# Is the standard enthalpy of formation always non-positive?

If the standard enthalpy of formation is defined as

the energy change when 1 mole of a substance is formed from its elements in their standard states

and forming chemical bonds is an exothermic process, then does this mean all standard enthalpies of formation are non-positive?

Why do some compounds such as $\ce{NO}$, $\ce{NO2}$ have positive standard enthalpies of formation? Is this because to form $\ce{NO}$, we must first break the bonds in $\ce{N2}$ and $\ce{O2}$ (endothermic) to form the bond in $\ce{NO}$ (exothermic, smaller energy produced than the energy required to break the bonds of original standard state elements)?

• Well, you've pretty much answered your own questions. Commented Apr 23, 2016 at 6:15
• Yes i agree right answer is no Commented May 20, 2021 at 2:36
• All combustion reaction are??????? Commented May 20, 2021 at 2:41
• Another good example is ozone (O3), whose $\Delta H^{\circ}_f = 142.7 \,\mathrm{ kJ/mol}$. It's also interesting to consider monatomic N(g), which has $\Delta H^{\circ}_f = 473 \,\mathrm{ kJ/mol}$. That's because, for every mole of N(g) you form, you have to break a half mole of N≡N triple bonds, which have a bond dissociation energy of 2 x 473 = 946 kJ/mol, and you don't form any bonds to compensate. Commented May 20, 2021 at 5:36
• @Fatima Please note your comments did not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented May 20, 2021 at 5:42

The net sum can be positive or negative, and it depends on what the reactants and products were. As you correctly said yourself, $\ce{NO}$ happens to have a positive standard enthalpy of formation because, as experimentally determined, more energy went into breaking the bonds of reactants than what we got back from forming the products.