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Calculate the enthalpy change accompanying the conversion of 10 g of graphite into diamond if the heats of combustion of C (graphite) and C (diamond) are -94.05 and -94.50 kcal respectively. |
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Answer» Solution :Given that, (i) `C("GRAPHITE")+O_(2)(g) to CO_(2)(g),DeltaH=-94.05"kcal"` (ii) `C("diamond")+O_(2)(g) to CO_(2)(g),DeltaH=-94.05"kcal"` Thus, applying the inspection method, [Eqn. (i)- Eqn. (ii)], we get, `C("graphite")+O_(2)(g)-C("dimond")-O_(2)(g) to CO_(2)(g)-CO_(2)(g),DeltaH=-94.05-(-94.50)` or `C("graphite") to C("dimond"),DeltaH=+0.45"kcal"` SINCE this enthalpy CHANGE is only for conversion of 1 "mole", i.e., 12 g of `C("graphite")`to `C("diamond")`, therefore, for the conversion of 10 g of C ("graphite") to C("diamond") `DeltaH=0.45xx(10)/(12)=0.375"kcal"` |
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