Question

Heat of combustion of carbon momoxide at constant volume and at 17 c is

Answer 1

The enthalpy of combustion of a solid carbon to form carbon dioxide is −393.7 kJ/mol−393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is −283.3 kJ/mol CO−283.3 kJ/mol CO. Use these data to calculate ΔHΔH for the reaction

2C(s)+O2(g)⟶2CO(g)2C(s)+OX2(g)⟶2CO(g)

Attempt at solution: From the description, I first wrote down my two equations:

1) C+O2⟶CO22) CO+O⟶CO21) C+OX2⟶COX22) CO+O⟶COX2

I want to have a 2CO2CO as a product. In the second equation I see COCO as reactant, so I multiply that equation by two and reverse it

2CO2⟶2O+2CO2COX2⟶2O+2CO

The enthalpy thereby changes by −2(−283.3 kJ/mol)=566.6 kJ/mol−2(−283.3 kJ/mol)=566.6 kJ/mol. Now I somehow need to get rid of that 2O2O on the reactant side, but I don't see how to do it. I tried adding the first equation but that doesn't work.

Any help please?

Edit: The two correct equations are

1) C+O2⟶CO22) 2CO+O2⟶2CO21) C+OX2⟶COX22) 2CO+OX2⟶2COX2

We reverse the second equation to get

2CO2⟶2CO+O22COX2⟶2CO+OX2

The enthalpy thereby changes to 283.3 kJ/mol283.3 kJ/mol. Adding the first equation to the preceding one gives

C+O2+2CO2⟶CO2+2CO+O2C+OX2+2COX2⟶COX2+2CO+OX2

or after deleting common species

C+CO2⟶2COC+COX2⟶2CO

We still need an O2OX2 at the reactant side, so we add equation 1 again to obtain

2C+O2⟶2CO