Consider the reaction `2N_(2)O_(5)rarr 4NO_(2)+O_(2)` If the conc. Of `N_(2)O_(5)` is reduced from `2.33 M to 2.08M` after 184 minutes, the rate of production of `NO_(2)` during this period will be ------- mol `L^(-1) min^(-1)`.

Options

A. `2.72xx10^(-3)`
B. `1.36xx10^(-3)`
C. `0.68xx10^(-3)`
D. `8.16xx10^(-3)`

Answer

A

Explanation

A verage rate of concentration of `N_(2)O_(3)` :
`-(Delta[N_(2)O_(5)])/(Detat)=-((2.08-2.33)molL^(-1))/(184min) =1.36xx10^(-3)molL^(-1)min^(-1)`
The stoichiometry of the reation show that the rate of appearance of `NO_(2)` is rwist the rate of appearnce of `NO_(2)` is twice the rate of disappearnce of `N_(2)O_(3)` .
Thus
`(Delta[NO_(2)])/(Deltat)=2(-(Delta[N_(2O_(5)])/(Deltat))`
`2(1.36xx10^(-3)mole^(-1)mim^(-1))`
`=2.73xx10^(-3)molL^(-1)mim^(-1)`