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At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 25.0. A reaction mixture in a 10.00-L flask contains 0.33 moles each of hydrogen and fluorine gases plus 0.30 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

Solutions

Expert Solution

   H2(g)   + F2(g)    <===> 2HF(g)

initial          0.33/10 = 0.033 M      0.033 M                   0.03 M

change          x                                         x                            2x

equil             0.033-x                         0.033 - x                       0.03+2x

Reaction quotient(Q) = [HF]^2/[H2][F2]

                                           = 0.03^2/(0.033^2)

                                           = 0.826

   K = 25

Q<K .so that froward reaction is favourable.

at equilibrium

                  K = [HF]^2/[H2][F2]

                  25 = (0.03+2x)^2/(0.033-x)^2

     x = 0.02

concentration of H2 at equil = 0.033 - x

                                                 = 0.033- 0.02 = 0.013 M

queen_honey_blossom answered 1 year ago
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