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)

Given (I) C(diamond) +O_(2)(g)rarrCO_(2)(g),DeltaH^(@)=-92.0Kcal mol^

1.	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)
Answer» `2.907KcalK^(-1)` `2.013KcalK^(-1)` `305.4calK^(-1)` `-2.013KcalK^(-1)` SOLUTION :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)`