At a particular temperature, K = 1.00×102 for the reaction:
What is the new equilbrium concentration (in mol/L) of HF?
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)
The final concentration of H2 is 1.52-x=1.51-1.27=0.25
0.25mol/L is the answer
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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