Consider the reaction: 2ClF 3 (g)+ 2NH 3 (g)$\to$N 2 (g)+ 6HF(g)+ Cl 2 (g) When calculating the $\Delta$H° rxn ,why is the $\Delta$H f ° for N 2 not important?

The standard enthalpy of formation for an element in its standard state, like N2, is zero. Therefore, it does not contribute to the enthalpy change of the reaction.

Consider the reaction: 2ClF 3 (g)+ 2NH 3 (g)$\to$N 2 (g)+ 6HF(g)+ Cl 2 (g) When calculating the $\Delta$H° rxn ,why is the $\Delta$H f ° for N 2 not important?

The standard enthalpy of formation for an element in its standard state, like N2, is zero. Therefore, it does not contribute to the enthalpy change of the reaction.

Solved

Consider the Reaction: 2ClF₃(g)+ 2NH₃(g) $\to$ N₂(g)+ 6HF(g)+ Cl₂(g)
When Calculating The

Question 67

Multiple Choice

Consider the reaction: 2ClF₃(g) + 2NH₃(g) $\to$ N₂(g) + 6HF(g) + Cl₂(g)
When calculating the $\Delta$ H°_rxn,why is the $\Delta$ H_f° for N₂ not important?

A) Because nitrogen is in its standard elemental state and no energy is needed for this product to exist.
B) Because any element or compound in the gaseous state requires a negligible amount of energy to exist.
C) Because the products are not included when calculating $\Delta$ H°_rxn.
D) Because nitrogen is in its elemental state and does not contribute to the reaction itself.
E) Two of the above statements explain why N₂ is not important when calculating $\Delta$ H°_rxn.