whatever the value of K is. In this situation ΔG = 0.

Your equations (ΔG = ΔG° + RTlnQ and ΔG° = -RTlnK) are indeed correct, based on what we learnt in university. I learnt those equations in my university education. However, this is a high school chemistry forum and this is the equation (ΔG=-RTlnK) that the students are taught in the UK syllabus. We do agree that ΔG = 0 at equilibrium. Hence, by using his equation of ΔG=-RTlnK and equating ΔG = 0, K must be 1 "at equilibrium" although it is not technically sound. In fact, some of their papers use the formula ΔG° = -RTlnK and ΔG=-RTlnK
interchangeably. Hence, by asking this fellow student "Do you appreciate the difference between ΔG and ΔG°?" might be too harsh on him.

Perhaps, we can check with this student which equation(s) has he learnt in the school. Here are my thoughts: Although some equations are absolutely correct, sometimes we are taught the "wrong" method because it is "easier" to understand it that way at that particular stage of our learning curve. For example, when we were young we were taught that there are only 3 states of matter. As we become more knowledgeable, we realise this is not true.

Thanks for pointing out the correct equations. Hopefully, the student would be able to appreciate both sides of our discussion. Cheers.
« Last Edit: June 17, 2019, 05:15:48 AM by David Tan »
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#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #7 on: June 17, 2019, 08:22:12 AM »
Hi @David Tan

I appreciate your conciliatory tone, and I'm sorry if my earlier post appeared harsh, but some things are not merely "simpler" or "easier", but just wrong. "3 states of matter" is a simplification; if we were told dogmatically that these were the only states, and there were definitely no others, that would be a falsehood. "K = 1 at equilibrium" is not a simplification, but (except coincidentally, when ΔG° ≈ 0) is just plain false. The OP is being confused by being taught wrong things, whether by mistake or for the sake of "simplicity".

The key point to grasp is that the quantity that is equal to -RTlnK is not the quantity that is equal to 0 at equilibrium. Calling them both "ΔG" is asking for trouble, and confusion persists through university studies, as many posts on these forums will show. K, and ΔG°, are constants for a given temperature, irrespective of the state of any particular system. Q and ΔG for a system vary with the state of the system, and at equilibrium Q = K and ΔG = 0. I realise that these points are not always made as clear as they should be, even at university, but until they are understood there will always be confusion.

I'd be interested to know how OP's teacher would answer his/her initial question.

#### David Tan

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• Gender: • I am a Chemistry teacher + author. ##### Re: Gibb's Free Energy and Equilibrium Constant Relation
« Reply #8 on: June 17, 2019, 08:38:01 AM »
Hi @David Tan

I appreciate your conciliatory tone, and I'm sorry if my earlier post appeared harsh, but some things are not merely "simpler" or "easier", but just wrong. "3 states of matter" is a simplification; if we were told dogmatically that these were the only states, and there were definitely no others, that would be a falsehood. "K = 1 at equilibrium" is not a simplification, but (except coincidentally, when ΔG° ≈ 0) is just plain false. The OP is being confused by being taught wrong things, whether by mistake or for the sake of "simplicity".

The key point to grasp is that the quantity that is equal to -RTlnK is not the quantity that is equal to 0 at equilibrium. Calling them both "ΔG" is asking for trouble, and confusion persists through university studies, as many posts on these forums will show. K, and ΔG°, are constants for a given temperature, irrespective of the state of any particular system. Q and ΔG for a system vary with the state of the system, and at equilibrium Q = K and ΔG = 0. I realise that these points are not always made as clear as they should be, even at university, but until they are understood there will always be confusion.

I'd be interested to know how OP's teacher would answer his/her initial question.

No worries. We are here to help one another. After all, we are not obliged to answer their questions. We chose to help them out of kindness and self initiatives. I am very positive that you are a very kind soul and hence, this explains your "disappointment" at how things were taught in high school and even university level. I agree with you. I am also interested to know how OP's teacher would answer that initial question. Perhaps, it would be better if we know what the full question is. Cheers.
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Master's degree in Chemistry from National University of Singapore