Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: scientizscht on 06/09/2019 18:02:56
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What energy is released from mixing two bottles with 1M of HCl and 0.5M of HCl?
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Not much.
I think you already calculated it. It was something like a hundred meV (Not MeV) per hydrogen ion.
A litre of molar HCl contains 1 mole of hydrogen ions.
https://en.wikipedia.org/wiki/Faraday_constant
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Not much.
I think you already calculated it. It was something like a hundred meV (Not MeV) per hydrogen ion.
A litre of molar HCl contains 1 mole of hydrogen ions.
https://en.wikipedia.org/wiki/Faraday_constant
Where is that calculation please?
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Not much.
I think you already calculated it. It was something like a hundred meV (Not MeV) per hydrogen ion.
A litre of molar HCl contains 1 mole of hydrogen ions.
https://en.wikipedia.org/wiki/Faraday_constant
Where is that calculation please?
Here
https://www.thenakedscientists.com/forum/index.php?topic=77645.msg581986#msg581986
and also here
http://www.sciencemadness.org/talk/viewthread.php?tid=153432#pid621609
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Which calculation exactly measures that energy?
Is there an online calculator with it to experiment?
I read about the Nerst equation but it is about a redox reaction and electron transfer, which is not the case I described. Also, it gives you an energy but not sure how you can use that value.
Here someone calculates that:
E = -0.05916 V × pH
First, Energy is not measured in Volts. Second, what that pH is? Is it the difference or the absolute of one side and which one? Last, how does this apply to a bottle of 1L and a bottle of 1mL?
https://socratic.org/questions/how-does-ph-affect-the-nernst-equation
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Differences in energy are conserved.
So if you have two redox cells say H+ <--> H2 at a Pt electrode but where the concentration of H+ is different they will have different voltages.
Conversion from voltage to KJ/ mol is pretty simple.
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So you convert the V to kJ/mol, mol of what? protons? or HCL?
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So you convert the V to kJ/mol, mol of what? protons? or HCL?
Try both and see if you notice anything.
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What energy is released from mixing two bottles with 1M of HCl and 0.5M of HCl?
Hi scientizscht,
I notice the others are taking an electrochemical viewpoint.
However, unless this was your explicit setup, which I don't think is the case, then this is more of a 'heat' question.
Or rather, a change in enthalpy.
If we assume equivolume addition from both bottles (adding the same amount from both sources):
Final solution would be approx. 0.75M HCl;
thus, we can treat the total system change as either an Enthalpy of Dilution scenario, ΔHdil, or as an Enthalpy change of solution.
In such a scenario, we next determine whether the solution is ideal. In an ideal solution,
solvent-solvent, solvent-solute, and solute-solute interactions are equivalent. In this case, the change in enthalpy between the initial and end states is zero (0).
For our system, interactions are NON-ideal, and so |ΔHdil| > 0.
We can use a quantity called the Enthalpy of Dilution to Infinite Dilution.
By finding values for the two initial states, the 1.0M and the 0.5M HCl solution, going to infinite dilution, we can interpolate the value for the 0.75M system. Subtracting the two values from each other should work.
(NOTE: Most literature reports these relative to solution molality, making the conversion to a molarity system tedious at best)
If someone else is more familiar with how to treat these physical systems, I would like to know how we should view the enthalpy changes.
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Thanks for your analysis.
Isn't the difference in enthalpy the same as the electric voltage?
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Isn't the difference in enthalpy the same as the electric voltage?
Strictly speaking, there is neither a potential source nor a potential sink here, so I don't think there is any voltage to speak of. The enthalpy of mixing is probably not directly comparable to voltage as a result.
However, if we wanted to treated this as a system of two solutions making contact (but before mixing), we could pretend to have a short lived exchange cell.
In that case, there would effectively be a liquid junction between the two solutions.
For such a scenario, a junction potential exists. A liquid junction where the same electrolyte is present in both phases, but at different concentrations, has a potential equivalent to:
Ej = (t+ + t-) * (RT/F) * ln(a1/a2)
Where Ej is the junction potential, t+/- are ionic transfer coefficients (for H+ and Cl-, respectively), R is the ideal gas constant, T is temperature, F is Faraday's Constant, and a1/2 are the mean activity coefficients of the two solutions.
(see Electrochemical Methods, 2nd ed., by Bard & Faulkner for more detail).
Whether this view is viable is uncertain.
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Imagine I set up an electrochemical cell with two hydrogen electrodes, the two different concentrations of acid and a membrane- say filter paper- between them.
I can calculate the voltage.
And, if I connect the electrodes to a resistor I can get the cell to produce heat in that resistor.
If I leave it long enough the battery will "run flat"
At that point the concentrations of the acids will be equal.
