For the reaction `N_2O_(5(g))rarr2NO_(2_(g))=1//2O_(2(g))` the value of rate of disappearance of `N_2O_5` is given as `6.25xx10^(-3)"mol "L^(-1)s^(-1)`. The rate of formation of `NO_2` and `O_2` is given respectively as

Options

A. `6.25 xx 10^(-3)" mol "L^(-1)s^(-1)` and `6.25 xx 10^(-3)" mol "L^(-1)s^(-1)`
B. `1.25 xx 10^(-2)" mol "L^(-1)s^(-1)` and `3.125 xx 10^(-3)" mol "L^(-1)s^(-1)`
C. `6.25 xx 10^(-3)" mol "L^(-1)s^(-1)` and `3.125 xx 10^(-3)" mol "L^(-1)s^(-1)`
D. `1.25 xx 10^(-2)" mol "L^(-1)s^(-1)` and `6.25 xx 10^(-3)" mol "L^(-1)s^(-1)`

Answer

B

Explanation

`N_2O_(5(g))rarr2NO_(2_(g))=1//2O_(2(g))`
For the given reaction the rate law may be written as:
`(-d[N_2O_5])/(dt)=1/2(d[NO_2])/(dt)=(2d[O_2])/(dt)`
given that `(-d[N_2O_5])/(dt)=6.25xx10^(-3)" mol "L^(-1)s^(-1)`
`therefore (d[NO_2])/(dt)=2xx6.25xx10^(-3)=1.25xx10^(-2)" mol "L^(-1)s^(-1)`
and `(d[O_2])/(dt)=(6.25xx10^(-3))/(2)=3.125xx10^(-3)` mol `L^(-1)s^(-1)`