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Determine direction of mass transfer
How can I find out the concentration of chlorine in an acid bath?How to use molality and molarity to determine the molar mass of a compound?Calculating Molar Mass of GasHeat Transfer during Phase ChangeHow do phase transfer catalysts bring anions to organic phase?EQPt pH of a Monoprotic Acid-Base Titration: Product of Molarites Over Their SumHow to without calculations understand the liquid-liquid extraction efficiency?Diffusion across a thin filmDirection of mass transfer in two film theory
$begingroup$
For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.
But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.
Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!
concentration phase diffusion
asked May 20 at 12:54
lotte07lotte07
825
$endgroup$
add a comment |
$begingroup$
For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.
But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.
Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!
concentration phase diffusion
asked May 20 at 12:54
lotte07lotte07
825
$endgroup$
2
$begingroup$
You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
$endgroup$
– Buck Thorn
May 20 at 13:43
$begingroup$
What's the source of what you copy/pasted?
$endgroup$
– Mithoron
May 20 at 22:52
add a comment |
$begingroup$
For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.
But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.
Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!
concentration phase diffusion
asked May 20 at 12:54
lotte07lotte07
825
$endgroup$
For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.
But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.
Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!
concentration phase diffusion
concentration phase diffusion
asked May 20 at 12:54
lotte07lotte07
825
asked May 20 at 12:54
lotte07lotte07
825
asked May 20 at 12:54
lotte07lotte07
825
asked May 20 at 12:54
lotte07lotte07
825
asked May 20 at 12:54
lotte07lotte07
825
825
2
$begingroup$
You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
$endgroup$
– Buck Thorn
May 20 at 13:43
$begingroup$
What's the source of what you copy/pasted?
$endgroup$
– Mithoron
May 20 at 22:52
add a comment |
2
$begingroup$
You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
$endgroup$
– Buck Thorn
May 20 at 13:43
$begingroup$
What's the source of what you copy/pasted?
$endgroup$
– Mithoron
May 20 at 22:52
2
2
$begingroup$
You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
$endgroup$
– Buck Thorn
May 20 at 13:43
$begingroup$
You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
$endgroup$
– Buck Thorn
May 20 at 13:43
$begingroup$
What's the source of what you copy/pasted?
$endgroup$
– Mithoron
May 20 at 22:52
$begingroup$
What's the source of what you copy/pasted?
$endgroup$
– Mithoron
May 20 at 22:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Could anyone please explain how to determine the direction of the mass transfer [...]
In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.
[...]and how it can be the same at t1 and t2
The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
$endgroup$
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
add a comment |
$begingroup$
You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:
- When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
- When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).
In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
$endgroup$
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
add a comment |
Your Answer
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2 Answers
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oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
Could anyone please explain how to determine the direction of the mass transfer [...]
In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.
[...]and how it can be the same at t1 and t2
The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
$endgroup$
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
add a comment |
$begingroup$
Could anyone please explain how to determine the direction of the mass transfer [...]
In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.
[...]and how it can be the same at t1 and t2
The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
$endgroup$
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
add a comment |
$begingroup$
Could anyone please explain how to determine the direction of the mass transfer [...]
In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.
[...]and how it can be the same at t1 and t2
The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
$endgroup$
Could anyone please explain how to determine the direction of the mass transfer [...]
In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.
[...]and how it can be the same at t1 and t2
The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
edited May 20 at 14:57
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
answered May 20 at 14:51
Karsten TheisKarsten Theis
6,6161050
6,6161050
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
add a comment |
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
$begingroup$
+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant. If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way. All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down. However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.
$endgroup$
– KalleMP
May 20 at 21:56
add a comment |
$begingroup$
You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:
- When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
- When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).
In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
$endgroup$
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
add a comment |
$begingroup$
You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:
- When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
- When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).
In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
$endgroup$
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
add a comment |
$begingroup$
You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:
- When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
- When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).
In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
$endgroup$
You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:
- When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
- When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).
In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
edited May 20 at 15:09
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
answered May 20 at 13:50
Buck ThornBuck Thorn
4,017627
4,017627
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
add a comment |
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
$begingroup$
I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.
$endgroup$
– lotte07
May 20 at 14:36
add a comment |
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You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.
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– Buck Thorn
May 20 at 13:43
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What's the source of what you copy/pasted?
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– Mithoron
May 20 at 22:52