gibbs energía libre y constante de equilibrio

[Angelgarcia12]

Hi guys!

I just did a test question and I don't understand why a statement is not correct, I hope someone can help me. 

If the ∆G´º of a reaction is 0:
to.

a. Keq must be 1
b. The reaction will be in equilibrium
c. Concentration of reactants must be equal to concentration of products
d. The reaction is spontaneous

The right statement is a. although I don't know what c it isn´t

 

Thanks

Edited November 6, 2023 by Angelgarcia12
e

[sethoflagos]

a) is correct by definition. dG = RT ln(K_eq) so dG=0 implies K_eq=1

Remembering that K_eq is also by definition the ratio of reaction rate constants for forward and reverse reactions, we have to remember that the rate constants are functions not only of the concentrations of chemical species (either reactants or products respectively) but also the reaction mechanisms. In general the mechanisms for the forwards and reverse reactions will be different, and therefore simultaneous equality of both reactant and product concentrations and forwards and reverse reaction rates could only occur under very special conditions (if ever in real life).

Therefore, in general, statement c) is incorrect.

I hope this is what you were looking for. Your question isn't perfectly phrased in English.

54 minutes ago, Angelgarcia12 said:

Hi guys!

I just did a test question and I don't understand why a statement is not correct, I hope someone can help me. 

If the ∆G´º of a reaction is 0:
to.

a. Keq must be 1
b. The reaction will be in equilibrium
c. Concentration of reactants must be equal to concentration of products
d. The reaction is spontaneous

The right statement is a. although I don't know what c it isn´t

 

Thanks

 

Edited November 6, 2023 by sethoflagos
not perfectly phrased in English

[exchemist]

1 hour ago, Angelgarcia12 said:

Hi guys!

I just did a test question and I don't understand why a statement is not correct, I hope someone can help me. 

If the ∆G´º of a reaction is 0:
to.

a. Keq must be 1
b. The reaction will be in equilibrium
c. Concentration of reactants must be equal to concentration of products
d. The reaction is spontaneous

The right statement is a. although I don't know what c it isn´t

 

Thanks

I think the point abut c being wrong may be that the expression for K involves the product of concentrations of product species, raised to the appropriate power, divided by the product of concentrations of reactant species raised to the appropriate power. So K is not simply one concentration divided by another.

(Also, strictly speaking K should involve activities rather than concentrations, though this may be too subtle for the test question you are describing.)

[sethoflagos]

1 hour ago, exchemist said:

I think the point abut c being wrong may be that the expression for K involves the product of concentrations of product species, raised to the appropriate power, divided by the product of concentrations of reactant species raised to the appropriate power. So K is not simply one concentration divided by another.

(Also, strictly speaking K should involve activities rather than concentrations, though this may be too subtle for the test question you are describing.)

I bow to your greater experience in this area, but I think we're saying the same thing aren't we?

In passing, I don't see the cases for the falsity of b) and d) being quite so clear cut. Does b) fall on the isothermal constraint and d) on catalyst dependency, or am I barking up the wrong tree(s).

[exchemist]

31 minutes ago, sethoflagos said:

I bow to your greater experience in this area, but I think we're saying the same thing aren't we?

In passing, I don't see the cases for the falsity of b) and d) being quite so clear cut. Does b) fall on the isothermal constraint and d) on catalyst dependency, or am I barking up the wrong tree(s).

Yes we are I think. I'm rusty on all this so you may be just as likely to be right as me these days, but I think b is off-target as we don't know whether the reaction has been allowed to reach equilibrium. If you start with just left hand side reactants, say, it will take some time to generate enough right hand side products before equilibrium is reached. These thermodynamic quantities tell about what the equilibrium state will look like, not whether or not equilibrium has been reached. I think d is off target because a "spontaneous" reaction is generally said to be one in which ΔG is -ve. 

[Angelgarcia12]

Thank you so much

I would like to ask you another question:

Are the following statements True and False:

1. Hydrophobic molecules clump together in water thanks to hydrophobic forces

2. In stereoisomerism, the different D or L configurations indicate the deviation of the polarized light by the molecules.

