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?

Question

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?

1 Answer

Impact of this question

3652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-04T05:41:29

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

Explanation:

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.

Science

Math

Humanities

... and beyond

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?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Consider the reaction: 2ClF_3 (g) + 2HN_3(g) -> N_2(g) + 6HF(g) + Cl_2(g)2ClF_3 (g) + 2HN_3(g) -> N_2(g) + 6HF(g) + Cl_2(g). When calculating DeltaH^(o)rxnDeltaH^(o)rxn, why is the DeltaH_f^oDeltaH_f^o for N_2N_2 not important?

1 Answer

Explanation:

Related questions

Impact of this question

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?