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Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 3.8 L nickel reaction...

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Chemistry

Consider the formation of hydrogen fluoride:

H2(g) + F2(g) ↔ 2HF(g)

If a 3.8 L nickel reaction container (glass cannot be used becauseit reacts with HF) filled with 0.0054 M H2 is connectedto a 3.6 L container filled with 0.036 M F2. Theequilibrium constant, Kp, is 7.8 x1014 (Hint, this is a very large number, whatdoes that imply?) Calculate the molar concentration of HF atequilibrium.

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4.0 Ratings (420 Votes)

Very large equilibrium constant means its product dominant reaction or reaction goes to completion .

After connection , Total volume = 3.8 + 3.6 = 7.4 litres.

Moles of Hydrogen = 3.8 x 0.0054 = 0.02052 moles

   Moles of F2   = 3.6 x 0.036 = 0.1296 moles

New molarity after connecting ,

[H2]     = 0.02052/7.4 = 0.00277 M    limiting Reagent

[F2]     = 0.1296/7.4 = 0.0175 M

   H2(g)     +               F2(g)        ↔          2HF(g)  

         0.00277                        0.0175                             0

      (0.00277-0.00277 )           (0.0175 - 0.00277)            2 x 0.00277= 5.54 x 10-3 M

           
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