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

Consider the formation of hydrogen fluoride:

H2(g) + F2(g) ↔ 2HF(g)

If a 3.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0054 M H2 is connected to a 3.6 L container filled with 0.036 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

Solutions

Expert Solution

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

           

queen_honey_blossom answered 2 years ago
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