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At a particular temperature, K = 1.00×102 for the reaction:

 H2(g) + F2(g) 2HF(g)
To the mixture above, an additional 9.50×10-1 mol of H2 is added.

What is the new equilbrium concentration (in mol/L) of HF?

Feb 2nd, 2015

Initial concentrations: H2: 1.9mol/1.25L=1.52mol/L F2: 1.9mol/1.25L=1.52mol/L

Using ICE Table: H2             F2            2HF

I: 1.52           1.52           0

C: x                 x               2x

E: 1.52-x       1.52-x         2x

Since K= Product/ reactant

100=(2x)^2/ (1.52-x) ^2

To solve for x take square root of both sides: 10=2x/(1.52-x)

15.2-10x=2x

15.2=12x

x=1.27

The final concentration of H2 is 1.52-x=1.51-1.27=0.25

adding 0.95: H2 is now 1.2mol/L now F2 is 0.25mol/L and is limiting

so all F2 react giving 0.25*2 more HF=0.5

adding 2x form before 2.54+0.5 =3.04 :)

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Feb 2nd, 2015

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Feb 2nd, 2015
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Feb 2nd, 2015
Dec 5th, 2016
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