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Below was a question in the 2012 Australian Chemistry Olympiads:

The enthalpy change of formation ($\Delta_\mathrm{f}H^\circ$) for a species at $\pu{298 K}$ is defined as the enthalpy change that accompanies the formation of one of the following of one mole of a substance from its constituent elements in their standard states. Which of the following species has $\Delta_\mathrm{f}H^\circ = \pu{0 kJ/mol}$?

a. $\ce{H2O(l)}$
b. $\ce{Na(g)}$
c. $\ce{Na(s)}$
d. $\ce{CO2(g)}$
e. $\ce{O3(g)}$

Original

Solutions stated the answer is b, but isn't the standard state of $\ce{Na}$ solid?

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    $\begingroup$ Clearly a typo. The correct answer is (c) just like you think it is. $\endgroup$
    – Zhe
    Jul 1, 2017 at 0:32
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    $\begingroup$ The correct answer is (C) due to the fact that it is an element to its standard state $\endgroup$
    – Samuel.E
    Sep 4, 2017 at 14:21

2 Answers 2

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The correct answer is indeed c. $\ce{Na(s)}$

Since for the heat of formation to be zero, the molecule should be the must abundant chemical formula of the element.

$\ce{Na}$ is a metal and is predominantly found in solid state (I dont think it exists as gas anywhere in the atmosphere of earth).

$\Delta_\mathrm{f}H^\circ$ is a quantity which is taken relative to the heat of formation of the most abundant naturally occuring form of an element.

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    $\begingroup$ I dare you to name one place on Earth where sodium exits naturogenically in a solid state. $\endgroup$
    – Linear Christmas
    Sep 7, 2017 at 16:47
  • $\begingroup$ Linear, the issue you describe is not the fault of sodium, but of the abundance of water. Sodium metal is perfectly happy when it isn't getting electrons ripped from its valence shell to form hydrogen gas. ; ) $\endgroup$
    – Pete
    Sep 8, 2017 at 19:57
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    $\begingroup$ @Pete I think sodium is an electron pusher. Poor water didn't even want that electron! $\endgroup$
    – SendersReagent
    Sep 8, 2017 at 20:00
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    $\begingroup$ @Pete I don't think you can blame one or the other in a reaction; that's the definition of the pot calling the kettle black :}. In any case, what I meant is that the predominant state in nature does not have direct bearing on what is taken as the relative zero state. IUPAC defines it generally as what is thermodynamically most stable given a standard state with some well-defined pressure, and additionally a temperature. Sometimes, kinetically stable (metastable) is used but has to be made clear. In short, need not be most abundant form in nature. $\endgroup$
    – Linear Christmas
    Sep 8, 2017 at 20:32
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Of course it must be a typo. Correct answer is "c".

The $\Delta_\mathrm{f}H^\circ$ of $\ce{Na(s)}$ is $\pu{0 J}$ and the $\Delta_\mathrm{f}H^\circ$ of $\ce{Na(g)}$ is $\pu{107.8 kJ}$.

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