HI/RP reduction general info and calculations

[lutesium]

General comments about HI reductions

Addition of a reactant that will react with HI to get the I from HI and release H2O will considerably drop the concentration and might bring it to a level which is too low to be useful to continue the rxn and this will block everything.

 To avoid this happening, using more HI solution is a good way to go. But adding more I2 will also bring the concentration higher if enough RP is present. If 2nd route is choosen the rxn will release HI gas as it proceeds (reduces the substrate)

This can be explained as follows:
The substrate will use I2 (from HI),release H2O, you will add some more I2 to bring the conc. of the acid back to %57 and when the substrate gives its iodine back to be reduced the rp will convert the given I to HI which will be excess and escape from the condenser as the rxn proceeds. This might be a good sign of a complete rxn. When fuming stops the reducing
is complete (30-48 hours) 


Also using freebase reactants will considerably push the yields up because the HCl ions will harden the formation of HI due to the Cl- ion which is a better nuclepohile than I-

Now lets get to the calculations:


T: Amount of H2O put at the start
L: Amount of H2o left from the HI synthesis
U: Amount of H2o used for the HI synthesis

  T=L+U
--------------------------
G: Amount of HI dissolved  (in the leftover H2O(L))
S: Amount of the total solution

(1) S=G+L

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For a max %57 solution

(2)  (S.57)/100=G  (The amount of dissolved HI is equal to the %57 of the final solution)

---------------------
In the above equation put (G+L) in the place of S  (S=G+L)

(3)  ((L+G).57)/100=G

-------------------
To find U we can use the following equation
 
3H2O + PI3 -> 3HI + H3PO3
This means that: For every mole of HI synthesised we waste 1 mole of H2O (1:1)


U= Nu . MwH2O (Nu: Mole amount of wasted H2O , MwH2O: Molecular wt of h2O=18)
Nu=Ng (Ng: Mole amount of synththesised HI)       

(4)  U=Ng.18

-------------
T=L+U =>
L=T-U
 ==acc to 4=>

(5)  L=(T-(Ng.18))
-------------

To find G (Amount of HI in the sol.) 

G= Ng . MwHI =>

(6) G=(Ng . 128)

NOW TO (3) PUT
G AS (6)
L AS (5)
 
   |----L----||---G----|             |---G----|
( ((T-(Ng.18))+(Ng.128)) .57) /100 = (Ng . 128)
 
(T+110Ng)57=12800Ng

When you solve the eqation the result will be

 
(7)        T = 114 . Ng

So what does this mean?

this means that the amount of water that will be put in the reaction from the start
should be 114 times of the mole amount of the HI that will be synthesised

Or this can be tweaked as

Ng = Mg / MwHI
Ng=  Mg / 128

If you put the above value in the place of Ng at (7)
and solve the equation

( T= 0,89 . Mg

Mg means the weight of HI in the reaction mixture.

Because we use 1 mole of Iodine for every mole of HI created the results will be identical.


AND THE FINAL RESULT IS 

 T= 0,89 . Mg

THIS MEANS THAT WE SHOULD START WITH H2O WEIGHING 0,89 TÝMES OF THE I2 ADDED.

(T IS H2O ADDED AT THE START  -  Mg ÝS THE WEÝGHT OF HI WHÝCH IS TAKEN EQUAL TO I2)

-------------------------------------------------
In other words the GOLDEN EQUATION is:

A: The weight of the water put in from the start
B: The amount of I2 that should be added
 
 A= 0,89 x B


----------------
Example

For 10mls H2O

10=0,89.B
B=11,2
 
11,2 g I2 should be added

And the result will be 21,2 gms of %57 HI solution
---------------------------------------------------------

These are golden equations that will help anyone interested in HI/RP reactions. All this hard work will be best rewarded only if it helps some bee here.

Thanx

[geezmeister]

This would suggest an approximate ratio of I:H2O of 1.2:1.0. This ratio works very well if one refluxes for 48 hours and uses lab grade precursors. I found refluxing for 48 hours at this ratio gave better yields than 36 hours.

I also found during the course of a number of experiments that the quality of the product and the yield from a 36 hour reflux done using a 1.2:0.8 ratio were equivalent to the longer reflux with the higher ratio of water. These refluxes were done with a balloon on the condenser to contain any excess HI, although the balloon was considered as helpful only for the first half of the reflux. If I had no reason to open the reaction, which was the norm for this reaction, the balloon was left in place and the reflux was stopped with the same pressure measured by balloon size as was present at the start of the reaction. Whether the slight pressure provided by the balloon had the effect of increasing the concentration of the HI above 57% in the reaction I do not know, but I assumed there was some small increase because of the pressure.

