[blah2370]

Thank you

[Yggdrasil]

I would probably start from a different approach to solve this problem.  If you have no solid iodine left at the end, what is your final partial pressure of iodine.  What is your final partial pressure of HI at the end?

[blah2370]

I guess after reading this response i am a bit lost on how to start this problem. Is there anyone that can give me an idea on where to start? I thought of starting with the number of moles of solid I2 and creating an ice table that way, but it seems i come up with two unknowns (being the initial amount of H2 (the answer to the probleme?) and the x or alpha)

[Yggdrasil]

I would work backwards on the ICE table.  Figure out the amounts of each substance at equilibrium, then work backwards to find the initial amounts.  For example, you know that 12.7g of solid iodine need to be reacted away.  How much of this becomes iodine vapor?  How much becomes HI?  What partial pressure of HI does this correspond to?

[AWK]

This is much more a stoichiometric problem than an equilibrium one.
moles I2 in an equilibrium (0.1 bar , 10L, 313 K) convert to moles of I2 from the ideal gas law. 12.7 g of solid I2 also convert to moles. The  difference of these two numbers is equal to moles of I2 that should be converted to HI = moles of H2 bonded in HI (moles HI=moles of I2 bonded x 2).
The equilibrium H2 + I2 = 2HI can be calculated in moles, concentrations, pressures - in all cases K is the same!
Having known moles of I2 and HI in the equilibrium, you can calculate moles of H2 in this equilibrium.
Sum of H2 moles in the equilibrium and moles H2 used to formation of HI is the answer to your question (eventually convert them to mass)

[richter85]

Hmm I am also particularly interested in this problem. AWK Your method is confusing. I don't seem to follow what you are trying to say. Specifically, on the part where you mention the sum of H2 calculations.

Can a more clarified explanation can be posteD?

[AWK]

Hmm I am also particularly interested in this problem. AWK Your method is confusing. I don't seem to follow what you are trying to say. Specifically, on the part where you mention the sum of H2 calculations.

Can a more clarified explanation can be posteD?

unfortunately Blah2370 removed the problem:

The equilibrium constant for the reaction H2(g)+ I2(g) <--> 2HI(g) is 20.0 at 313K. The vapor pressure of solid iodine is 0.10 bar at 313K. If 12.7g of solid iodine are placed in a 10-L vessel at 313K, what is the minimum amount of hydrogen gas that must be introduced to remove all the solid iodine?

12.7 g of I2 = 0.05 mole
 moles of I2 in gas phase n=pV/RT=0.1 x 10 /(0.0831447 x  313) = 0.038426 mole
moles of I2 converted to HI 0.05 - 0.0381447 =0.011574
from which 0.023149 mole of HI were produced
k=nHI^2/(nH2 x nI2)
nH2= nHI^2/(nI2 x K)=5.3586E-4 / (20 x 0.011574) = 0.0023149 at the equilibrium
total H2 =0.0023149 + 0.011574 = 0.01389 moles = 0.0278 grams

[richter85]

Awk I got one more question;

I don't really agree to your answer (0.01389 moles of H2)

This is how I tried to do the problem.

My ICE Tables goes

        I2       H2       HI
I     0.05     x           0
C     -b       -b         +2b
E    0.0001   x-.4999   2(.4999)

I don't know if you get my logic on this, but i'll try to explain it. The problem asked to find a concentration of H2, in which I2 is headed to zero on equilibrium. I chose a small number for equilibrium of I (something close to zero). With this number, using the equation

Keq = [HI]^2/[H2][I2]
where H2 = x-.4999 and I2 = .0001, I solved for X. This is the initial concentration of H2 that will be needed to drive reactant (I2) to zero.

I noticed that as I2 approaches zero, you need to increase H2 concentration. IN fact, at 50,000 mol H2, I2 is very very small, but still some present. The answer I came to is about 7 moles of H2. At this concentration, there are only 0.0001 mole of I2 left (still a lot).



Is my logic right or am I missing something?


Using your answer of 0.01389 moles of H2;

20 = [2x]^2/[0.01389-x][0.05-x]

x = 0.06688 or 0.01297

0.06688 is discarded, because 0.05-0.06688 is not possible. therefore, at H2 = 0.01389, there is still 0.05-0.01389 = 0.03611 moles of I2 left.

[AWK]

You have information how much I2 is in the equilibrium!. You cannot set I2 as 0.0001

[DrCMS]

Is my logic right or am I missing something?

No you've missed the part of the question that said:
what is the minimum amount of hydrogen gas that must be introduced to remove all the solid iodine?

[richter85]

I seem to interpret 0.05 as the whole solid portion of iodine instead of part of it being the solid iodine.

and i didn't set 0.0001 as I2, For the E section, its actually 0.05 - B, but i wrote 0.0001 for short.

however, I still think my answer is wrong now. Thanks for the heads up.

[AWK]

Awk I got one more question;

ICE tables are useful when you start from amounts of reagents before an equilibrium is  reached. In your problem  you can find amounts of reagents in the equilibrium from stoichiometry (except one), not from ICE tables. Since K is known, you should only rearrange expression for K.