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

Answer

C

Explanation

For the given reaction
`N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g)`
the rate can be expressed in terms of any of the three species (reactant or products):
Rate `=(-d[N_(2)O_(5)])/(dt)=(d[NO_(2)])/(2dt)=(d[O_(2)])/(1//2dt)`
Thus, rate of information of `NO_(2)` is given as:
`(d[NO_(2)])/(dt)=2((-d[N_(2))_(5)])/(dt))`
`=2.(6.25xx10^(-3)molL^(-1)s^(-1))`
`=12.5xx10^(-3)molL^(-1)s^(-1)`
`=1.25xx10^(-2)molL^(-1)s^(-1)` The rate of formation of `O_(2)` is given as
`(d[O_(2)])/(dt)=(1)/(2)(-(d[N_(2)O_(5)])/(dt))`
`=(1)/(2)(6.25xx10^(-3)molL^(-1)s^(-1))`
`=3.125xx10^(-3)molL^(-1)s^(-1)`