At some temperature, K = 158 for the gas phase reaction What is the concentratio

Question

At some temperature, K = 158 for the gas phase reaction

What is the concentration of HF in an equilibrium mixture established by adding 4.03 mol each of H2 and F2 to a 1.00 L container at this temperature?

What would be the equilibrium concentration of HF if 3.48 mol HF were removed from the above equilibrium mixture?

At some temperature, K = 158 for the gas phase reaction What is the concentration of HF in an equilibrium mixture established by adding 4.03 mol each of H2 and F2 to a 1.00 L container at this temperature? What would be the equilibrium concentration of HF if 3.48 mol HF were removed from the above equilibrium mixture?

Explanation / Answer

H2 + F2 <----> 2HF

also you should probably tell us (since this is gas) if the equilibrium constant, K , is a concentration equilibrium constant or a pressure equilibrium constant.

assuming with concentration, Kc, since not enough info to assume the K is Kp.

4.03 mol/1 L = 4 M of each H2 and F2

K= [HF]^2/[H][F]

H2 + F2 <----> 2HF
4.03 M 4.03 M 0 (initial condition)
-x -x +2x (change)
4.03-x 4.03-x 2x

158=(2x)^2/[(4.03-x)*(4.03-x)] use quadratic formula
x= 3.47 Moles

[HF]= 2x= 2*3.47 = 6.95 M

part B:

H2 and F2 concentration decreased by 3.47 M
remaining = (4.03-x)=(4.03-.3.47)= 0.56 Moles

we take out 4.03 moles, in a 1 L space, it's the same as 4.03 M
so 6.95 M-4.03M= 2.92 M

H2 + F2 <----> 2HF
0.56 M 0.56 M 2.92M
-x - x + 2x
0.56-x 0.56-x 2.92+2x

158= (2.92+2x)^2 / [(0.56-x)*(0.56-x)]

158=(2.92+2x)^2 / (0.56-x)^2=[(2.92+2x) / (0.56-x)]^2

square root both sides:
12.56 = (2.92+2x) / (0.56-x)
x=0.282 moles

[HF] = 2.92+2x = 3.48 M

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