For the reaction at `273 K`
`NO(g) + O_(3)(g) rarr NO_(2)(g) + O_(2)(g)`
It is observed that the pressure of `NO(g)` falls form `700 mm Hg` to `500 mm Hg` in `250 s`. Calculate the average rate of reaction in (a) `atm s^(-1)` , (b) `mol L^(-1) s^(-1)`.

Explanation

`NO(g) + O_(3)(g) rarr NO_(2) (g) + O_(2) (g)`
(a) `"Average rate" = ("Change of pressure")/("Change of time") = (Delta P)/(Delta t)`
`= (700 - 500)/(250) = (200)/(250)`
`= (200)/(760 xx 250) atm s^(-1)`
`= 1.053 xx 10^(-3) atm s^(-1)`
(b) `(Delta(n//V))/(Delta t) mol L^(-1) t^(-1) = (Delta P)/(Delta t) xx (1)/(RT)`
`= (1.053 xx 10^(-3) atm s^(-1))/((0.082 L atm K^(-1) mol^(-1))(273 K))`
`= 4.7 xx 10^(-5) mol L^(-1) s^(-1)`