Henry's law constant


by moleman1985
Tags: constant, henry
moleman1985
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#1
Mar25-07, 04:14 PM
P: 20
I need to draw an equilibrium curve for a process the equation being
y*=F(x)
y*=Hx, where H is henry's coefficient or constant, but I carnt find H and do not know which H I need (for which substance).
It is for the absorption of vapour water into sulfuric acid so I think I need to find the Henry's value for the water vapour and preferably at different temperatures and different pressures, the system is gas phased controlled.

thankyou
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daysyworld
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#2
Apr5-07, 02:10 PM
P: 12
Quote Quote by moleman1985 View Post
I need to draw an equilibrium curve for a process the equation being
y*=F(x)
y*=Hx, where H is henry's coefficient or constant, but I carnt find H and do not know which H I need (for which substance).
It is for the absorption of vapour water into sulfuric acid so I think I need to find the Henry's value for the water vapour and preferably at different temperatures and different pressures, the system is gas phased controlled.

thankyou
Hello, this is my first post here and I have to improve my English level, so sorry if I can't answer clearly your question.

Firstly, If you say that you want to draw an equilibrium curve.

The expresion y*=Hx, it's not a curve. It is a line. Usually if we work with a not concentrate solutions we can use the expresión because the equilibrium curve is like a line, and if you want to estimate the stages with McAbe-Thiele for the absorption it will be easy.

Then, if the solution is high concentrated you have to use a table with the sulfuric acid-water data, because the equilibrium will be a curve. In this case, if you want to aply McAbe-Thiele method you can work with a "soluto free basis" to transform your curve in a line.

The values you need are the equilibrium values for the Sulfuric acid - water system.
moleman1985
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#3
Apr5-07, 03:27 PM
P: 20
hi yes it is a dilute solute, I have all the mole fractions in and out of the absorption system and have drawn an operating line, is this operating line also infact the equilibrium line? I understand that this operating line gives me, y and x values (the solute fraction in the gas and liquid), but what else can I possible get from this line. Thankyou.

daysyworld
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#4
Apr5-07, 03:43 PM
P: 12

Henry's law constant


If I understand you correctly, the line you have dranw is the Equilibrium Line,
to draw the operation line you have to aply this equation:

G(Y1 - Y) = Ls(X1 - X)

being 1 the bottom of the tower.

Anyway you can obtain the same line joining the points (X1,Y1) [ bottom ] and (X2,Y2) [top]
moleman1985
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#5
Apr5-07, 03:52 PM
P: 20
ah yes that what my lecturer was saying to me, but if I presume that the mass transfer throughout the column is constant, won't the equilibrium line also be the operating line?
daysyworld
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#6
Apr5-07, 04:10 PM
P: 12
Quote Quote by moleman1985 View Post
ah yes that what my lecturer was saying to me, but if I presume that the mass transfer throughout the column is constant, won't the equilibrium line also be the operating line?
If the operation line is also the equilibrium line, your tower have no stages and this never happens.
( To explain that clearly you can see this graphic. The stages are calculated always between the two lines )

If you think about you have said, you'll see that you have calculated the operation line to specific concentration in the top, usuallư defined by performance. If you change the performance of the tower, the points (x1,Y1) (x2,Y2) will be diferent , and the operation line too.
Attached Thumbnails
torre.JPG  
moleman1985
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#7
Apr5-07, 04:29 PM
P: 20
Think I understand but that graph shows a straight line for the operating line which I have already, and a curve for the equilibrium line, I know how to extrapolate data from the two lines but cannot draw an equilibrium line I think. I constructed my operating line using (y1,x1) and (y2,x2) or should that be my operating line, if so how do I construct the other line.
thankyou.
daysyworld
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#8
Apr6-07, 03:19 AM
P: 12
Could you copy here the wording of the absorption , and the dates you have to resolve it ?

