The rate of the reaction 2 N2 O5 arrow 4 NO2 + O2 can be written in three ways: (-d [N2 O5]/dt) = k [N2 O5] (d[NO2]/dt) = k' [N2 O5] (d[O2]/dt) = k" [N2 O5] The relationship between k and kâ² and between k and kâ²â² are-

Question

The rate of the reaction 2 N2 O5 arrow 4 NO2 + O2 can be written in three ways: (-d [N2 O5]/dt) = k [N2 O5] (d[NO2]/dt) = k' [N2 O5] (d[O2]/dt) = k" [N2 O5] The relationship between k and kâ² and between k and kâ²â² are- (A) k' = k, k′′= k (B) k′= 2k; k′′= k (C) k′= 2k, k′′= k/2 (D) k′ = 2k; k′′= 2k

Tardigrade Admin · Accepted Answer

Rate = - (1/2) (d[N2O5]/dt) = + (1/4) (d[NO2]/dt) = (d[O2]/dt) (1/2) K[N2 O5] = (1/4) K' [N2O5] K' = 2 K and K" = (K/2)

Q. The rate of the reaction 2N2​O5​→4NO2​+O2​ can be written in three ways: dt−d[N2​O5​]​=k[N2​O5​] dtd[NO2​]​=k′[N2​O5​] dtd[O2​]​=k"[N2​O5​] The relationship between k and k′ and between k and k′′ are-

k' = k, k′′= k

k′= 2k; k′′= k

k′= 2k, k′′= k/2

k′ = 2k; k′′= 2k

Rate =−21​dtd[N2​O5​]​=+41​dtd[NO2​]​=dtd[O2​]​ 21​K[N2​O5​]=41​K′[N2​O5​] K′=2K and K"=2K​

Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = + \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$
$\frac{1}{2} K [N_{2} O_{5}] = \frac{1}{4} K^{'} [N_{2} O_{5}]$
$K^{'} = 2 K$ and $K " = \frac{K}{2}$