IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

shreyaarya

the following reaction was studied in a closed vessel:

2N₂O₅ $\rightarrow$ 4NO₂ + O₂

It was found dat d conc. of NO₂ increases by 2*10^ -2mol L^-s^- in 5 seconds.

calculate the rate of change of conc. of N₂O₅

MUDIT

ok for us only this reaction is important
2N2O5 gives 4NO2
so d[N205]/dt = (1/2)d[NO2]/dt i.e thr rate of change of conc of NO2 is twice the rate of change of conc of N2O5
now rate of change of conc of NO2= (2*10^-2)/5=4*10^-3
so rate of change of conc of N2O5= 2*10^-3 (ans)

please feel free to nudge me if you have any doubts

Ask & Discuss Questions with Community & Experts