The burning enthalpy (exothermic) of $\ce{NO +O2 ->NO2}$ is $\pu{57 kJ mol^-1}$ and $\ce{O=O}$ bond energy is $\ce{494 kJ mol^{-1}}$, then what is the bond energy of $\ce{N-O}$ ?

So I balanced the equation

$$\ce{2NO + O2 ->2NO2}$$

wrote down Lewis structures and got

$$(2\Delta H(\ce{N=O}) + \Delta H(\ce{O=O}) - 2(\Delta H(\ce{N-O}) + \Delta H(\ce{N=O})) = \pu{-57 kJ mol^-1}$$

($\pu{-57 kJ mol^-1}$ since it is an exothermic reaction), so we are left only with

$$\Delta H(\ce{O=O}) - 2\Delta H(\ce{N-O}) = \pu{-57 kJ mol^-1}$$


$$\Delta H(\ce{N-O}) = \frac{494+57}{2}\pu{kJ mol^-1} = \pu{275.5 kJ mol^-1}$$

However the right answer is

$$\frac{494-2\cdot 57}{2} \pu{kJ mol^-1}= \pu{190 kJ mol^-1}$$

and I really don't get what am I doing wrong?

  • 1
    $\begingroup$ Have a look at a very similar question asked recently: Calculating bond dissociation enthalpy of F-F bond. Also, I've never seen square brackets used to denote bond energies and I'm wondering what textbook does that (in chemistry "[" and "]" are commonly used to denote either a coordination complex, or equilibrium concentration). It also appears that there is something off with the unit symbols or lack thereof. I already corrected Kj to kJ, but it looks like those should be per mole (?). $\endgroup$
    – andselisk
    Commented Jul 22, 2019 at 4:46
  • 1
    $\begingroup$ @andselisk Thanks, I edited the units and the notation. For the given link - I used the same method and got the correct answer, which make me think there is a catch in this exercise which I don't get. $\endgroup$ Commented Jul 22, 2019 at 5:04
  • $\begingroup$ Consider the reaction enthalpy of $$\ce{2NO + O2 ->2NO2}$$ and $$\ce{20 NO + 10 O2 ->20 NO2}.$$ They are different, and also different from the reaction enthalpy of burning one mole of NO. $\endgroup$
    – Karsten
    Commented Jul 22, 2019 at 13:39
  • $\begingroup$ @KarstenTheis How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO + O2 -> NO2}$ enthalpy? $\endgroup$ Commented Jul 23, 2019 at 10:35

1 Answer 1


How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -> NO2}$ enthalpy?

The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of formation, it is understood that the substance in question is the only product, the stoichiometry coefficient is one, and that the reactants are elements with coefficients so that the equation is balanced. I am not aware of any such convention for the enthalpy of combustion.

What is the enthalpy of $\ce{2NO(g) + O2(g) -> 2NO2(g)}$?

We can look up the enthalpies of formation for the nitrogen-containining compounds:

$$\Delta H_f(\ce{NO}) = \pu {90.25 kJ/mol}$$ $$\Delta H_f(\ce{NO2}) = \pu {31.18 kJ/mol}$$

From this, we can calculate the reaction enthalpy:

$$\Delta H_r = 2 \cdot \pu {31.18 kJ/mol} - 2 \cdot \pu {90.25 kJ/mol} = \pu {-118.14 kJ/mol}$$

Which combustion reaction has an enthalpy of -57 kJ/mol?

If we divide the balanced chemical equation by 2, we get:

$$\ce{NO + 1/2 O2 -> NO2}$$

The reaction enthalpy is one half of that of the original reaction, -59.07 kJ/mol. So for the reaction describing the combustion of one mole of NO, the reaction enthalpy is about -57 kJ/mol.

What about the bond dissociation energy?

Everything else in the OP's calculation is correct. If you plug in the correct combustion enthalpy, the calculated bond energy will be correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.