Given... N_2(g) + 3H_2(g) rarr 2NH_3(g) + 92.4*kJ... ..what is DeltaH_f^@ NH_3(g)?

1 Answer
anor277
Oct 17, 2016

DeltaH_f^@ NH_3(g) = -46.2*kJ*mol^-1.

Explanation:

By definition, DeltaH_f^@ is the enthalpy change associated with the formation of 1 mole of product from its constituent elements in their standard states at 298*K.

I think you have quoted the enthalpy change for the reaction:

N_2(g) + 3H_2(g) rarr 2NH_3(g), which of course is TWICE the enthalpy of formation for NH_3, because this reaction formed TWO moles of ammonia.

Finally, we can write:

1/2N_2(g) + 3/2H_2(g) rarr NH_3(g) ;DeltaH_"rxn"^@=DeltaH_f^@"ammonia"=-46.2*kJ*mol^-1.

Related questions
See all questions in Enthalpy
Impact of this question
9031 views around the world
You can reuse this answer
Creative Commons License