Given the ammonia synthesis: #N_2(g) + 3H_2(g) rarr 2NH_3(g)# #DeltaH_"rxn"^@=-92.6kJmol^-1#, what reaction enthalpy is associated with the formation of #8.55xx10^4*g# of ammonia?

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Answer 1 · 2017-06-30T04:37:59

How? We use the enthalpy as a variable in the reaction..........and get #DeltaH_"rxn"=-232725*kJ........#

We got...

#N_2(g) + 3H_2(g) rarr 2NH_3(g)+92.6*kJ#

Which tells that the formation of #34*g# ammonia releases #2xx92.6*kJ#. Agreed?

But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.

#"Moles of ammonia"=(8.55xx10^4*g)/(17.01*g*mol^-1)=5026.5*mol#.

And thus #DeltaH_"rxn"=(-5026.5*molxx92.6*kJ*mol^-1)/2#

#=??*kJ#

Why is #DeltaH_"rxn"# negative?

Given the ammonia synthesis: #N_2(g) + 3H_2(g) rarr 2NH_3(g)# #DeltaH_"rxn"^@=-92.6*kJ*mol^-1#, what reaction enthalpy is associated with the formation of #8.55xx10^4*g# of ammonia?