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. `4k_(1)=k_(1)=2k_(1)`
D. None of these

Answer

D

Explanation

None of the given options is correct
`k_(1)[N_(2)O_(5)]=(k_(1))/(2)[N_(2)O_(5)]"and"k_(1)[N_(2)O_(5)]=2k_(1)[N_(2)O_(5)]`
Equating these rate constants
`k_(1)[N_(2)O_(5)]=(k_(1))/(2)[N_(2)O_(5)]=2k_(1)[N_(2)O_(5)]`
`2k_(1)=k_(1)=4k_(1)`