The decomposition of `N_(2)O_(5) " in CCl"_(4)` at 318K has been studied by monitoring the concentration of `N_(2)O_(5)` in the solution . Initially , concentration of `N_(2)O_(5) " is "2.33"mol.L"^(-1)` & after 184 minutes , it is reduced to `2.08"mol.L"^(-1)` . The reaction takes place according to the equation `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` . Calculate the average rate of this reaction in terms of hours, minutes & seconds . What is the rate of production of `NO_(2)` during this period ?

Explanation

Average rate `=(1)/(2)(-(Delta[N_(2)O_(5)])/(Deltat))=-(1)/(2)((2.08-2.33)/(184))`
`=6.79xx10^(-4)"mol.L"^(-1)."min"^(-1)=4.07xx10^(-2)"mol.L"^(-1).h^(-1)`
`=1.13xx10^(-5)"mol.L"^(-1).s^(-1)`
Now, average rate `(1)/(4)=[(Delta[NO_(2)]]/(Deltat)]`
`therefore(Delta[NO_(2)])/(Deltat)=6.79xx10^(-4)xx4=2.72xx10^(-3)"mol.L"^(-1)"min"^(-1)`