MaxL221

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What is the equilibrium constant, K, at 400K if 3.00 moles of CO2(g) introduced into a 2 liter container are 20.0% dissociated at equilibrium?
2 CO2(g) « 2CO(g) + O2(g)




A. (0.30)2(0.15) / (1.20)2 mol/liter
B. (1.20)2 / (0.30)2(0.15) mol/liter
C. (0.60)2(0.30) / (2.40)2 mol/liter
D. (2.40)2 / (0.60)2(0.30) mol/liter
E. (0.30)2(0.15) / (2.40)2 mol/liter


Answer is A can anyone explain this step by step...i'm having trouble with the mole conversion..thanks!
 

RockstarDMD

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MaxL221 said:
What is the equilibrium constant, K, at 400K if 3.00 moles of CO2(g) introduced into a 2 liter container are 20.0% dissociated at equilibrium?
2 CO2(g) « 2CO(g) + O2(g)




A. (0.30)2(0.15) / (1.20)2 mol/liter
B. (1.20)2 / (0.30)2(0.15) mol/liter
C. (0.60)2(0.30) / (2.40)2 mol/liter
D. (2.40)2 / (0.60)2(0.30) mol/liter
E. (0.30)2(0.15) / (2.40)2 mol/liter


Answer is A can anyone explain this step by step...i'm having trouble with the mole conversion..thanks!
Earlier today I explained an almost identical question. Here is the link.

http://forums.studentdoctor.net/showthread.php?t=297118
 

cryptozoologist

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disregard the temperature number. treat it like a simple equilibrium constant problem with an ICE table

3 mols CO2 / 2 L = 1.5 M concentration

20% dissociation results in teh following concentrations

2CO2 --> 2CO + O2
I 1.5 0 0
C -.3 +.3 +.15
E 1.2 +.3 +.15

answer A given the exponents
 
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