The correct option is C 0.375 kcal
Given that:
(i) C(graphite)+ O2(g) →CO2(g) ΔH1=−94.05 kJ mol−1
(ii) C(diamond)+ O2(g) →CO2(g) ΔH2=−94.50 kJ mol−1
now, [eqn. (i) - eqn. (ii)] we get :
C(graphite)+ O2(g)−C(diamond) −O2(g) →CO2(g) −CO2(g)
For this reaction ΔH=ΔH1−ΔH2
Simplifying the above equation we get,
C(graphite) → C(diamond)
Also putting values of ΔH1and ΔH2
ΔH= −94.05−(−94.50)=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)
ΔH=0.45×1012=0.375 kcal