 

Thanks

[exchemist]

5 hours ago, Angelgarcia12 said:

Thank you so much

I would like to ask you another question:

Are the following statements True and False:

1. Hydrophobic molecules clump together in water thanks to hydrophobic forces

2. In stereoisomerism, the different D or L configurations indicate the deviation of the polarized light by the molecules.

 

Thanks

This sounds like homework. Before we respond, what are your thoughts about the answers?

[Angelgarcia12]

13 hours ago, exchemist said:

This sounds like homework. Before we respond, what are your thoughts about the answers?

It's been a while since I had any homework haha.
The reality is that I have seen its in exams as false but I think they are true and there is an error.
But the explanation is what I ask for, because in my notes the statement says without deep explanation.
Thank you

[exchemist]

32 minutes ago, Angelgarcia12 said:

It's been a while since I had any homework haha.
The reality is that I have seen its in exams as false but I think they are true and there is an error.
But the explanation is what I ask for, because in my notes the statement says without deep explanation.
Thank you

Well both are indeed false.

1) There is no such thing as a "hydrophobic force".

For a non-polar molecule to dissolve into water, it has to get between the water molecules. There will be attractive London forces between water molecules and the non-polar molecule, but its presence between water molecules will reduce their mutual hydrogen bonding and thus raise the energy of the solution, so it is energetically unfavourable - and will only happen to a slight extent. (Dissolution is still favoured entropically, so the free energy change won't be determined purely by the enthalpy change. The influence of entropy will be greater as the temperature goes up: ΔG = ΔH-TΔS. ) So there is no "repulsion" of any kind: it's just a reduction in net attractive force.

2) I had actually forgotten this 🙂, but D and L stereoisomers are a naming convention, relating chiral molecules to the enantiomers of glyceraldehyde, i.e. to the way those rotate the plane of polarised light. It does not mean and given D or L stereoisomer will rotate polarised light in a particular way.

More here: https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_22._Carbohydrates/22.03%3A_The_D_and_L_Notation

 

[studiot]

Here is a counter example to (1)

Lycopodium powder.

This was once used to demonstrate brownian motion, where the hydrophobic powder particles definitely do not clump together.

[exchemist]

7 hours ago, studiot said:

Here is a counter example to (1)

Lycopodium powder.

This was once used to demonstrate brownian motion, where the hydrophobic powder particles definitely do not clump together.

Why do you say lycopodium powder grains are hydrophobic?  

[studiot]

2 hours ago, exchemist said:

Why do you say lycopodium powder grains are hydrophobic?  

 

https://www.facebook.com/watch/?v=323855678960326

 

Quote

Lycopodium powder is the spores from a club moss. They are extremely hydrophobic so create a barrier between the fingers and water 

 

[exchemist]

10 hours ago, studiot said:

 

https://www.facebook.com/watch/?v=323855678960326

 

 

Thanks - something I did not know. But that raises a question in my mind about its use to demonstrate Brownian motion. My understanding is this is done in a water droplet on a microscope slide. But if the grains are hydrophobic, then presumably they will not be in the bulk liquid but clinging to the glass of the slide and being nudged this way and that, along the surface of the glass. Is that your understanding?  Or is a surfactant used to to get them to disperse into the water? 

[studiot]

2 hours ago, exchemist said:

Thanks - something I did not know. But that raises a question in my mind about its use to demonstrate Brownian motion. My understanding is this is done in a water droplet on a microscope slide. But if the grains are hydrophobic, then presumably they will not be in the bulk liquid but clinging to the glass of the slide and being nudged this way and that, along the surface of the glass. Is that your understanding?  Or is a surfactant used to to get them to disperse into the water? 

I checked an my memory of long ago was correct.

I thought brownian motion was where I had first heard of lycopodium powder.