The wetter refluxes were done without a balloon after the initial HI production stage, with no noticed loss in yield or quality. I hope these comments may be of some help in evaluating your conclusion.

[lutesium]

Increased pressure should increase the solublity of gases. But a baloon does not
increase the pressure due to its ability to increase in volume at the same time.
I know that you already know this but I want to explain it

P.V = n.R.T

P: Pressure
V: Volume
n: Mole number
R: Gas constant(22,4/273)
T: Absolute temprature (Kelvin) ( 0C= 273K) (C+273=K)


In your case you increased n (by creating HI) but this caused a proportional
increase in volume (baloon). The resistance of the membranes of the baloon
is the only way to compress the HI inside the system and this will create some
pressure but I dont think that this resistance is so noteworthy to increase
the solubiliy of the gas. So I think that a baloon is not worth the hassle
other than security reasons.


And sonething important to note here is that gas solubility decreases with increasing
temprature. So rxns done at lower tempratures should dissolve more HI.


>I also found during the course of a number of experiments that the quality of the >product and the yield from a 36 hour reflux done using a 1.2:0.8 ratio were >equivalent to the longer reflux with the higher ratio of water

This is an interesting result. Theoritically using more HI solution seems safer
than little because with little HI the HI can be used to a level that the concentration falls below what's needed to accomplish the reduction. I think the gaseous HI escaping from the top of the condenser is a good indicator of the
concentration of the solution. In many rxns HI fumes can be seen escapeing even at the 30th hour of reflux. This means the iodoE is reduced to mtmph and given its
I. This I reacted w/P then H2O to create more HI which escapes from the rxn due to the already saturated (%57) rxn mixture

Hope this creates new thoughts

[geezmeister]

The increase in pressure depends on the balloon, at least in part. I used a bicycle inner tube for a time, until I had a flask explode on me mid-reaction. The increase in pressure using a single balloon may be small. It might also be hard to call it insignificant, particularly if you use two of them. Your comment assumes the balloon's resistance to expansion is constant. In my experience it is not. IT increases with expansion. A non-elastic bladder would have no effect on pressure, while still containing the gas.

My balloon use isn't for the purpose of increasing the pressure or the concentration of the HI. It to keep what HI I make in the flask as much as possible.

You start from a premise that 57% concentration of HI is needed to drive the reaction. I'm not sure I agree with you. I think as long as you have a 50% concentration, the reaction will proceed just fine. Certainly nothing wrong with a higher concentration, though.

[ahgreich]

that the aq HI is created before the addition of the pfed?

[lutesium]

Geez:  I made my comments theoritically. Of course in real world a baloons resistance
incerase as it expands. But using an inner tube of a bicycle is nothing different than stoppering the glassware and watching it explode because of the strong resistance of the meterial that the tube is made

>You start from a premise that 57% concentration of HI is >needed to drive the reaction. I'm not sure I agree with >you. I think as long as you have a 50% concentration, the >reaction will proceed just fine. Certainly nothing wrong >with a higher concentration, though.

I didn't mean that %57is essential  for the rxn to complete. Of course a lower conc. will do. But the most important point here is the mole amount of total HI present in the solution. In other words, say you reduce 1gms E. The rxn will go smoother and will possibly never lock up if you used 10mls H2O and 11,2g I2. This is because the large excess of HI present in the solution. But trying to reduce 10g E with the same amounts of H2O and I2 will create many problems due to the large HI absorbtion (to form IodoE) and large H2O release. These may diminish the conc. up to %20, drop the total amount of HI present in the solution. These will bring the rxn to a lockup.

And this is the main problem of wet refluxes. At dry rxns the rxn is continuously dried by the forming PI3 and the HI conc is always at the max value possible.


And what is your point to prevent the excess Hý escape? As I mentioned earlier, after the iEph has formed and HI conc dropped you add more I2 and conc. gets up to 57 again. And when Ieph gives its I, this I is reduced to HI again and this HI is excess because the solution is already %57. So  I dont think there is a reason to keep this already excess HI.


Ahgreich: Yes. The aq HI is created before the addition of pfed. This is the only acceptable way to do a HI/RP reaction. Addition of pfed before the HI is created causes many problems due to the exhothermic rxn between RP and I2.