Question on Le Chatelier's principle

[sgstudent]

When I have a reaction CO(g)+2H₂(g) <=> CH₃OH if I increas the concentration of CO, the equilibrium will be shifted to the right so more methanol will form. However, in that process won't the concentration of Hydrogen gas decrease? So won't that cause a change in equilibrium as well? If so, how would it affect the equilibrium

Also for ionic Salts like NaCl(s) <=> Na+ (aq)+Cl- (aq), how will increasing the NaCl affect the equilibrium? Because I'm not sure what's the concentration of pure solids. Also, I think this explains why solubility decreases with concentration but I can't seem to understand why that happens using Le Chatelier's principle here.

Hope you guys can help me out here :)

 

[Borek]

When I have a reaction CO(g)+2H₂(g) <=> CH₃OH if I increas the concentration of CO, the equilibrium will be shifted to the right so more methanol will form. However, in that process won't the concentration of Hydrogen gas decrease? So won't that cause a change in equilibrium as well? If so, how would it affect the equilibrium

Net effect is still shift to the right.

Also for ionic Salts like NaCl(s) <=> Na+ (aq)+Cl- (aq), how will increasing the NaCl

What does it mean "increasing the NaCl"?

 

[sgstudent]

Net effect is still shift to the right.



What does it mean "increasing the NaCl"?

Oh I understand the first one already. By that I meant putting in more solid sodium chloride into the solution. Would that increase the concentration of the solid sodium chloride in water shifting the equilibrium to the right?

Also for reactions like HA(g)+H2O(l)->H3O+(aq)+A-(aq) and I increase the volume of water. So in this case the concentration of water increases while the concentration of the H3O+ and A- drop so can I use that to explain that the equilibrium is shifted to the right? What about if the HA is already aqueous? Cos in that case HA's concentration would drop too and the right hand side products' concentration would also drop.

Thanks for the help :)

 

[Borek]

Oh I understand the first one already. By that I meant putting in more solid sodium chloride into the solution. Would that increase the concentration of the solid sodium chloride in water shifting the equilibrium to the right?

No, adding solid won't shift the equilibrium (assuming solid was already present). When talking about equilibrium we deal with concentrations of dissolved substances, amount of solids doesn't matter.

Also for reactions like HA(g)+H2O(l)->H3O+(aq)+A-(aq) and I increase the volume of water. So in this case the concentration of water increases

I would not say so. You are trying to extend meaning of the word "concentration" - while technically it makes some sense, in practice when we deal with separated phases we don't treat the systems in terms of concentrations of a phase in the mixture.

Please remember that LeChatelier's principle is just a rule of thumb, and as each rule of thumb, its applicability is limited. Trying to use it to predict more complicated systems you risk getting incorrect answers (even if - with some more insight - it can be still useful).

 

[sgstudent]

No, adding solid won't shift the equilibrium (assuming solid was already present). When talking about equilibrium we deal with concentrations of dissolved substances, amount of solids doesn't matter.



I would not say so. You are trying to extend meaning of the word "concentration" - while technically it makes some sense, in practice when we deal with separated phases we don't treat the systems in terms of concentrations of a phase in the mixture.

Please remember that LeChatelier's principle is just a rule of thumb, and as each rule of thumb, its applicability is limited. Trying to use it to predict more complicated systems you risk getting incorrect answers (even if - with some more insight - it can be still useful).

Hi Borek :)

Oh i understand the first one. So actually why does adding more solid NaCl into the water causes it to dissolve more? And why would there be a limit to that dissolving?

For the second question, what do you mean by different phase? Because I was thinking of it like this number of moles of H2O/Volume of the solution.

Thanks again Borek :)

 

[Borek]

Oh i understand the first one. So actually why does adding more solid NaCl into the water causes it to dissolve more? And why would there be a limit to that dissolving?

You are talking about different scenarios. One is when the solution is not saturated and there is no solid present - then, adding more solid will mean it dissolves. But when the solution is saturated, amount of solid present doesn't matter.

This is a little bit confusing case, as dissolution equilibrium requires solid to be present. When there is not enough solid to saturate the solution we can't talk about dissolution equilibrium.

For the second question, what do you mean by different phase? Because I was thinking of it like this number of moles of H2O/Volume of the solution.

Your reaction equation contained both a liquid and a gas, these are different phases.

 

[sgstudent]

You are talking about different scenarios. One is when the solution is not saturated and there is no solid present - then, adding more solid will mean it dissolves. But when the solution is saturated, amount of solid present doesn't matter.

This is a little bit confusing case, as dissolution equilibrium requires solid to be present. When there is not enough solid to saturate the solution we can't talk about dissolution equilibrium.



Your reaction equation contained both a liquid and a gas, these are different phases.

