Question

For the reaction, $N _{2} O _{5} \longrightarrow 2 NO _{2}+ O _{2}$;
Given $-\frac{d\left[ N _{2} O _{5}\right]}{d t}=k_{1}\left[ N _{2} O _{5}\right], \frac{d\left[ NO _{2}\right]}{d t}=k_{2}\left[ N _{2} O _{5}\right]$ and $\frac{d\left[ O _{2}\right]}{d t}=k_{3}\left[ N _{2} O _{5}\right]$
The relation in between $k_{1}, k_{2}$ and $k_{3}$ is:

• A. $2 k_{1}=k_{2}=4 k_{3}$
• B. $k_{1}=k_{2}=k_{3}$
• C. $2 k_{1}=4 k_{2}=k_{3}$
• D. none of these.

Answer 1

Answer: (A)

$\frac{-d\left[ N _{2} O _{5}\right]}{d t}=\frac{1}{2} \frac{d\left[ NO _{2}\right]}{d t}=\frac{2 d\left[ O _{2}\right]}{d t}$
$\therefore k_{1}\left[ N _{2} O _{5}\right]=\frac{k_{2}}{2}\left[ N _{2} O _{5}\right]$
$=2 k_{3}\left[ N _{2} O _{5}\right]$