please help me within five munite

label Chemistry
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particular temperature, K = 1.00×102 for the reation

H2(g) + F2(g) 2HF(g)

 experiment,temperature,1.90 mol of H2 and1.90 mol of F2 are introduced into1.25-L flask and allowed  react.equlibrium all specie remain gas phase.Wht's the K.C(in mol/L) of H2?

Feb 1st, 2015

Please note that at Equilibrium, we have:

[H2] = y

[F2] = y

[HF] = 3.8 -2y

since we have 3.8 mol of HF. So substituting those quantities into the expression of the Equilibrium constant, we have:

100 = [(3.8 -2y)^2] / (y^2), or:

10 = (3.8-2y) /y, from which I get:

[H2] = y= 0.316 mol/liter, more precisely we got a dissociation of about:

(2 *y)/ [HF] = (2*0.316)/ 3.8= 17% (approximated value) of HF

Feb 1st, 2015

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Feb 1st, 2015
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