Ohh so say for H2O(l)<=>H2O(g), adding more water wouldn't cause the concentration of water to increase right? And since Le Chatelier's principle can only work in a closed system (right?), now that the volume of the vapour is smaller, the pressure would increase. So for it to reduce the pressure, it would go back into a liquid state. Am I right?

Lastly, was thinking about this for a while. My sister told me that the principle stated that it would not be able to completely override the stress on the system. What does this mean?

Would this mean that in my water to liquid equilibrium, when increasing the pressure it would reduce the pressure by reverting into a liquid. But it cannot revert it such that the pressure drops to the same amount? This seems to contradict this link I found http://www.articlesbase.com/college...the-vapor-pressure-of-a-substance-647877.html where it states that the vapour pressure would remain the same after adding more liquid. So I'm not too sure why this contradicts.

Thanks Borek :)

 

[Borek]

Ohh so say for H2O(l)<=>H2O(g), adding more water wouldn't cause the concentration of water to increase right?

Concentration of water in what? You are again dealing with two separate phases, and you are again trying to apply concentration to mixture containing more than one phase.

And since Le Chatelier's principle can only work in a closed system (right?), now that the volume of the vapour is smaller, the pressure would increase. So for it to reduce the pressure, it would go back into a liquid state. Am I right?

You final conclusion is OK, but as the starting point was rather shaky I am not convinced the path you used to arrive to that conclusion was correct.

Lastly, was thinking about this for a while. My sister told me that the principle stated that it would not be able to completely override the stress on the system. What does this mean?

Would this mean that in my water to liquid equilibrium, when increasing the pressure it would reduce the pressure by reverting into a liquid. But it cannot revert it such that the pressure drops to the same amount? This seems to contradict this link I found http://www.articlesbase.com/college...the-vapor-pressure-of-a-substance-647877.html where it states that the vapour pressure would remain the same after adding more liquid. So I'm not too sure why this contradicts.

There are two separate problems here.

In a given temperature saturated water vapor has a well known (measured) value - and nothing is going to change it. So if you do any change to the system, water will either condense till the vapor is saturated again, or it will evaporate till it is saturated again. But - if there is not enough water - it will be not able to saturate whole volume even after it will evaporate to the end.

In the simple, classic case - say mixture of CO, O₂ and CO₂ - when you add O₂ pressure goes up and reaction proceeds to the right, and the final pressure will be higher than it was before O₂ was introduced, but lower than it was after O₂ was introduced. That's what your sister means - some of the stress (additional pressure) was relieved, but not all (final pressure is not identical to the pressure from the very beginning).

I already told Le Chatelier's principle can be tricky, don't try to use it for complicated systems. Once you will understand how they work, you will be able to see what Le Chatelier's principle predicts for them, unfortunately it doesn't work the other way around - Le Chatelier's principle is not a good starting point to learn behavior of complicated systems.

 

[sgstudent]

Concentration of water in what? You are again dealing with two separate phases, and you are again trying to apply concentration to mixture containing more than one phase.



You final conclusion is OK, but as the starting point was rather shaky I am not convinced the path you used to arrive to that conclusion was correct.



There are two separate problems here.

In a given temperature saturated water vapor has a well known (measured) value - and nothing is going to change it. So if you do any change to the system, water will either condense till the vapor is saturated again, or it will evaporate till it is saturated again. But - if there is not enough water - it will be not able to saturate whole volume even after it will evaporate to the end.

In the simple, classic case - say mixture of CO, O₂ and CO₂ - when you add O₂ pressure goes up and reaction proceeds to the right, and the final pressure will be higher than it was before O₂ was introduced, but lower than it was after O₂ was introduced. That's what your sister means - some of the stress (additional pressure) was relieved, but not all (final pressure is not identical to the pressure from the very beginning).

I already told Le Chatelier's principle can be tricky, don't try to use it for complicated systems. Once you will understand how they work, you will be able to see what Le Chatelier's principle predicts for them, unfortunately it doesn't work the other way around - Le Chatelier's principle is not a good starting point to learn behavior of complicated systems.

Ohh so for 2CO(g)+O2(g)<=>2CO2 if I add in more O2, the pressure would increase so relieve that stress it would react to form CO2. However, the final pressure would still be greater than before any additional oxygen was added. But why is it like this? Cos if I had a case of a weak acid HA(g)+H2O(l)<=>H3O+(aq)+A-(aq) if some of the H3O+ gets reacted, in order to relieve the stress more HA and water would react. So how would we say that it cannot completely relieve the stress here?

 

[Borek]

But why is it like this?

Because that's the way it is, period.

It can be calculated using equilibrium constant - and you would see it is this way. It can be calculated from the thermodynamic properties of the system - and you would see it is this way. The principle is just a poor man's version of these approaches.