1.

\( 2 N _{2} O _{5} \longrightarrow 4 NO _{2}+ O _{2} \) if rate of Appearance of NO2 is \( 2 mol L ^{-1} \) find.i) Rate of Reactionii) Rate of Disappearance of \( N _{2} S \)

Answer»

\(2N_2O_5 \longrightarrow 4NO_2 + O_2\)

Overall rate of reaction r = \(\frac{-1}2 \frac{d[N_2O_5]}{dt}\)

\(= +\frac14\frac{d[NO_2]}{dt}\)

\(= + \frac{d[O_2]}{dt}\)

∵ Given, rate of appearance of NO2\(= \frac{d[NO_2]}{dt} = 2 mol/L\)

(i)  ∴ rate of reaction r = \(+\frac14\frac{d[NO_2]}{dt}\)

\(= \frac14\times2 \)

\(= \frac12 mol/L.s\)

(ii) rate of disappearance of N2O\(- \frac{d[N_2O_5]}{dt}\)

∵ \(r =\frac{-1}2 \frac{d[N_2O_5]}{dt}\)

\(- \frac{d[N_2O_5]}{dt} = 2 \times\frac12\)

\(= 1 mol/L.s\)

Hence, rate of disappearance of N2O5 is 1 mol/L.s.



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