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Chemistry Question on Enthalpy change

The enthalpies of combustion of carbon and carbon monoxide are −393.5-393.5 and −283 kJ mol−1-283\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}} respectively. The enthalpy of formation of carbon monoxide per mol is

A

110.5 kJ110.5\,kJ

B

676.5 kJ676.5\,kJ

C

−676.5 kJ-676.5\,kJ

D

−110.5 kJ-110.5\,kJ

Answer

−110.5 kJ-110.5\,kJ

Explanation

Solution

(i) C(s)+O2(g)→CO2(g);C(s)+{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);
ΔH=−393.5 kJ\Delta H =-393.5\,kJ (ii)
CO(g)+12O2(g)→CO2(g);CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);
ΔH=−283.0 kJ\Delta H=-283.0\,kJ (i) and (ii) gives (iii)
C(s)+12O2(g)→CO(g);C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g);
ΔH=−110.5 kJ\Delta H=-110.5\,kJ
Equation III also represents the enthalpy of formation of 1 mole of CO and thus, enthalpy change is the heat of formation of CO(g).CO(g).