Question

At 25oc, Kc = 0.090 for the following reaction. ​H2O(g) + Cl2O(g) 2 HOCl(g) Calculate the...

At 25oc, Kc = 0.090 for the following reaction.
​H2O(g) + Cl2O(g) 2 HOCl(g)
Calculate the equilibrium concentration of all species if the initial concentration of HOCl = 1.50 M.
​H2O(g) + Cl2O(g) 2 HOCl(g)
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Answer #1

                        ​H2O(g) + Cl2O(g) -------------> 2 HOCl(g)

Start                  0                0                                1.5

Change             +x              +x                                 -2x

Equil.                  +x             +x                               1.5-2x

                   Kc     = [HOCl]^2/[H2O][Cl2O]

                   0.09    = (1.5-2x)^2/x*x

                  0.09    = (1.5-2x/x)^2

                0.3        = 1.5-2x/x

              0.3x         = 1.5-2x

                   x   = 0.652

            [H2O]   = x   = 0.652M

            [Cl2O]   = x   = 0.652M

           [HOCl]    = 1.5-2x   = 1.5-2*0.652    = 0.196M

               

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