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I'm a little confused by this question, perhaps my understanding of enthalpy isn't as good as it should be.

For the compound $\ce{KNO3}$, which of the following reactions is used to determine the standard enthalpy of formation, $\Delta H^\circ_\mathrm f$?

$$\ce{K(s) + N (g) + O3(g) -> KNO3 (s)\\ 2K + N2 + 2O3 -> 2KNO3\\ 2K + N2 +3O2 -> 2KNO3\\ K + 1/2 N2 + 3/2 O2 -> KNO3}$$

I know that the $\Delta H^\circ_\mathrm f(\ce{KNO3})= \pu{−494.6 kJ/mol}$ from reading a table, I'm just confused about how do I get from that number to determine the formation of a compound.

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  • $\begingroup$ The textbook is asking which of the 4 equations is used to determine the heat of formation (only one is correct). Since the standard state of oxygen is $O_2$, we know the first 2 equations are not used to calculate the heat of formation. $\endgroup$ – LDC3 Apr 6 '14 at 18:11
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    $\begingroup$ Since the standard enthalpy of formation refers to the formation of 1 mole of product, equation 4 should be correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H$. $\endgroup$ – Jannis Andreska Apr 6 '14 at 18:36
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The standard enthalpy of formation $\Delta H_\mathrm f$ refers to the formation of 1 mole of product, with the educts and product being in their respective standard states. The fourth equation is therefore correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H_\mathrm f$. The first two equations can be ruled out because they do not contain the educts nitrogen and/or oxygen in their standard states, which are $\ce{N2}$ and $\ce{O2}$, respectively.

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