Consider the reaction `N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` Which was carried out in liquid `"CCI"_(4) " at" 48^(@)C` . The concentration of `N_(2)O_(5) " in CCl" _(4)` solution at the start of the reaction was 2.05(M) and it reduced to 1.80 (M) after 170 min . Determine the average rate and the rate of formation of `NO_(2)` during the given interval.

Explanation

The average rate of the reaction,
`= -(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)=(1)/(4)(Delta[NO_(2)])/(Deltat)=(Delta[O_(2)])/(Deltat)`
The rate of consumption of `N_(2)O_(5)` in the interval of 170 min.
= `-(Delta[N_(2)O_(5)])/(Deltat)=([N_(2)O_(5)]_(2)-[N_(2)O_(5)]_(1))/(t_(2)-t_(1))`
`=-((1.80-2.05))/(170-0)=1.47xx10^(-3)"mol.L"^(-1)."min"^(-1)`
The average rate `=(1)/(2)xx` the rate of consumption of `N_(2)O_(5)`
= `(1)/(2)xx1.47xx10^(-3)`
`=7.35xx10^(-4)"mol.L"^(-1). "min"^(-1) `
and the rate of formation of `NO_(2)=4xx((1)/(2)xx` the rate of consumption of `N_(2)O_(5))=2.94xx10^(-3)"mol .L"^(-1)."min"^(-1)`