The hydrogenation of vegetable ghee at `25^(@)C` reduces the pressure of `H_(2)` form `2 atm` to `1.2 atm` in `50min`. Calculate the rate of reaction in terms of change of
(a) Pressure per minute
(b) Molarity per second

Explanation

(a) `"Rate of reaction" = ("Change in pressure")/("Time in min") = (2-1.2)/(50)`
`= (0.8)/(50) = 1.6 xx 10^(-2) atm min^(-1)`
(b) `because` Changein molartiy may be derived by `PV = nRT`
`:. (n)/(V) = (P)/(RT)`
`(n)/(V) = (0.8)/(0.0821 xx 298) = 0.0327`
`"Rate of reaction" = ("Change in molarity")/("Time in second")`
`= (0.0327)/(50 xx 60) = 1.09 xx 10^(-5) mol L^(-1) s^(-1)`