ICE chart, Equilibrium question

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SUMMARY

The equilibrium constant (Kc) for the reaction H2(g) + F2(g) <=> 2HF(g) is calculated using the ICE chart method. Given initial concentrations of 0.010 mol H2 and 0.050 mol F2 in a 1.00 L container, and 0.0125 mol HF at equilibrium, the correct calculation yields Kc = 9.51. The confusion arose from an incorrect initial calculation of Kc as 9.3 x 10^-6, which was clarified through peer assistance in the forum discussion.

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Homework Statement



What is the value of Kc for the reaction H2(g) + F2(g) <=> 2HF(g) given that when 0.010 mol H2 and 0.050 mol F2 are added to a 1.00 L container, 0.0125 mol HF is present at equilibrium.

Homework Equations


The Attempt at a Solution



I'm confused about how to set up the ICE chart in this scenario. We have

H2(g) + F2(g) <=> 2HF(g)

.01... (.05) ... 0
-x ...-x ... 2x
.01-x...(.05-x) ... (.0125)

2x = .0125, x = .00625,

[.0125] ^2 / [.01 - .00625] [ .05-.00625] = 9.3 x 10 ^-6

Which is the incorrect answer, 9.51 ^-1 is right but I'm not sure how to get there. Any help would be appreciated.
 
Last edited:
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0.1 or 0.01?
 
Borek said:
0.1 or 0.01?

Oops, I'm sorry it's .01, .05, Any ideas :D?
 
Check your math. Formula you derived looks OK to me and its value is not 9.3x10-6. Or at least that's not what I got just by keying it into a calculator.
 
Borek said:
Check your math. Formula you derived looks OK to me and its value is not 9.3x10-6. Or at least that's not what I got just by keying it into a calculator.

Ugh...That feeling when you stress out over a question just to type in the right answer 45 minutes later. Thanks you sir!
 

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