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Ammonia can be oxidized in oxygen as shown: 4 NH3(g) + 3 O2(g) <=> 2 N2(g)...

Ammonia can be oxidized in oxygen as shown: 4 NH3(g) + 3 O2(g) <=> 2 N2(g) + 6 H2O(g). When 0.541 mol of NH3 and 0.595 mol of O2 are placed in a 1.00 L container at a certain temperature, the equilibrium {N2} is 0.1347. Calculate the value of Kc for the reaction. Give your answer to 4 decimal places.

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Answer #1

The molarity of NH3   = no of moles/volume in L

                                   = 0.541/1   = 0.541 M

The molarity of O2 = no of moles/volume in L

                               = 0.595/1   = 0.595M

             4 NH3(g) + 3 O2(g) <=> 2 N2(g) + 6 H2O(g)

I           0.541            0.595              0              0

C        -4x                   -3x                2x             6x

E         0.541-4x        0.595-3x         2x          6x

               [N2]   = 2x   = 0.1347M

                           2x   = 0.1347

                            x     = 0.06735

    [NH3]    = 0.541-4x    = 0.541-4*0.06735   = 0.2716M

   [O2]        = 0.595-3x    = 0.595-3*0.06735   = 0.39295M

   [N2]        = 2x                = 2*0.06735            = 0.1347M

   [H2O]       = 6x               = 6*0.06735             = 0.4041M

            Kc               =   [H2O]^6[N2]^2/[NH3]^4[O2]^3

                                 = (0.4041)^6*(0.1347)^2/(0.2716)^4(0.39295)^3   = 0.2393 >>>>>answer

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