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Offline madscientist

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Mole fraction help (going nuts)
« on: August 13, 2008, 11:33:34 PM »
Hi all,

I am having lots of trouble working something out (I am waiting on a prescribed text book which was ordered 6 weeks ago which doesn't help)
any help would be greatly appreciated.

The question reads as follows:

The standard Gibbs energy of reaction for the decomposition of water

H2O(g) --> H2(g) + 1/2 O2(g)

is 135.2 kJ/mol at 2000K.  Suppose that steam at 350 kPa is passed through a furnace tube at 2000K. Calculate the mole fraction of H2(g) in the output gas stream.


My working so far:

delta Gorxn = 135.2 kJ/mol (at 2000K), (i dont understand how this can be the standard gibbs energy when the temperature and pressure is non-standard (2000K)??? make sense or am i just dense?)

Ptotal = PH20 = 350 kPa = 3.5 bar

T = 2000K

mole fraction of H2(g) = XH2 = nH2 / ntotal = VH2 / Vtotal

(n = moles)

PH2 = (XH2)*(Ptotal)

XH2 = PH2 / Ptotal = PH2 / PH2O

All I could think of to do was to calculate the equilibrium constant for this reaction from the Gibbs energy value given and then evaluate the constant in terms of partial pressures...

delta Gorxn = - RT ln K

and

K = e(delta Gorxn / -RT)
   = e(135200 J.mol-1 / (- 8.3145 J.K-1.mol-1 * 2000 K))
   =2.945*10-4

The equilibrium constant expression in terms of partial pressures....

K ~ Kp = (((PH2)(PO2)1/2) / PH2O)(Po)-delta n

Po = standard pressure = 1 bar

delta n for the reaction:

aA --> bB + cC

is equal to: c + d - a

for the decomposition of water: delta n = 1 + 0.5 - 1 = 0.5

Here is where I get stuck because i don't have all the variables needed in order to calculate PH2, the closest i can get is as follows:

(PH2)(PO2)1/2 = (PH20)(K) / ((Po)- delta n)
                   = (3.5 bar)(2.945*10-4) / ((1 bar)-1/2)
                   =1.031*10-3 bar1.5

I'm tempted to just put:

XH2 = (1 - (XO2 + XH2O))

But i don't think it would go down too well...

Any ideas or hints would be extremely helpful.

Cheers,

Mad
The only stupid question is a question not asked.

Offline Yggdrasil

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Re: Mole fraction help (going nuts)
« Reply #1 on: August 14, 2008, 10:44:33 AM »
Use an ICE table.

Offline Hunt

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Re: Mole fraction help (going nuts)
« Reply #2 on: August 19, 2008, 01:37:21 PM »
Standard Gibb's free energy can be defined at any temperature as long as the pressure is 1 bar for a gas. That's how the standard state is defined. But at 350 KPa this is no more standard , so you should calculate the corresponding non-standard delta G. First you should use ICE tables or something like that to express the mole fraction in terms of the advancement alpha ( P / PT ). You should have no trouble there. I think the expression for the mole fraction of hydrogen is :

XH2 = 2 alpha / 2 + alpha

To find alpha , you need the usual expression in terms of Kp :

Kp = Po1/2 alpha3/2 / 21/2 ( 1 - alpha )

You already know the relation between Kc and Kp. However you cant use the value of standard delta G ( at 1 bar ) to find Kc. You need to find the new delta G at 350 KPa. I think an approximate expression for this is :

Delta G = Delta Go + (delta n ) RT Ln P

Where delta G is the new delta G at pressure P ( 350 KPa ) . Both delta G's are at 2000 K. delta n is the difference in stoichiometric coeffcients ( 1 +1/2 -1 = 1/2 )

Once delta G is found, Kc is found and then Kp is deduced . Then alpha and finally the mole fraction is found. All of this can be reduced to one equation , if u have time you can try it. It might look messy though ...

I didnt give the proof of the relations but incase of doubt you can come back and i'll post when i have time. Good luck . 

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