# I can't find reaction enthalpy ∆H for water formation reaction

$$\ce{1/2 O2(g) + H2(g) → H2O (l)}$$

The tabulated value is $$\Delta H = \pu{−285.8 kJ/mol}$$

I'm using the bond energies values for each molecule and for some reason, I am getting a different result, not even close to $$\pu{-285 kJ/mol}$$.

Bond dissociation energies:

$$\ce{H-H} = \pu{436 kJ/mol}$$

$$\ce{O=O} = \pu{499 kJ/mol}$$

$$\ce{O-H} = \pu{464 kJ/mol}$$

My attempt to calculate from bond dissociation energies:

(500.1/2) + 436 → 464.2

250 + 436 → 928

($$∆H$$ found is positive)

• Well, i wanted to see someone doing the correct way, so i could see my mistakes and fix them – Bruno Machado Jul 11 at 16:20
• Bond formation is exothermic and bond breaking is endothermic. – Zhe Jul 11 at 16:25
• i know that, i want to know how to get the result using the bond energies. – Bruno Machado Jul 11 at 16:38
• @BrunoMachado Thanks for showing your attempt. Zhe guessed correctly that you had trouble with plus and minus. I guessed incorrectly that you where concerned about the difference between -285 and -245 kJ/mol. Now we know. – Karsten Theis Jul 11 at 16:53

Here is the tally of bonds:

1. One $$\ce{H-H}$$ bond broken
2. One half $$\ce{O=O}$$ double bond broken
3. Two $$\ce{O-H}$$ bonds made

Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $$\ce{O-H}$$ bond depends on what else is attached to the oxygen. For H-H and O=O, this is no problem because there is only one molecule, respectively, with that kind of bond.

$$\Delta H = \pu{436 kJ/mol} + \pu{249.5 kJ/mol} - \pu{928 kJ/mol} = \pu{-242.5 kJ/mol}$$

We got the right sign and are not too far off in magnitude.

Why is the physical state important?

In solids and liquids, there are intermolecular interactions that contribute to the enthalpy. This is why the enthalpy of sublimation and boiling is positive - it takes energy to break the intermolecular forces. Here are the values for enthalpy of formation for water in the two states at room temperature:

$$\Delta H_\mathrm{f}(\ce{H2O(l)}) = \pu{-285.8 kJ/mol}$$

$$\Delta H_\mathrm{f}(\ce{H2O(g)}) = \pu{-241.8 kJ/mol}$$

And here is the enthalpy of vaporization

$$\Delta H_\mathrm{vap}(\ce{H2O(g)}) = \pu{43.99 kJ/mol}$$

• But ∆H is products - reagents, right? – Bruno Machado Jul 11 at 16:53
• why you did reagents - products? – Bruno Machado Jul 11 at 16:54
• Products - reactants is for calculating reaction enthalpies from $\Delta H_f$ values. Bonds broken - bonds made (i.e. bonds dissociation energies for the reactants - bond dissociation energies for the products) is what you need for this kind of exercise. – Karsten Theis Jul 11 at 16:56
• $\Delta H_f$ for H2 is zero by definition. $\Delta H_f$ for the H-atom would be positive, 218 kJ/mol. If you do products - reactants for the enthalpies of formation, you get a negative value - the expected exothermic reaction. Or you could simply say you are making an H-H bond and not breaking anything. With the bond dissociation energy of +436 kJ/mol, you would come to the same conclusion (bond formation: exothermic: negative change in enthalpy). – Karsten Theis Jul 11 at 17:20
• @KarstenTheis Sorry, I don't think I made it clear that that comment was to Bruno... :( Probably best for me just to delete it at this point... – Zhe Jul 12 at 0:00

Karsten Theis' answer is a perfect one for your question. He also did his best to explain why is he used the bond-making bond-breaking sign conversion but you are still in mind set of product-reactant explanation. So I decided to explain it in different direction.

Remember the law of conservation of energy, which states that the total energy of an isolated system remains constant? So it applies here as well. As Karsten Theis has explained, for the reaction, $$\ce{H2 (g) + 1/2 O2 (g) -> H2O (g)}$$:

$$\text{energies of \ce{H2} and \ce{1/2 O2} } = \text{energy of \ce{H2O} + energy absorbed or released}$$

Thus, $$\pu{436 kJ/mol} + \pu{\frac{499}{2} kJ/mol} = \pu{928 kJ/mol} + \Delta H$$ $$\therefore \; \Delta H = \pu{436 kJ/mol} + \pu{\frac{499}{2} kJ/mol} - \pu{928 kJ/mol} = \pu{−242.5 kJ/mol}$$

Rest of the explanation to get $$\Delta H_f (\ce{H2O(l)})$$ is exactly same as Karsten Theis'. I hope it is clear to you now.