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Chemistry Question on Thermodynamics

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

A

-676.9 kJ

B

676.5 kJ

C

110.5 kJ

D

-110.5 kJ

Answer

-110.5 kJ

Explanation

Solution

C(s)+O2(g)⟶CO2(g)C (s)+ O _{2}(g) \longrightarrow CO _{2}(g) ΔH=−393.5 kJmol−1\Delta H=-393.5\, kJmol ^{-1} CO(g)+12O2(g)→CO2(g)CO (g)+\frac{1}{2} O _{2}(g) \rightarrow CO _{2}(g) ΔH=−283 kJmol−1\Delta H=-283\, kJmol ^{-1} On subtracting equation (ii) from equation (i), we get C(s)+O2(g)→CO(g);ΔH=−110.5 kJmol−1C (s)+ O _{2}(g) \rightarrow CO (g) ; \Delta H=-110.5\, kJmol ^{-1} The enthalpy of formation of carbon monoxide per mole =−110.5 kJmol−1=-110.5\, kJmol ^{-1}