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 secondA. `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)`

The hydrogenation of vegetable ghee at `25^(@)C` reduces the pressure

1.	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 secondA. `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» Correct Answer - A `-(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)`

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