For the reaction,
`2N_(2)O_(5) (g) rarr 4NO_(2)(g) + O_(2)(g)`, the rate of formation of `O_(2)` is `0.032 g h^(-1)`.
(a) Calculate the rate of converison of `N_(2)O_(5)` in `g h^(-1)`.
(b) Calculate the rate of formation of `NO_(2)` in `g h^(-1)`.

Explanation

Refer Section "Relation Between Rate of Formation/Decompoistion".
`(1)/(2)(-d[N_(2)O_(5)])/(dt) = (1)/(4)(d[NO_(2)])/(dt) = (d[O_(2)])/(dt)`
(a) `:. (-d[N_(2)O_(5)])/(dt) = 2xx (d[O_(2)])/(dt) xx ("Mw of" N_(2)O_(5) "in g")/("Mw of" O_(2) "in g")`
`= 2 xx 0.032 xx (98)/(32) = 0.196 g h^(-1)`
(b) `:. (d[NO_(2)])/(dt) = 4 xx (d[O_(2)])/(dt) xx("Mw of" NO_(2) "in g")/("Mw of" O_(2) "in g")`
`= 4 xx 0.032 xx (46)/(32) = 0.184 g h^(-1)`