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Given (I) `C`(diamond) `+O_(2)(g)rarrCO_(2)(g),DeltaH^(@)=-92.0Kcal mol^(-1)` (II) `C`(graphite) + `O_(2)(g)rarrCO_(2)(g),DeltaH^(@)=-96.0Kcal mol^(-1)`A. `2.907KcalK^(-1)`B. `2.013KcalK^(-1)`C. `305.4calK^(-1)`D. `-2.013KcalK^(-1)` |
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Answer» Correct Answer - A Eq . `(I)` and Eq . `(II)` give C(diamond)`rarr`(graphite) `DeltaH^(@)=-02-(-96)=+4.0Kcal mol^(-1)` Moles of diamond `=(2.6xx1000g)/(12gmol^(-1))=216.6mol` `DeltaH^(@)` (to convert `200mol` of diamond into graphite) `=216.6xx4.0Kcal=866.4Kcal` `:. DeltaS=(DeltaH^(@))/(T)=(866.4Kcal)/(298K)=2.907KcalK^(-1)` |
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