1
$\begingroup$

I realised the other day that I'm not sure why the reactions of strong acids with metals are exothermic. Any reaction is exothermic if there's more energy released from bonds being made than is used in breaking bonds, of course (using the word 'bonds' in a sufficiently loose sense).

In these reactions, metallic bonds are broken; covalent bonds are formed in hydrogen, but the hydrogen is in a gaseous state, so its molecules must be less attracted to each other than the hydrogen ions were to each other in the liquid. I'd vaguely guessed that the latter fact would make this an endothermic reaction (it works for sodium carbonate, after all) but evidently not!

Is there a rigorous way of calculating the total enthalpy change, or an easy explanation that I'm missing?

I guess I can calculate the enthalpy change of every one of the state changes involved, but it's not obvious that that's always going to come out negative (which I gather it does) so I feel like there's probably a less mathematical way of explaining this observation.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Keep in mind, the fact that hydrogen is a gas at room temperature implies that its intermolecular forces are weak, but says nothing about its intramolecular forces. A H-H covalent bond is actually quite strong, with a bond enthalpy on par with C-H bonds. $\endgroup$
    – Sean Doris
    Feb 24, 2016 at 23:16
  • $\begingroup$ Also, you can't really overlook entropic driving forces in reactions that produce gasses. $\endgroup$
    – Lighthart
    Feb 25, 2016 at 1:21
  • $\begingroup$ Lighthart, yes, but this question seems to be about exothermic vs. endothermic, not exergonic vs. endergonic. $\endgroup$
    – Curt F.
    Feb 25, 2016 at 1:22

1 Answer 1

Reset to default
3
$\begingroup$

Yes there is a rigorous way of calculating the enthalpy change.

First write down an example reaction. It might not be as easy as you think. What species are really formed when zinc chloride (for example) dissolves in water?

The simple version below...

$$\ce{Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(g)}$$

...won't do, solvated zinc ions in hydrochloric acid are present in a variety of tetrahedral chloro-aqua complexes.

So maybe something like

$$\ce{Zn(s) + 2 H+(aq) + 2 Cl-(aq) + 2H2O(aq) -> ZnCl2(H2O)2(aq) + H2(g)}$$

This at least shows which bonds are being broken and which are being formed:

  • Zn-Zn bonds (metallic) are broken
  • H-H bonds are being formed
  • Zn-Cl and Zn-aqua (actually zinc-oxygen) bonds are being formed

To figure out the enthalpy change, you can figure out the approximate enthalpy of formation of each bond type. H-H bonds are pretty "strong", i.e. the bond enthalpy is relatively high at 436 kJ/mol. So when those bonds are formed, that enthalpy is released. Zinc-water (i.e. zinc-oxygen) and zinc-chlorine bonds are also pretty strong at about 284 and 229 kJ/mol, respectively. Zinc-zinc bonds enthalpies are only ~29 kJ/mol.

Add it all up, and you get a lot of enthalpy being released!

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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