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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)` |
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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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