calculate Kc and Kp for the reaction at 295K, N2O4 ------------>2NO2
if the equilibrium concentration are [N2O4]=0.75 and [NO2]=0.062M, R=0.08206L at K-1 mol-1

                                  N2O4  2NO2Eqm concentration   0.75       0.062Kc = NO22[N2O4] = 0.06220.75 = 0.0051Kp = Kc × (RT)n   = 0.0051 × (0.0821 ×295)1    (n= np - nr = 2 - 1 = 1)    = 0.1241

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