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Q1. How can we comment on the thermodynamic stability of the following benzene derivatives?

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Q2. What does thermodynamic stability mean, in general? What does it mean to compare the thermodynamic stability of more than compound?

Q3. How do we compare the thermodynamic stability of benzene derivatives? Is there a specific rule, or way to go about it? Would MOT (frost circles, to be specific) help?

I looked at the extent of resonance, and other such stabilizing phenomena/interactions. However, that only tells the extent of stabilisation of the compound due to those phenomena, not as a whole.

Edit: I wish to predict the order of stability using knowledge of chemistry, and not check enthalpy of formation data and then establish the order.

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Wikipedia defines thermal stability as .

Thermodynamic stability occurs when a system is in its lowest energy state, or chemical equilibrium with its environment. This may be a dynamic equilibrium, where individual atoms or molecules change form, but their overall number in a particular form is conserved.

For the particular problem in hand the lowest energy state definition is more relevant for us .

Enthalpy of formation can be used for judging the energy states.

A negative enthalpy change implies that the product has released heat thus lowering the amount of energy present in the product

This implies.

More negative the enthalpy of formation, more stable the compound.


Enthalpies of formation

  • Phenol
    $\ce{ -> -165.0 kJ/mol}$.

  • Hydroquinone
    $\ce{-> -371.1 ± 1.3 kJ/mol}$

  • P-Nitrophenol
    $\ce{-> -207.1 ± 1.1 kJ/mol}$

All data has been taken from http://webbook.nist.gov , for solid state.


The order can be calculated now as

Hydroquinone $>$ paranitrophenol $>$ phenol

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  • $\begingroup$ Great! How can I PREDICT the order of enthalpy of formation, though? $\endgroup$ – cogito_ai Feb 26 '18 at 10:38
  • $\begingroup$ @schrodinger_16 That's a question by itself, but I would go about seeing the number of bonds and how electronegative the constituent atoms are. $\endgroup$ – Avyansh Katiyar Feb 26 '18 at 10:40
  • $\begingroup$ Why, though? How does that help? Is it a general approach? $\endgroup$ – cogito_ai Feb 26 '18 at 10:42
  • $\begingroup$ Creating a bond releases energy , more electronegative the atoms greater the lattice enthalpy . It's a very general approach. Read chemical bonding again it's all there . $\endgroup$ – Avyansh Katiyar Feb 26 '18 at 10:45
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Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.

Among the above given compounds, considering any extra stability of the compounds, their thermodynamic stability can be guessed.

For phenol, when 1 mole of it is formed there can be slightly weak intermolecular Hydrogen bonding($\ce{C6H5-OH...OH-C6H5)}$. The hydrogen bonding is slightly weak because the lone pair of oxygen atom is delocalised with the benzene ring thus creating a $\delta$+ charge on the $\ce O$ atom, due to which the dipole dipole interaction between $\ce H$ of one $\ce{OH}$ and $\ce O$ of other phenol molecule is weakened. Thus, phenol enjoys only weak intermolecular hydrogen bonding.

For 4-nitrophenol, there is stronger intermolecular hydrogen bonding, which can exist between two phenol molecules as:

enter image description here

Due to enhanced no. of hydrogen bonds, the compound enjoys more stability than ordinary phenol and therefore its enthalpy of formation is more than phenol.

For hydroquinone, the number of hydrogen bonding is even more. One molecule can have a hydrogen bond with four other molecules through four hydrogen bonds as shown:

enter image description here

Thus the molecule of hydroquinone enjoys much more stability due to large number of hydrogen bonding, making its thermodynamic stability even more than p-nitro phenol.

Thus the stability order will be $$\text{hydroquinone > 4-nitrophenol > phenol}$$

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  • $\begingroup$ That is alright, but isn't that just the extent of hydrogen bonding? $\endgroup$ – cogito_ai Feb 26 '18 at 11:11
  • $\begingroup$ How are you using that to compare thermodynamic stability on an absolute scale? $\endgroup$ – cogito_ai Feb 26 '18 at 11:11
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    $\begingroup$ Extent of Hydrogen bonding can give us only the RELATIVE STABILITY between them. I am only saying that if you have such compounds you can JUST COMPARE their stability based on these observations. But It will not help to produce those numbers as said in another answer. The best way to logically deduce might be what I have written. But I myself dont really know what the values of$\ce { $\Delta $ H_f} $ are. I tried to logically figure out what might be their orders. $\endgroup$ – Soumik Das Feb 26 '18 at 11:23

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