How do you calculate the standard enthalpy of reaction, #DeltaH_(rxn)^@# for the following reaction from the given standard heats formation (#DeltaH_f^@#) values: #4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)#?

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Answer 1 · 2017-06-02T03:31:44

#DeltaH_"rxn"^@=DeltaH_f^@("products")-DeltaH_f^@("reactants")#

And here, #DeltaH_"rxn"^@=..................#

#(4xx90.3-6xx241.8-{4xx-45.9})*kJ*mol^-1#

#=-906*kJ*mol^-1#.

The formation of water clearly dominates the enthalpy change, it is a thermodynamic sink.

ANd of course, #DeltaH_f^@=0# for an element (here dixoygen) in its standard state (they even spoon-feed you this fact).