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### Topic: equilibrium  (Read 310 times)

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#### alwaysalone

• New Member
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• Mole Snacks: +0/-0 ##### equilibrium
« on: July 11, 2019, 07:31:03 AM »
The problem: For T= 1500 K and V=1L : CaCO3(s) :requil:CaO (s) + CO2(g) with   K= 0.05mol/L  and CO2(g) ::equil::CO (g) + O2(g) with XO2= 0.15 in equilibrium. There is 1 mol of CaCO3at the begining, then moles CaCO3in equilibrium.
I try this way:

CaCO3(s) =CaO (s) + CO2(g)    K= 0.05mol/L
initial n       1 mol          0            0
equil n        1-x             x            x

x/1L =K=0.05M = [CO2] and then I supossed that this quantity is the begining quantity for the other equation:
CO2(g) =CO (g) + 1/2O2(g)
initial n   0.05 M          0           0
equil n   0.05 -x         x          1/2x

Xo2: 0.15 = 1/2x/(0.05-x+x+0.5x)   x=0.017 M
Now, I've lost myself. ¿What can i do with this x?

#### Corribus

• Chemist
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• Gender: • A lover of spectroscopy and chocolate. ##### Re: equilibrium
« Reply #1 on: July 11, 2019, 10:35:41 AM »
I don't believe you can solve the problem by treating each equilibrium as isolated from the other, because the two equilibria are interconnected.  E.g., at equilibrium, the concentration of CaO and CO2 are not identical, because CO2 is consumed in the second reaction. Therefore you cannot solve for x in the first table as though these two species have the same concentration.

I would set up two ICE tables with different variables x and y for changes. The change in CO2 would then be x-y (gained in first equilibrium and lost in second). Then you can solve for x because you know y = 0.15 mol. A quick solve gives me moles of CaCO3, CaO and CO2 at equilibrium equal to 0.721, 0.279, and 0.129, respectively. This does return the equilibrium constant for the first reaction of 0.05, so at least that checks out.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman