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

1 Answer
Oct 17, 2016

Answer:

#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#.