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Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-A. `+43.76`B. `+40.66`C. `+37.56`D. None of these |
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Answer» Correct Answer - D `H_(2)O(l)toH_(2)O(g)` `DeltaH^(@)=DeltaE^(@)+Deltan_(g)RT` `40.66=66DeltaE^(@)+(1xx8.314xx373)/(1000)` `DeltaE^(@)=+37.56KJ//mol` for 2 mole `DeltaE^(@)=2xx37.26=75.12KJ` |
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