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C("graphite")+O(2)(g)rarrCO(2)(g) delt...

`C_("graphite")+O_(2)(g)rarrCO_(2)(g)`
`deltaH=-94.05Kcalmol^(-1)`
`C_("diamond")+O_(2)(g),DeltaH=-94.50Kcalmol^(-1)
Therefore,

A

`C_(("graphite")) rarr C_(("diamond")) , Delta H_(298K)^(@)= - 450cal mol^(-1)`

B

`C_(("diamond")) rarr C_(("graphite")), Delta H_(298K)^(@) = +450 cal mol^(-1)`

C

Graphite is the stabler allotrope

D

Diamond is harder than graphite

Text Solution

Verified by Experts

The correct Answer is:
C
C(graphite) `+ O_(2)(g) rarr CO_(2)(g), Delta H = -94.05` kcal/mole
C(diamond) `+O_(2)(g) rarr CO_(2)(g), Delta H= -94.50` kcal/mole
C(graphite) `rarr C` (diamond), `Delta H= -94.05+ 94.50 = +450` cal/mol
So graphite is the stabler allotrope. Hardness of two compounds cannot be predicted from the given equations
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