Maybe this way I can resolve your problem. Thanks.
moleman1985
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#9
Apr7-07, 11:37 AM
P: 20
hello after doing some further calculations I have obtained this graph which I hope shows an operating line and a equilibrium line, it looks like it.
The operating line is the straight one and was drawn using (x1,y1) (x2,y2), just those two points and they were joined and think has a gradient of approx. L/G.
The curve I hope is the equilibrium line, I constructed this by spliting the column into 10 equal stages, and presumed a constant mass transfer rate, the begging and end point are also (x1,y1) (x2,y2), but has the intermediate sections in it too.

Would this be correct? thankyou.
moleman1985
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#10
Apr7-07, 11:42 AM
P: 20
sorry this is the graph i drew
Attached Files
File Type: doc op. vs. eq..doc (23.5 KB, 23 views)
daysyworld
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#11
Apr7-07, 11:53 AM
P: 12
Hi, he looks good to me a priori.

My first option to draw the equilibrium line is taking real equilibrium dates, no spliting the tower, because the most important reason to do the operating line- equilibrium lin graphic is to obtain the real stages of the tower, but if they don't give you equilibrium dates, table or Henry's constant you can't do in another way.

Cheers.
siddharth
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#12
Apr7-07, 12:10 PM
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I need to draw an equilibrium curve for a process the equation being
y*=F(x)
y*=Hx, where H is henry's coefficient or constant, but I carnt find H and do not know which H I need (for which substance).
It is for the absorption of vapour water into sulfuric acid so I think I need to find the Henry's value for the water vapour and preferably at different temperatures and different pressures, the system is gas phased controlled.

thankyou
If you assume that the solute is dilute, then you can find the henry's coefficient using the Kremser Equation, if you set up your operating line in the solvent free basis.
daysyworld
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#13
Apr7-07, 12:19 PM
P: 12
Quote Quote by siddharth View Post
If you assume that the solute is dilute, then you can find the henry's coefficient using the Kremser Equation, if you set up your operating line in the solvent free basis.
The kremser equation for how many stages ? 10 ? It's not a date, it's assumption.
siddharth
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#14
Apr7-07, 12:21 PM
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Quote Quote by daysyworld View Post
The kremser equation for how many stages ? 10 ? It's not a date, it's assumption.
Yes, you're right, didn't notice that.

I suggest that the OP posts the question in full, exactly as it was given to him/her.
moleman1985
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#15
Apr7-07, 01:07 PM
P: 20
I've tried to find data to draw an equilibrium curve but cannot find it although that doesn't mean its not out there. The absorption tower consists of a gas stream containing water vapour and it needs to be reduced, the solvent used is 98% wt sulfuric acid, the acid at the bottom becomes 75% wt, due to its absorption.
So if I was to get equilibrium data would I have to get it for water vapour? as I am presuming that it is gas phase driven.
So was I wrong about my equilibrium curve then?
siddharth
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#16
Apr7-07, 01:22 PM
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Quote Quote by moleman1985 View Post
I've tried to find data to draw an equilibrium curve but cannot find it although that doesn't mean its not out there. The absorption tower consists of a gas stream containing water vapour and it needs to be reduced, the solvent used is 98% wt sulfuric acid, the acid at the bottom becomes 75% wt, due to its absorption.
Do you know the number of stages?
moleman1985
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#17
Apr7-07, 03:04 PM
P: 20
No i dont know the stages, I thought I could get that off the graph if I ever get an equilibrium line.
All I have to go off is the compositions of the liquid and gas in and out, everything else I have to do, although I know I have to use packed tower and plates, and a mid-way recycle which is cooled.
siddharth
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#18
Apr9-07, 09:04 AM
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Quote Quote by moleman1985 View Post
No i dont know the stages, I thought I could get that off the graph if I ever get an equilibrium line.
All I have to go off is the compositions of the liquid and gas in and out, everything else I have to do, although I know I have to use packed tower and plates, and a mid-way recycle which is cooled.
In that case, as daysyworld pointed out, I don't think you can find it.

Anyway, why do you want to calculate the Henry's law constant this way ? It's actually much easier to separately get the equilibrium data for this system at the given temperature and pressure. For example, look up a chemical engineering handbook.


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