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Chemistry Question on Chemical Kinetics

The rate of the reaction 2N2O5−>4NO2+O2 {2 N_2 O_5 -> 4 NO_2 + O_2} can be written in three ways: −d[N2O5]dt=k[N2O5]\frac{-d [N_2 O_5]}{dt} = k [N_2 O_5] d[NO2]dt=k′[N2O5]\frac{d[NO_2]}{dt} = k' [N_2 O_5] d[O2]dt=k"[N2O5]\frac{d[O_2]}{dt} = k" [N_2 O_5] The relationship between k and k' and between k and k'' are-

A

k' = k, k''= k

B

k'= 2k; k''= k

C

k'= 2k, k''= k/2

D

k' = 2k; k''= 2k

Answer

k'= 2k, k''= k/2

Explanation

Solution

Rate =−12d[N2O5]dt=+14d[NO2]dt=d[O2]dt= - \frac{1}{2} \frac{d\left[N_{2}O_{5}\right]}{dt} = + \frac{1}{4} \frac{d\left[NO_{2}\right]}{dt} = \frac{d\left[O_{2}\right]}{dt} 12K[N2O5]=14K′[N2O5]\frac{1}{2} K\left[N_{2} O_{5}\right] = \frac{1}{4} K' \left[N_{2}O_{5}\right] K′=2K K' = 2 K and K"=K2K" = \frac{K}{2}