"Given that the bond enthalpy of the carbon-oxygen bonds in carbon monoxide and carbon dioxide are 1073 kJ/mol and 743 kJ/mol respectively, and that of the bond in the oxygen molecule is 496 kJ/mol, calculate the enthalpy change for the combustion of 1 mole of carbon monoxide."
So I have:
2CO2(g) + O2 (g) -> 2CO2(g)
Delta H = (2 mol)(1073 kJ/mol) + (1 mol)(496 kJ/mol)-[(4 mol)(743 kJ/mol)]
Delta H = -330 kJ/mol
But I know that that is not right. It's part of the correct process...what am I missing?
5 answers
It looks ok to me except (4 mol)(743 kJ/mol) should be (2 mol)(743 kJ/mol)
Really? I'm pretty sure part of the process lead to the value -330 kJ/mol...
We're looking for 1 mole of carbon monoxide...is the value we've calculated for 2 moles?
We're looking for 1 mole of carbon monoxide...is the value we've calculated for 2 moles?
Hm...if I use (2 mol)(743 kJ/mol), I end up with a positive delta H, which doesn't make sense (combustion -> exothermic).
yeah that's right now all you need to do is divide it by 2 because the question says enthalpy change for the combustion of 1 mole of carbon monoxide.
The balanced combustion process for carbon monoxide is as follows,
2CO + O2 --> 2CO2
delta H = delta H (products) - delta H (reactants)
= 2(743) - [2(1073) + 1(496)]
= -1156 kJ/mol of reaction of CO
for 1mol of CO = -1156/2 = -578 kJ/mol
2CO + O2 --> 2CO2
delta H = delta H (products) - delta H (reactants)
= 2(743) - [2(1073) + 1(496)]
= -1156 kJ/mol of reaction of CO
for 1mol of CO = -1156/2 = -578 kJ/mol
3 answers