For the reaction, 2N2O5 → 4NO2 + O2, the rate equation can be expressed in two ways - (d[N2O5]/dt) = k [N2O5] and +(d[NO2]/dt) = k' [N2O5] k and k' are related as :

Question

For the reaction, 2N2O5 → 4NO2 + O2, the rate equation can be expressed in two ways - (d[N2O5]/dt) = k [N2O5] and +(d[NO2]/dt) = k' [N2O5] k and k' are related as : (A) k = k' (B) 2k = k' (C) k = 2k' (D) k = 4k'

Tardigrade Admin · Accepted Answer

2N2O5 →4NO2+O2 -(d/d t) [N2O5]=k[N2O5] Now ⇒ -(1/2) (d/d t) [N2O5]=(1/4) × K1 [N2O5] ⇒ 2k=k'

Q. For the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , the rate equation can be expressed in two ways $- \frac{d [ N _{2} O _{5} ]}{d t} = k [N_{2} O_{5}]$ and $+ \frac{d [ N O _{2} ]}{d t} = k^{'} [N_{2} O_{5}]$
k and k' are related as :

k = k'

2k = k'

k = 2k'

k = 4k'

$2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$

$- \frac{d}{d t} [N_{2} O_{5}] = k [N_{2} O_{5}]$

Now

$\Rightarrow - \frac{1}{2} \frac{d}{d t} [N_{2} O_{5}] = \frac{1}{4} \times K^{1} [N_{2} O_{5}]$

$\Rightarrow 2 k = k^{'}$

Q. For the reaction, 2N2​O5​→4NO2​+O2​, the rate equation can be expressed in two ways −dtd[N2​O5​]​=k[N2​O5​] and +dtd[NO2​]​=k′[N2​O5​] k and k' are related as :