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Topic: equilibrium problems  (Read 4914 times)

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Offline 0053

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equilibrium problems
« on: July 13, 2008, 08:20:48 PM »
1. finding initial amount.

"At a certain temp, kc= 4.0 for the following rxn;
2HF(g) <=> H2(g)  +  F2(g)
A 1.0 L rxn vessel contained 0.045 mol of F2(g) at equilibrium. What is the initial amount HF in the rxn vessel?"

I tried doing:
     2HF(g) <=> H2(g)  +  F2(g)
I    0             0.045+x       0.045+x     
C    +2x             -x             -x
E     +2x           0.045           0.045

4= [0.045][0.045]/[2x]^2

but it didnt work out...the answer is suposed to be HF=0.11mol


2. Combined gas concentration

"solid ammonium carbamate, NH4CO2NH2, decomposes into ammonia gas and carbon dioxide. calculate the value of kc for this system at 40*C if the combined gas concentration at equilibrium is 1.41 x 10^-2 mol/L."

I tried setting it up as:
    NH4CO2NH2    <=>  2NH3    +      CO2
I     0                     7.05x10^-3 +x        7.05x10^-3 +x
C     +x                        -x                        -x
E      +x                        7.05x10^-3             7.05x10^-3

im wondering if its correct to divide the "combined gas concentration" equally between the products. and im also stuck because i cant figure out the equilibrium concentration for NH4CO2NH2. ???

Offline enahs

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Re: equilibrium problems
« Reply #1 on: July 13, 2008, 10:40:41 PM »
In your first ICE table (for question 1), your initial conditions are X HF, 0 H2 and 0 F2.



Question 2:


Do solid enter into the Equilibrium constant? If so, what value do they get? Goes that change the math?


Look at your reaction stoichiometry. If you have that many mol's of gas, what fraction of it is NH3 and what fraction is CO2? So do you just device by two?

Offline Borek

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Re: equilibrium problems
« Reply #2 on: July 14, 2008, 03:39:44 AM »
In your first ICE table (for question 1), your initial conditions are X HF, 0 H2 and 0 F2.

Not necesarilly. ICE table seems to be constructed correctly, Think in terms of reverse reaction, slightly convoluted, but sound. It gives 0.011 as an answer.

0053, are you sure it should be 0.11 and not 0.011?
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Offline enahs

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Re: equilibrium problems
« Reply #3 on: July 14, 2008, 08:44:59 AM »
I just assumed he did the math wrong when solving for X (I did not check it), so I thought setting it up another way might lead him to the correct answer.


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