A table of standard enthalpies of formation (ΔH°f) gives a value of −467.9 kJ/mol for NaNO3(s). Which reaction has a ΔH° value of −467.9 kJ?
(a) Na+ (aq) + NO3−(aq) → NaNO3(s)
(b) Na(s) + N(g + O3(g) → NaNO3(s)
(c) Na(s) + 1/2 N2(g) + 3/2 O2(g) → NaNO3(s)
(d) 2 Na(s) + N2(g) + 3 O2(g) → 2 NaNO3(s)
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