The decomposition of `N_(2) O_(5)` is represented by the equation
`2N_(2)O_(5(g)) to 4NO_(2(g)) + O_(2(g))`
(a ) How is the rate of formation of `NO_(2)` related to the rate of formation of `O_(2)`?
(b ) How is the rate of formation of `O_(2)` related to the rate of consumption of `N_(2)O_(s)?`

Explanation

Given : `2N_(2)O_(5(g)) to 4NO_(2(g)) + O_(2(g))`
(a) Rate of formation of `O_(2)` at time `t= (d[O_(2)])/(dt)`
They are related to each other through rate of reaction
`:. ` Rate of reaction `=(1)/(4) (d[NO_(2)])/(dt)= (d[O_(2)])/(dt)`
( b) Rate of consumption of `N_(2) O_(5)` at time `t= - (d[N_(2)O_(5)])/(dt)`
Rate of reaction `=- (1)/(2) (d[N_(2)O_(5)])/(dt) =(d[O_(2)])/(dt)`
In general
Rate of reaction `= -(1)/(2) (d[N_(2)O_(5)])/(dt)= (1)/(4) (d[NO_(2)])/(dt)= (d[O_(2)])/(dt)`