Question

The reaction NO2 (g) + NO (g) ⇌ N2O (g) + O2 (g) reached equilibrium at a certain high temperature. Originally, the reaction vessel contained the following initial concentration: [NO2]i = 0.0560 M, [NO]i = 0.294 M, [N2O] = 0.184 M, and [O2] = 0.377 M. The concentration of the NO2, the only colored gas in the mixture, was monitored by following the intensity of the color. At equilibrium, the NO2 concentration had become 0.118 M. What is the value of Kc for this reaction at this temperature?

Answer 1

                               NO2 (g) +    NO (g)      ⇌      N2O (g)   +    O2 (g)

Initial                     0.056                0.294                   0.184             0.377

at equilibrium        0.056 +x           0.294 +x              0.184-x         0.377-x

So,

equilibrium NO2 concentration = 0.056 +x

But given that equilibrium NO2 concentration = 0.118 M

Hence,

                   0.056 +x = 0.118

                               x = 0.062 M

Hence,

equilibrium concentrations are

[NO2] = 0.118 M

[NO] = 0.294 +x = 0.294 + 0.062 = 0.356 M

[N2O] = 0.184-x = 0.184 -0.062 = 0.122 M

[O2] = 0.377-x = 0.377-0.062 = 0.315 M

Therefore,

     Kc = [ N2O] [O2]/ [NO2] [NO]

            = 0.122 x 0.315 / 0.118 x 0.356

             = 0.912

Therefore,

    Kc = 0.912