What is #DeltaH_f^@# for ammonia, and how is it formulated?

1 Answer
Jun 26, 2017

Answer:

The #"standard enthalpy of formation"# of ammonia is #-46*kJ*mol^-1#.

Explanation:

What does this mean...? Well the given equation is associated with the RELEASE of a given quantity of heat to the surroundings........

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g) +46*kJ#

We write enthalpies (and other thermodynamic quantities) #"per mole of reaction"# as written. And thus when we report #DeltaH_"rxn"^@=DeltaH_f^@(NH_3)=-46*kJ*mol^-1#......The negative sign indicates that heat is RELEASED to the environment rather than absorbed. The enthalpy of the #3xxNH# formed is GREATER than the enthalpy of the #H-H# and #N-=N# bonds broken.

And so if there are #2*mol# ammonia formed, i.e. #34*g# rather than #17*g#, the enthalpy is DOUBLED......... Are you clear on what I am saying?