3O2(g) + 4NO(g) ---->
2N2O5(g)
Rate of the reaction = rate of disappearance of reactants = rate
of appearance of products
  r = -(1/3)xd[O2]/dt = -(1/4)xd[NO]/dt =
+(1/2)xd[N2O5]/dt      Â
-----(1)
Given d[O2]/dt = 0.057 mol/L.sec
From Eqn(1) , -(1/3)xd[O2]/dt = -(1/4)xd[NO]/dt
                          Â
d[NO]/dt = (4/3)xd[O2]/dt
                                        Â
= (4/3) x 0.057
                                        Â
= 0.076 mol/L.sec
-(1/3)xd[O2]/dt =
+(1/2)xd[N2O5]/dt  Â
d[N2O5]/dt  =
(2/3)xd[O2]/dt
                Â
= (2/3) x 0.057
               Â
= 0.038 mol/L.sec