Consider the reaction: #2ClF_3 (g) + 2HN_3(g) -> N_2(g) + 6HF(g) + Cl_2(g)#. When calculating #DeltaH^(o)rxn#, why is the #DeltaH_f^o# for #N_2# not important?

Answer 1

Because #DeltaH_f""^@# for any element in its standard state is defined to be ZERO.

Of course, in that reaction we also ignored #DeltaH_f""^@# for #Cl_2(g)#, another element in its standard state. This definition may seem to be arbitrary, but since we measure the change in enthalpies, not absolute enthalpies, the treatment is consistent.

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Answer 2

Noah Aldrich

The enthalpy change ((\Delta H^\circ_{rxn})) for a reaction is calculated using the standard enthalpies of formation ((\Delta H_f^\circ)) for the reactants and products. Since nitrogen ((N_2)) is in its standard state (gaseous form) and is not being formed or decomposed in the reaction, its standard enthalpy of formation ((\Delta H_f^\circ)) is not considered because its value is zero by definition for elements in their standard states. Therefore, the contribution of (N_2) to the enthalpy change of the reaction ((\Delta H^\circ_{rxn})) is negligible, and it does not need to be included in the calculation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Consider the reaction: #2ClF_3 (g) + 2HN_3(g) -&gt; N_2(g) + 6HF(g) + Cl_2(g)#. When calculating #DeltaH^(o)rxn#, why is the #DeltaH_f^o# for #N_2# not important?

Consider the reaction: #2ClF_3 (g) + 2HN_3(g) -> N_2(g) + 6HF(g) + Cl_2(g)#. When calculating #DeltaH^(o)rxn#, why is the #DeltaH_f^o# for #N_2# not important?