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^(-3)mol L^(-1)S^(-1) and 3.125xx10^(-3)mol L^(-1)S^(-1)`and
C. `6.25xx10^(-3)mol L^(-1)S^(-1) and 3.125xx10^(-3)mol L^(-1)S^(-1)`and
D. `1.25xx10^(-3)mol L^(-1)S^(-1) and 6.25xx10^(-3)mol L^(-1)S^(-1)`and

Answer

B

Explanation

Given `-(-d[N_(2)O_(5)])/(dt)=6.25xx10^(-3)molL^(-1)S^(-1)`
For the reaction
`N_(2)O_(5) to 2NO_(2)+(1)/(2)O_(2)`
`(-d[N_(2)O_(5)])/(dt)=(1)/(2)(d[NO_(2)])/(dt)=(2d[O_(2)])/(dt)`
`therefore (d[NO_(2)])/(dt)=-(2d[N_(2)O_(5)])/(dt)=1.25xx10^(-2)molL^(-1)S^(-)`
`therefore (d[O_(2)])/(dt)=-(1)/(2)(d[N_(2)O_(5)])/(dt)`
`=3.12510^(-3)molL^(-1)S^(-1)`