`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
If `(-d[N_(2)O_(5)])/(dt) = k_(1)[N_(2)O_(5)]`
`(d[NO_(2)])/(dt) = k_(2)[N_(2)O_(5)]`
`(d[O_(2)])/(dt) = k_(3)[N_(2)O_(5)]`
What is the relation between `k_(1), k_(2)`, and `k_(3)`?

Options

A. `k'=k, k"=k`
B. `k'=2k, k"=k`
C. `k'=2k, k"= k//2`
D. `k'=2k, k"=2k`

Answer

C

Explanation

We can express the rate of the given reaction in terms of any of the species (reactant or product):
`Rate=(-d[N_(2)O_(5)])/(2dt)=(d[NO_(2)])/(4dt)=(d[O_(2)])/(dt)`
Relationship between `k` and `k'` :
`(-d[N_(2)O_(5)])/(2dt)=(d[NO_(2)])/(4dt)`
`(k)/(2)[N_(2)O_(5)]=(k')/(4)[N_(2)O_(5)]`
or `(k)/(2)=(k')/(4)`
Hence, `k'=(4)/(2)k=2k`
Relationship between `k` and `k"` :
`(-d[N_(2)O_(5)])/(2dt)=(d[O_(2)])/(dt)`
`(k)/(2)[N_(2)O_(5)]=k"[N_(2)O_(5)]`
`(k)/(2)=k"`