For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.

Options

A. `6.25xx10^(-3)"mol"L^(-1)s^(-1)and 6.25xx10^(-3)"mol"L^(-1)s^(-1)`
B. `1.25xx10^(-2)"mol"L^(-1)s^(-1)and 3.125xx10^(-3)"mol"L^(-1)s^(-1)`
C. `6.25xx10^(-3)"mol"L^(-1)s^(-1)and 3.125xx10^(-3)"mol"L^(-1)s^(-1)`
D. `1.25xx10^(-2)"mol"L^(-1)s^(-1)and 6.25xx10^(-3)"mol"L^(-1)s^(-1)`

Answer

B

Explanation

`(-d[NO_(2)])/(dt)=+(1)/(2)(d[NO_(2)])/(dt)=+2(d[O_(2)])/(dt)`
`:. (-d[NO_(2)])/(dt)=2{(-f[N_(2)O_(5)])/(dt)}=2xx6.25xx10^(-3)`
`1.25xx10^(-2)"mol" L^(-1)d^(-1)`
` (-d[O_(2)])/(dt)=(1)/(2){(-f[N_(2)O_(5)])/(dt)}=(1)/(2)xx6.25xx10^(-3)`
`=3.125xx10^(-3)"mol" L^(-1)s^(-1)`