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

Options

A. `1.09xx10^(-5)mol L^(-1)s^(-1)`
B. `2.67xx10^(-4)mol L^(-1)s^(-1)`
C. `8.94xx10^(-7)mol L^(-1)s^(-1)`
D. `3.25xx10^(-3)mol L^(-1)s^(-1)`

Answer

A

Explanation

`-(Deltap_(H_(2))/(Deltat)=-((1.2-2.0)atm)/(50min)xx(min)/(60s)`
`=2.67xx10^(-4)s^(-1)`
Assuming ideal gas behavior, we have
`PV=nRT`
`P=(n)/(V)RT=CRT`
Thus, `DeltaP=DeltaCRT`
Diving both sides by time interval, `Deltat` , we get
`(Deltap)/(DeltaT)=(Deltac)/(Ddeltat)RT`
or `(Deltac)/(Deltat)=(Deltap//Deltat)/(RT)`
`=(2.67xx10^(-4)atms^(-1))/((0.0821(Latm)/(mol.K))(298K))`
`=1.09xx10^(-5)molL^(-1)s^(-1)`