A table of standard enthalpies of formation (ΔH°f) gives a value ...

Question

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)

Accepted Answer

Hi everyone this problem reads the value of negative 1207.6 kg jewels per mole for calcium carbonate is based on the table of standard entropy of formation, Which of the following reactions has a standard entropy of formation equal to negative 1,207.6 killer jewels. So we want to know which of the following reactions is going to have the standard entropy of formation equal to the same value of negative 1207.6 kg per mole for calcium carbonate. Okay, so let's start off by writing out what do we mean by our standard entropy for the reaction? So our standard entropy of a reaction is equal to the standard heats of formation or the standard n therapies of formation of the products minus the standard entropy ease of formation of the reactant. Okay, so we have products minus react ints. And we can say when the standard entropy a formation for the reactant is equal to zero. This is going to lead our standard heat of formation or our standard entropy of the reaction equal to the standard heat of formation for our products. Okay, because in the problem we know that the value for the standard entropy for the reaction is negative 1207.6. And we know that the standard entropy for the products is the same value. So that means that our standard entropy ease of formation for reactant is going to equal zero. Okay. Which is the case for standard entropy of formation. So we have our calcium carbonate and the standard states for calcium carbonate are calcium solid, calcium solid carbon and oxygen gas respectively. Okay, so what that means is it means that the standard entropy of formation for calcium carbonate is going to equal are calcium solid plus our carbon solid plus three halves oxygen gas. Okay, this is going to lead us to our calcium carbonate solid. And this is going to have a standard. This is going to have a standard entropy of reaction. Let me write that correctly here. So we're gonna have a standard entropy of reaction equal to negative 1207.6 killer jewels. So that is going to be our reaction. So out of the answer choices given that best matches answer choice C. Okay, so that is it for this problem. I hope this was helpful.

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