Why is the heat of formation for NaOH negative?

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SUMMARY

The heat of formation for sodium hydroxide (NaOH) is -425.6 kJ/mol, indicating that the formation process is exothermic. When NaOH dissolves in water, it undergoes dissociation, which is distinct from formation, leading to confusion regarding enthalpy changes. While the enthalpy change for dissociation appears to be +425.6 kJ/mol, the dissolution process is exothermic, releasing heat to the environment. Understanding the difference between enthalpy of formation and enthalpy of dissolution is crucial for accurate thermodynamic analysis.

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Nanomaly
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Note: This isn't an actual homework question, and I tried asking others (including teachers), but no one could answer it.

Now, the heat of formation for sodium hydroxide is -425.6 kJ/mol (so it's exothermic).

When you dissolve NaOH in water, the reaction is exothermic. But this is dissociation, the opposite of formation. So logically, you would flip the sign around for the heat of formation to get the enthalpy change of the dissociation, right? But if you do, the enthalpy change becomes +425.6 kJ/mol, which means that it is endothermic. However, the dissolving of NaOH in water releases heat to the environment, so it's exothermic!

What's going on? This has been baffling me for a long time now.
 
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Nanomaly said:
(snip) But this is dissociation, the opposite of formation. (snip)

Add NaOH to water and get elemental sodium, elemental oxygen, and elemental hydrogen? Guess again --- the words dissolution and dissociation both begin with "dis," but that does not make them equivalent. Formation is from the elements, ALWAYS. Things dissolve without dissociating. Things dissociate without being dissolved. Go back and learn the definitions.
 
yeah, they are different cases. enthalpy of formation is different from enthalpy of dissolution. The former is associated with the overall process of formation from its most stable elements (which have the enthalpy of formation of zero), the latter is regards the process where water solvates the compound to its corresponding cation and anion.
 

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