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Anonymous

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Right guys

We all know that H2SO4 in concentrated form attracts water. Conversely battery acid i.e. Dilute H2SO4 will evaporate down to much stronger acid if left in the open air.

So if I put a jug of 95% concentration H2SO4 in the same conditions as a jug of battery acid (approx 35%), where would the equilibrium point be?

Jon
 
spaceships said:
Right guys

We all know that H2SO4 in concentrated form attracts water. Conversely battery acid i.e. Dilute H2SO4 will evaporate down to much stronger acid if left in the open air.

So if I put a jug of 95% concentration H2SO4 in the same conditions as a jug of battery acid (approx 35%), where would the equilibrium point be?

Jon
Click to expand...
oh that just makes my head hurt.
 
where is butcher with some hard core math...... LOL

but my guess would be that they would never be equal due to the fact that left in the open environment at a ever changing percent of humidity that evaporation & absorption would be in constant flux.

put these 2 in a sealed container and maybe they come out equal in time.

It's Chaos Theory

EDIT: guess of % added 52% H2SO4 each in a sealed environment.
 
Well if I had the question, I guess I would do like the chemist would, and find the answer by experimentation, both solutions have an affinity for water, below the azeotrope of sulfuric acid the acid would not want to give up its water easily, I do not know the answer or if the more concentrated acid would even take water from the weaker acid, never though of it, or tried it, and have not studied it, so I guess if I wished to find out the answer, (I could act like I was a real chemist) and do some chemistry experiments to find out the answer to my question.
 
niteliteone said:
They will stay 95% and 35% respectively :shock:
You forgot to take the caps off the jugs in your example :lol:
Click to expand...


LOL, that is almost as bad as this.......

1. How would you put an elephant in a fridge ?
(A) open fridge door, put in elephant, close fridge door.

2. How would you put a zebra in a fridge ?
(A) open fridge door, remove elephant, put zebra in fridge, close fridge door.

3. The King of the Jungle is having a party. All of the animals come to attend the party except one. Which one is it ?
(A) the zebra, its still in the fridge

4. you have to desperately cross the jungle river to get to the other side. next to the river is a sign that reads "Caution deadly alligators & crocodiles"
How are you going to make it to the other side
(A) you swim, all the alligators & crocodiles are at the king of the jungles party.
 
Good points! :lol:
[stt]
I cannot see any reason, why they should meet halfway related to the concentration in wt%. The function even do not need to be linear, since H2SO4 builds different chemical structures of hydrates at different concentrations in a non-linear function as you can see in the graph: http://www.diss.fu-berlin.de/diss/s...ionid=70B25DCA44F80FDFA76B38878F7F89A1?hosts= ...at least at low temperatures, when it's crystallizing.

Edit: Okay, that is only about crystallized sulfuric, but look at graph 6.6 about vapor pressures as a function of temperature and concentration. It ist also non-linear.[/stt]

[stt]Edit: BUT no, we have a limited amount of water in the system, so it will be the same as if we just blended the two solutions. If it was 1 kg 30wt% and 1kg 80wt%, we have 300g+800g sulfuric in the total amount of liquid, the concentration will be 55wt%.[/stt]

[stt]Edit 3: BUT at 20°C the vapor pressure of 70wt% sulfuric is 100 mBar, so they will not blend equally in this case.

We have to know, how much of each solution we have. Then we need to know the given temperature and the exact concentration of the dilluted solution. And finally we have to presume a normal atmospheric pressure.[/stt]

(Source: graph 6.6 in http://www.diss.fu-berlin.de/diss/s...ionid=70B25DCA44F80FDFA76B38878F7F89A1?hosts=)
 
[stt]vessel A: 1kg 30wt% H2SO4
vessel B: 1kg 95wt% H2SO4
A and B are similar vessels with constant areal at all possible volumes
20°C
1000mBar


against the end of the experiment we will have

vessel A: 0,7386kg 70wt% H2SO4
vessel B: 1,2714kg 74,7wt% H2SO4

Vessel B will go on to attract H2O from vessel A until the forces will reach a final equilibrium between 70 and 74,7wt%, which I am not able to calculate, but ~72,35wt% are a good guess.

Please validate.[/stt]
 
mmhhh My calculation is a little off, I guess. Have to read about vapor pressure and think about it.... the vapor pressure is independant from the hygroscopic qualities of H2SO4, right?
 
