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^(-6)`B. `1.09xx10^(-5)`C. `1.09xx10^(-7)`D. `1.09xx10^(-9)`

Answer» Correct Answer - 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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