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The hydrogenation of vegetable ghee at 2...

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

A

`1.09xx10^(-6)`

B

`1.09xx10^(-5)`

C

`1.09xx10^(-7)`

D

`1.09xx10^(-9)`

Text Solution

Verified by Experts

The correct Answer is:
b

The change in molarity `=n/V=(DeltaP)/(RT)`
`=0.8/(0.0821xx273)=0.0327`
`:.` Rate of reaction=`0.0327/(50xx60)`
`=1.09xx10^(-5) mol litre^(-1) sec^(-1)`
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