But:

Not only is sulfuric acid hygroscopic in concentrated form but its solutions are hygroscopic down to concentrations of 10 Vol-% or below. http://en.wikipedia.org/wiki/Hygroscopy

Sulfuric is hygroscopic, because the dillution with water is a highly exotherm process. Also, we haven't got one chemical compound in sulfuric, but different blends of hydrates at different concentration, if I understood right. From practice we know, that the more concentrated, the more heat is produced by dillution. This might be the reason why the hygroscopicity drops with dillution to a point it is nearly equal to the forces of the vapor pressure of water. At room temperature the practice tells us, around 20-37,5% the hygroscopicity is negligable.

The point of azeotrophy can only be reached by forcing the water to vaporize off using heat (also endotherm processes must take part).

All this would explain, why flasks with sulfuric tend to "sweat" and the liquid that after some time may form around the flask seems to be dilluted sulfuric. The vapor pressure of sulfuric may be low, but it is existent, so there may gas some molekyles off. They meet water vapor outside the flask, attract it and condense. Then they attract more water vapor until a very dilluted sea of sulfuric forms around the bottom of the flask. I allways wondered about this phenomenon. Thanks for this thread!
 
Interesting question...I am now thinking about it nearly 30min...and came to the conclusion, that this question is really really difficult to answer. I think we need someone with a good thermodynamic knowledge. But I will tell my thoughts so far...

1. sulfuric acid is a solution of h2so4 and water with an azeotropic point at 98,3w%
2. every mixture of acid-water has a vapor pressure consisting of the partial vapor pressure of each component
3. therfore we have to use Raoult's law assuming a ideal mixture (non-ideal mixtures with positiv/negative deviation)
4. So for the question we have one mixture A(l) which produce a gasoid phase A(g) with a defined composition. And we have a mixture B(l) which produce a gasoid phase B(g) with a defined composition.
5. combining A(g) and B(g) leads to C(g)
6. C(g) is in an equlibrium with each mixture A(l) and B(l) and concentrate the first one and diminish the other one.
7. The great BUT is now, that with den concentration and diminishing the compostion C(g) is changing too. So the whole process is a complex differential equation.
8. with the additional assumption of a non-ideal mixture it would be easier to test it with an experiment instead of calculate it
 
Loito said:
Interesting question...I am now thinking about it nearly 30min...and came to the conclusion, that this question is really really difficult to answer. I think we need someone with a good thermodynamic knowledge. But I will tell my thoughts so far...

1. sulfuric acid is a solution of h2so4 and water with an azeotropic point at 98,3w%
2. every mixture of acid-water has a vapor pressure consisting of the partial vapor pressure of each component
3. therfore we have to use Raoult's law assuming a ideal mixture (non-ideal mixtures with positiv/negative deviation)
4. So for the question we have one mixture A(l) which produce a gasoid phase A(g) with a defined composition. And we have a mixture B(l) which produce a gasoid phase B(g) with a defined composition.
5. combining A(g) and B(g) leads to C(g)
6. C(g) is in an equlibrium with each mixture A(l) and B(l) and concentrate the first one and diminish the other one.
7. The great BUT is now, that with den concentration and diminishing the compostion C(g) is changing too. So the whole process is a complex differential equation.
8. with the additional assumption of a non-ideal mixture it would be easier to test it with an experiment instead of calculate it
Click to expand...


sounds like you already have a pretty good understanding of thermodynamics and a bit of chemical engineering knowledge :)
 
spaceships said:
Right guys

We all know that H2SO4 in concentrated form attracts water. Conversely battery acid i.e. Dilute H2SO4 will evaporate down to much stronger acid if left in the open air.

So if I put a jug of 95% concentration H2SO4 in the same conditions as a jug of battery acid (approx 35%), where would the equilibrium point be?

Jon
Click to expand...
Easy!

We have three different volumes/phases
- V1 jug with initially 95% concentration H2SO4
- V2 The atmosphere above the jugs
- V3 jug with 35% battery acid

The equilibrium point is when there is no changes in the system. The only way we can have that is when the partial pressure of water is the same in all three volumes and the same for the sulfuric acid.
In the beginning water will evaporate from V1 into V2. When the amount of water in the air increases it starts to be absorbed into the concentrated sulfuric acid. This will continue until the concentration in both jugs is the same.
As this is a process driven by the difference between concentrations the speed will slow down the to closer equilibrium it gets. A true equilibrium will probably never be reached since the sun puts a limit on the time for any experiment on the Earth. :mrgreen:

Bonus points for any calculation of the relative amounts of fluid in V1 and V2! (Hint, it's easy it just takes a little time!)

Göran

Göran
 

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