Consider the decomposition of `N_(2)O_(5)` as
`N_(2)O_(5)rarr2NO_(2)+1//2O_(2)`
The rate of reaction is given by
`-(d[N_(2)O_(5)])/(dt)=(1)/(2)(d[NO_(2)])/(dt)=2(d[O_(2)])/(dt)`
`=k_(1)[N_(2)O_(5)]`
Therefore `(-d[N_(2)O_(5)])/(dt)=k_(1)[N_(2)O_(5)]`
`(+d[NO_(2)])/(dt)=2k_(1)[N_(2)O_(5)]=k_(1)[N_(2)O_(5)]`
`(+d[O_(2)])/(dt)=(1)/(2)k_(1)[N_(2)O_(5)]=k_(1)[N_(2)O_(5)]`
Choose the correct option

Options

A. `k_(1)=k_(1)=k_(1)`
B. `k_(1)=2k_(1)=k_(1)`
C. `2k_(1)=k_(1)=4k_(1)`
D. `4k_(1)=2k_(1)=k_(1)`

Answer

C

Explanation

`-(d[N_(2)O_(5)])/(dt)=(1)/(2)(d[NO_(2)])/(dt)=2(d[O_(2)])/(dt)`
`k_(1)[N_(2)O_(5)]=(1)/(2)k'_(1)[N_(2)O_(5)[=2k''_(1)[N_(2)O_(5)]`
or `k_(1)=(1)/(2)k'_(1)=2k''_(1)`
or `2k_(1)=k_(1)=4k_(1)`.