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Author Topic: Calculating equilibrium product mass given initial reactant masses and Kp?  (Read 12720 times)

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o1ocups

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How do I calculate the mass of a product at equilibrium given initial reactant masses and Kp? I really don't mean to abuse this forum, but I am just stuck  >:(

So here is the question:

Hydrogen can be extracted from natural gas according to the following reaction:
CH4(g)+CO2(g) ::equil:: 2CO(g)+2H2(g)
Kp=4.5*102 at 825 K
An 85.0 L reaction container initially contains 22.3 kg of CH4 and 55.4 kg of CO2 at 825 K.
Assuming ideal gas behavior, calculate the mass of H2 (in g) present in the reaction mixture at equilibrium.

And here is what I did:

CH4CO2COH2
I16.3514.8100
C-x-x+2x+2x
E16.35-x14.81-x2x2x

(I got 16.35M and 14.81M with the moles (found with masses) and 85L)

Assume ideal behavior
Kp=Kc(RT)delta n
Kc=Kp/(RT)delta n
Kc=450/(0.08206*825)2
   =0.0982

Kc=[CO]2[H2]2/[CH4][CO2]
0.0982=(2x)2(2x)2/(16.35-x)(14.81-x)
4x4=(0.0982)(16.32-x)(14.81-x)
x=1.485M

Equilibrium concentration of H2 = 2x = 2(1.485) = 2.97M

And then I am stuck.

So I tried doing it as if it wasn't an equilibrium:
1259 mol CO2*2*2.016g H2 = 5.076 kg

But then that just doesn't make sense.

Where should I go from here? Can somebody give me a hint? Please?  :(
« Last Edit: February 16, 2009, 02:50:58 PM by o1ocups »
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Astrokel

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #1 on: February 16, 2009, 03:00:20 PM »

Quote
Kc=[CO]2[H2]2/[CH4][CO2]
0.0982=(2x)2(2x)2/(16.35-x)(14.81-x)
4x4=(0.0982)(16.32-x)(14.81-x)
x=1.485M

Check your math. After you got [H2]eqm, you could work out the moles with volume given.

ps: are you using GC to solve the power 4 equation? I think it is really complicated if you were to work it with Kc
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o1ocups

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #2 on: February 16, 2009, 03:07:54 PM »


After you got [H2]eqm, you could work out the moles with volume given.

That's where I am getting confused. How do I know the volume? It's not given.  ???
Would it just be 85 L since it's a gas?

Quote
I think it is really complicated if you were to work it with Kc

Is there an easier way to do it?  :-X

By the way, yeah, I'm not solving the equation by hand.
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Astrokel

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #3 on: February 16, 2009, 03:15:34 PM »

Quote
That's where I am getting confused. How do I know the volume? It's not given.

It is in the question :o

Quote
Is there an easier way to do it? 

By the way, yeah, I'm not solving the equation by hand.
If you are allowed to use GC or equation solver, then it is fine to work with Kc. Note power 4 equation has 4 solutions. I don't know if pressure way is better or not because i haven not try it yet. Anyway since you have work with Kc, might as well continue to see if you get the answer.
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o1ocups

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #4 on: February 16, 2009, 04:38:45 PM »

If I want to use the pressure instead, do I calculate the initial pressures of CH4 and CO2 using ideal gas law? I got huge numbers for them:
CH4: P=nRT/V=(1390)(0.0821)(825)/(8.5)=11076 atm
CO2: P=nRT/V=(1259)(0.0821)(825)/(8.5)=10032 atm

Then I just use ICE table and Kp and solve for x, right?
But my numbers seem to be wrong...

OK, say if they are right, once I solved for x, I would get the equilibrium pressure for H2, right? And then I just use that in ideal gas law again and solve for n (mole) then convert that to grams?
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Astrokel

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #5 on: February 16, 2009, 04:49:15 PM »

Yes. Careless mistake on volume and PH2 is 2x
.
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o1ocups

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #6 on: February 16, 2009, 05:19:46 PM »

OK. FINALLY.

But I still have to calculate for the percent yield.

Can I just take the 5076g that I calculated in the original post as my theoretical yield?
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Astrokel

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #7 on: February 16, 2009, 05:35:28 PM »

You are right theoretical yield is when you assume the reaction goes on completion - there is no equilibrium. But once you treat the reaction goes on completion, you are doing normal stoichiometry and have to be careful if there is any limiting reagent.
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o1ocups

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #8 on: February 16, 2009, 05:39:26 PM »

You are right theoretical yield is when you assume the reaction goes on completion - there is no equilibrium. But once you treat the reaction goes on completion, you are doing normal stoichiometry and have to be careful if there is any limiting reagent.

there are 1390 mols of CH4 but only 1259 mols of CO2, and since both of them is 1 mol (in stoichiometry), CO2 must be the limiting reagent, right? so that's what I used and I got 5076g, which is a LOT bigger than my actual yield. I ended up having an extremely small percent yield, and I can just instinctively feel that it's wrong  :'(
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o1ocups

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Re: Calculating equilibrium product mass given initial reactant masses and Kp?
« Reply #9 on: February 16, 2009, 05:50:06 PM »

Sorry, I actually got it right  ;D

And once again, thank you, thank you, thank you!!!  :)

Haha there are 4 guests viewing this. It's due in 6 minutes good luck.
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