 

Quote

Lycopodium powder - Wikipedia

Because of the very small size of its particles, lycopodium powder can be used to demonstrate Brownian motion. A microscope slide, with or without a well, is prepared with a droplet of water, and a fine dusting of lycopodium powder is applied.

 

 

Quote

https://www.vedantu.com/question-answer/what-is-the-role-of-lycopodium-powder-class-12-chemistry-cbse-60b06e074d283232b04d68

 

What is lycopodium powder used for in physics?

 

 

The powder has long been used in physics experiments to explain Brownian motion and other phenomena. During the experiment to estimate the molecular size of oleic acid, lycopodium material coats the entire water surface. Since oleic acid does not dissolve in water, a drop of the solution spreads on the water surface.

 

[exchemist]

59 minutes ago, studiot said:

I checked an my memory of long ago was correct.

I thought brownian motion was where I had first heard of lycopodium powder.

 

 

 

 

Yeah but I'm not questioning whether or not it is used to show Brownian motion. Evidently it is. It is just that my previous understanding of the demonstration, that the particles were in the body of the liquid and being buffeted in all directions by water molecules, can't be correct. They must be clinging to the glass and sliding around in 2 dimensions only.  

[studiot]

9 minutes ago, exchemist said:

Yeah but I'm not questioning whether or not it is used to show Brownian motion. Evidently it is. It is just that my previous understanding of the demonstration, that the particles were in the body of the liquid and being buffeted in all directions by water molecules, can't be correct. They must be clinging to the glass and sliding around in 2 dimensions only.  

Water has very strong Van Der Waals forces.

Hydrophobic particles are not influenced by these.

That leaves momentum/KE exchange for the buffeting.

[exchemist]

1 minute ago, studiot said:

Water has very strong Van Der Waals forces.

Hydrophobic particles are not influenced by these.

That leaves momentum/KE exchange for the buffeting.

This does not seem to address my point. 

I suppose one advantage of the particles clinging to the glass surface would be that they all stay in one plane and thus can be easily kept in focus when examined with the microscope.  I would have thought their motion might be inhibited somewhat by contact with the glass, but apparently it is not enough to prevent the demonstration from working. I suppose if they are hydrophobic they will probably not be strongly stuck to the glass either, since that too is a polar medium. 

Perhaps what I need to do is find some detailed instructions for how to set this demonstration up, and see what they say about it.  

[studiot]

12 minutes ago, exchemist said:

This does not seem to address my point. 

I suppose one advantage of the particles clinging to the glass surface would be that they all stay in one plane and thus can be easily kept in focus when examined with the microscope.  I would have thought their motion might be inhibited somewhat by contact with the glass, but apparently it is not enough to prevent the demonstration from working. I suppose if they are hydrophobic they will probably not be strongly stuck to the glass either, since that too is a polar medium. 

Perhaps what I need to do is find some detailed instructions for how to set this demonstration up, and see what they say about it.  

I look forward to your report.

Please note I am not expert in this  -  it was only a memory from long ago so I was gratified to find out that my memory had not failed me.

So I am having to do some fast thinking on the hoof to answer your excellent questions and if you come up with something new I would be glad to hear about it.

[exchemist]

1 hour ago, studiot said:

I look forward to your report.

Please note I am not expert in this  -  it was only a memory from long ago so I was gratified to find out that my memory had not failed me.

So I am having to do some fast thinking on the hoof to answer your excellent questions and if you come up with something new I would be glad to hear about it.

Well, it starts to get interesting. I did not find any instructions for demonstrating how to do this using lycopodium spores but instead came across this amateur site: http://www.microscopy-uk.org.uk/mag/indexmag.html?http://www.microscopy-uk.org.uk/mag/artoct20/ms-brownian.html which seems to claim the grains are too big to show Brownian motion themselves and it is only when they burst and release their contents that there are particles small enough to demonstrate the effect.

This site also says the hydrophobic properties of lycopodium spores are due to air trapped by hairs on their surface. If so, when they burst, this would not apply to their contents - which might then be able to disperse in the body of the liquid.

So maybe we are getting somewhere.