How do you calculate the enthalpy change in the following reaction #4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O#?

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Answer 1 · 2017-06-01T17:35:07

Well first you need some data............I eventually get #DeltaH_"rxn"=-1028.6*kJ*mol^-1#.

And I quote #DeltaH_f^@# values for the given substances. I use this site, and perhaps you should check that I have got the values right.......

#NH_3(g)#, #DeltaH_f^@=-80.8*kJ*mol^-1#

#NO(g)#, #DeltaH_f^@=+90.29*kJ*mol^-1#

#H_2O(l)#, #DeltaH_f^@=-285.8*kJ*mol^-1#

And of course #DeltaH_f^@# #O_2(g)# #=# #0#, by definition for an element in its standard state under standard conditions.

And #DeltaH_"reaction"^@=SigmaDelta_f^@("products")-SigmaDelta_f^@("reactants")#

And we do this summation for the given reaction:

#4NH_3(g) + 5O_2(g) rarr 4NO(g) + 6H_2O(l)#

#DeltaH_"rxn"^@={(6xx-285.5+4xx90.29)-(4xx-80.8)}*kJ*mol^-1#

#=-1028.6*kJ*mol^-1# (i.e. per mole of reaction as WRITTEN.....).

The enthalpy is strongly negative, i.e. exothermic, because the reaction forms water..........a thermodynamic sink.......