If I make the paper thick enough the diffusion will be slow and most of the transfer of chloride ions will be that driven by the cell reaction rather than passive diffusion.
And if I make the resistor high enough (i.e. the current low enough) then the energy dissipated will all be used heating the resistor.
So, if I integrate the product of the current passed and the voltage I will get the energy released.
And the conservation of energy says that, if I start with two different concentrations, but end up with two equal concentrations, and I extract the maximal energy from that process, then it doesn't matter how I do it.
It doesn't matter if I do complicated stuff with impractical cells, or just mix the liquids.
The energy released is the same in both cases.
And, in the case of the cells, I can calculate it.
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Imagine I set up an electrochemical cell with two hydrogen electrodes, the two different concentrations of acid and a membrane- say filter paper- between them.
I can calculate the voltage.
And, if I connect the electrodes to a resistor I can get the cell to produce heat in that resistor.
If I leave it long enough the battery will "run flat"
At that point the concentrations of the acids will be equal.
If I make the paper thick enough the diffusion will be slow and most of the transfer of chloride ions will be that driven by the cell reaction rather than passive diffusion.
And if I make the resistor high enough (i.e. the current low enough) then the energy dissipated will all be used heating the resistor.
So, if I integrate the product of the current passed and the voltage I will get the energy released.
And the conservation of energy says that, if I start with two different concentrations, but end up with two equal concentrations, and I extract the maximal energy from that process, then it doesn't matter how I do it.
It doesn't matter if I do complicated stuff with impractical cells, or just mix the liquids.
The energy released is the same in both cases.
And, in the case of the cells, I can calculate it.
Can you present the calculations here?
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And the conservation of energy says that, if I start with two different concentrations, but end up with two equal concentrations, and I extract the maximal energy from that process, then it doesn't matter how I do it.
That clears up my confusion. No need for stretching out to find weird system analogies.
So, can just use the tried and true Nernst equation (as applied to membrane potentials)?
E = (RT/zF) * ln([solution 1]/[solution 2])
^ This is the form I'm thinking of.
Also, I really liked your explanation of the system energy via circuit design.
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And the conservation of energy says that, if I start with two different concentrations, but end up with two equal concentrations, and I extract the maximal energy from that process, then it doesn't matter how I do it.
That clears up my confusion. No need for stretching out to find weird system analogies.
So, can just use the tried and true Nernst equation (as applied to membrane potentials)?
E = (RT/zF) * ln([solution 1]/[solution 2])
^ This is the form I'm thinking of.
Also, I really liked your explanation of the system energy via circuit design.
Thanks but what are the units of this equation?
Is it kJ/mole?
I do not see the input for the absolute amount of the HCL.
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E in that equation is an emf- the units are volts.
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Okay, I found this form of the Nerst equation:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.ph-meter.info%2Fimg%2Fnernst4.png&hash=c36ef743dea206426a924ddff30422c3)
How do I find Eo?
Is the pH(inside)-pH(outside) the difference between the two different HCL solutions?
How do I find the voltage?
Does this equation say that the energy I can get, is the same regardless if I have 1kg of HCL solutions or 10kg?
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How do I find Eo?
For a hydrogen electrode, it is zero by definition.
It doesn't matter anyway because you are looking at differences in E
Is the pH(inside)-pH(outside) the difference between the two different HCL solutions?
More or less
Does this equation say that the energy I can get, is the same regardless if I have 1kg of HCL solutions or 10kg?
No
Because a voltage is an energy per electron
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Okay then how do I use the volts from the above equation for calculate the energy for a 1kg solution?
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1 eV = 96.487 kJ/mol
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1 eV = 96.487 kJ/mol
Thanks so the 0.0591x(pH difference) calculates eV? So two solutions of 1mol of HCL each with pH difference equal to 1, can generate 0.0591x96.487 kJ of electricity?
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Thanks so the 0.0591x(pH difference) calculates eV
Near enough.
Strictly it's an integral of the voltage as the concentration changes.
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Thanks so the 0.0591x(pH difference) calculates eV
Near enough.
Strictly it's an integral of the voltage as the concentration changes.
Okay will have a look, and how do I calculate the proton concentration from pH? It's been ages since I did these at school.
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The pH is the log of the reciprocal of the hydrogen ion concentration (Strictly, the H+ ion activity- but they are pretty close).
For molar HCl the pH is close to zero; for 0.5M it's close to 0.3 .
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Oh then the energy is miniscule, but power is V x I, so even if voltage is small, cannot you extract more power if you have a bigger cable where protons flow in larger quantities?
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Oh then the energy is miniscule,
What energy is released from mixing two bottles with 1M of HCl and 0.5M of HCl?
Not much.