For the reaction, N 2 O 5 longrightarrow 2 NO 2+ O 2; Given -(d[ N 2 O 5]/d t)=k1[ N 2 O 5], (d[ NO 2]/d t)=k2[ N 2 O 5] and (d[ O 2]/d t)=k3[ N 2 O 5] The relation in between k1, k2 and k3 is:

Question

For the reaction, N 2 O 5 longrightarrow 2 NO 2+ O 2; Given -(d[ N 2 O 5]/d t)=k1[ N 2 O 5], (d[ NO 2]/d t)=k2[ N 2 O 5] and (d[ O 2]/d t)=k3[ N 2 O 5] The relation in between k1, k2 and k3 is: (A) 2 k1=k2=4 k3 (B) k1=k2=k3 (C) 2 k1=4 k2=k3 (D) none of these.

Tardigrade Admin · Accepted Answer

(-d[ N 2 O 5]/d t)=(1/2) (d[ NO 2]/d t)=(2 d[ O 2]/d t) ∴ k1[ N 2 O 5]=(k2/2)[ N 2 O 5] =2 k3[ N 2 O 5]

Q. For the reaction, $N_{2} O_{5} ⟶ 2 N O_{2} + O_{2}$ ;
Given $- \frac{d [ N _{2} O _{5} ]}{d t} = k_{1} [N_{2} O_{5}], \frac{d [ N O _{2} ]}{d t} = k_{2} [N_{2} O_{5}]$ and $\frac{d [ O _{2} ]}{d t} = k_{3} [N_{2} O_{5}]$
The relation in between $k_{1}, k_{2}$ and $k_{3}$ is:

$2 k_{1} = k_{2} = 4 k_{3}$

$k_{1} = k_{2} = k_{3}$

$2 k_{1} = 4 k_{2} = k_{3}$

none of these.

$\frac{- d [ N _{2} O _{5} ]}{d t} = \frac{1}{2} \frac{d [ N O _{2} ]}{d t} = \frac{2 d [ O _{2} ]}{d t}$
$∴ k_{1} [N_{2} O_{5}] = \frac{k _{2}}{2} [N_{2} O_{5}]$
$= 2 k_{3} [N_{2} O_{5}]$

Q. For the reaction, N2​O5​⟶2NO2​+O2​; Given −dtd[N2​O5​]​=k1​[N2​O5​],dtd[NO2​]​=k2​[N2​O5​] and dtd[O2​]​=k3​[N2​O5​] The relation in between k1​,k2​ and k3​ is:

2k1​=k2​=4k3​

k1​=k2​=k3​

2k1​=4k2​=k3​