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Chemistry Question on Thermodynamics

Given that C+O2→CO2;ΔH∘=−a kJC+O_{2} \rightarrow C O_{2} ; \Delta H^{\circ} = -a\,kJ. 2CO+O2→2CO2;ΔH∘=−b kJ2 CO + O _{2} \rightarrow 2 CO _{2} ; \Delta H ^{\circ}=- b \,kJ The heat of formation of COCO is

A

b - 2a

B

2a−b2\frac{2a-b}{2}

C

b−2a2\frac{b-2a}{2}

D

2a - b

Answer

b−2a2\frac{b-2a}{2}

Explanation

Solution

C+O2→CO2;ΔH∘=C + O_{2} \to CO_{2} ; \Delta H^{\circ} = -a ,kJ 2CO+O2→2CO2;ΔH∘=2CO + O_{2} \to 2CO_{2} ; \Delta H^{\circ} = -b, kJ Formation of COCO can be written as : C+12→CO;ΔH∘=?C+\frac{1}{2}\to CO ; \Delta H^{\circ} = ? Adding equation : C+O2→CO2;ΔH∘=−a kJC + O_{2} \to CO_{2} ; \Delta H^{\circ} = -a \,kJ CO2→CO+12O2;ΔH∘=b2CO_{2} \to CO + \frac{1}{2}O_{2} ; \Delta H^{\circ} = \frac{b}{2} kJ We get ΔHf∘(CO)=b2−a=b−2a2\Delta H^{\circ} _{f }\left(CO\right) = \frac{b}{2}-a = \frac{b-2a}{2}