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I'm doing a write-up for this for my students:

enter image description here
Source: Amirav Research Group (Facebook)

The energy of formation of isopropyl alcohol is $-318.2~\mathrm{kJ/mol}$ (kilojoules per mole), that of acetone is $-249.4~\mathrm{kJ/mol}$, so $68.8~\mathrm{kJ/mol}$ are gained on this end.

But I'm getting confused about the hydrogen end. I'm seeing that The energy of formation of $\ce{H+}$ is $0~\mathrm{J/mol}$ (it can be treated as a free proton), and that of $\ce{H2}$ is also $0~\mathrm{J/mol}$. Of course you would say that oxygen has $0~\mathrm{J/mol}$ as well, so since water has enthalpy of formation of $241.83~\mathrm{kJ/mol}$, I'd expect it to yield that much energy in the combustion.

But hydrogen gas has an “energy density” as a fuel of $71~\mathrm{kJ/mol}$. So the treatment that $\ce{H2}$ and $\ce{O2}$ have $0~\mathrm{J/mol}$ enthalpy of formation and water has enthalpy of formation of $242~\mathrm{kJ/mol}$ seems to not square with that. Where's the rest of the energy?

Particularly, in the diagram above you see the charge separation as the electron pair moves to the hydronium, producing hydrogen. But a charge separation (J = coulomb volt) is energy, so moving the electrons to the hydrogen has to be adding energy to them, which seems to indicate that the $\ce{H+} = 0~\mathrm{J}$ and $\ce{H2}$ also = $0~\mathrm{J}$ things is wrong. But I don't know where to go from here, does anyone feel confident helping with this?

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    $\begingroup$ @Nilay Please do not wrap every number you see in \ce{...}. Only use this environment for chemical formulae. I don't want to check every edit you make. $\endgroup$ – Martin - マーチン Mar 11 '16 at 5:29
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I'm seeing that The energy of formation of H+H+ is 0 J/mol0 J/mol (it can be treated as a free proton)

No. Nope. Noway. Formation of proton from hydrogen molecule is extremely endotermic process. It requires to 1) dissociate H-H bond, which is, btw, 432 kJ/mol, and then ionization of hydrogen atom, which is around 1300 kJ/mol.

However. In solution hydrogen ion is bound to water molecules (or other solvent), which results in serious energy discount, around 1100 kJ/mol for the case of water.

However again :) this is of no significance here, since the dive into small details is not required here.

The gross process is $\ce{ Me2CHOH \space =\space Me2CO + H2 }$, and since entalphy of formation of hydrogen is roughly zero, only change from isopropanol to acetone is relevant. Hence, the difference is 68.8 kJ/mol.

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  • $\begingroup$ if the 2H+ + 2e- -> H2 transition is worth the association of a 432kJ bond, isn't the [H+] in solution 1. a limiting factor of this reaction and also 2. actually the energy source? So this reaction works only in a maintained acidic environment is what I'm saying. $\endgroup$ – JCMontalbano Aug 24 '18 at 16:11
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I guess you switched a minus sign there; it's endothermic as the product acetone is less negative in energy than the reactant i-propanol. So the energy is going into the acetone rather than the proton-hydrogen conversion.

I'm quite surprised that the energy density is not equal to the combustion energy, but these values are not really important here since there is no combustion in the first place.

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  • $\begingroup$ yes but the combustion, and the [H+] consumption, is relevant because this is intended as the photocatalyst which supplies acetone for a fuel cell So combustion is the other half of this apparatus $\endgroup$ – JCMontalbano Aug 24 '18 at 16:14
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2H+ cannot be converted to $\ce{H2}$, because H+ has no electrons.

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It depend on how deeply you want to divide the chemical reaction cycle.

The overall reaction is:

Isopropyl alcohol -> acetone + hydrogen gas

Typically the internal energy of hydrocarbon compounds is measured by burning them to carbon dioxide and water in an excess of oxygen to find the heat of combustion. Since this reaction produces hydrogen too, the energy of the reaction for one mole would be the difference between burning isopropyl alcohol minus that of burning acetone and hydrogen.

User permeakra gave the chemical reaction cycle broken a different way in terms of the heat of formation, but the net result is the same. That is the beauty of chemistry.

Now you can divide the chemical reaction cycle into all sort of intermediate species, but the net reaction has to be what was outlined above. In fact the problem is usually worked backwards. By knowing what the energy of intermediate species like a solvated $\ce{H^+}$ cation is and knowing what the energy of a solvated $\ce{OH^-}$ anion is we can work out what the energy of some unstable intermediate organic "molecule" must be.


DavePhD is absolutely right in his answer. You can't just convert two $\ce{H^+}$ cations into $\ce{H2}$, because a $\ce{H^+}$ cation has no electrons. You have to get two electrons from some other chemical species and that will cost some chemical energy to free the electrons, not to mention the energy contained in the electron itself (which isn't chemistry but nuclear physics).

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There is a change in molar entropy as well as expansion work being done by the system in the transition from $\ce{H+}$(aq) to $\ce{H2}$(g). Assuming constant temperature and pressure, there will be a change in the internal energy of the system as well.

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