For the reaction `2N_(2)O_(5)rarr4NO_(2)+O_(2)` the rate equation can be express as
`(d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5)` and `(d[NO_(2)])/(dt)=k'[N_(2)O_(5)]` k and k' are related as

Options

A. `k=k'`
B. `2k=k'`
C. `k=2k'`
D. `k=4k'`

Answer

B

Explanation

`2N_(2)O+_(5)rarr4NO_(2)+O_(2)`
from reaction stoichiometry 2 mol of `n_(2)O_(5)` prodluce 4 mol of `NO_(2)` that is just twice the number
But in the question rate is defined in mass per unit volume per unit time so converting this rate into moles per unit volume per unit time
Rate of production of `NO_(2)`=2m mol `L^(-1) s^(-1)` x Molar mass of `NO_(2)=92 mg L^(-1) s^(-1)`