The correct option is D 0.375 kcal
Given that
(i) Cgraphite+O2(g)→CO2(g);ΔH=−94.05kcal
(ii) Cdiamond+O2(g)→CO2(g);ΔH=−94.50kcal
Thus applying the inspection method [Eq. (i) - Eq. (ii)], we get
⇒Cgraphite→Cdiamond;ΔH=0.45 kcal
Since this enthalpy change is only for conversion of 1 mol i.e. 12 g of Cgraphite to Cdiamond, therefore, for the converion of 10 g of Cgraphite to Cdiamond
ΔH=0.45×1012
=0.375 kcal