Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows:
4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(g)
Using the following bond energies, estimate the enthalpy change for the reaction.
BE(O–H) = 464 kJ/mol
BE(N–H) = 389 kJ/mol
BE(O=O) = 498 kJ/mol
BE(N–O) = 222 kJ/mol
BE(N=O) = 590 kJ/mol
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Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows: 4NH3(g) + 7O2(g) --->...
Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows: 4 NH3(g) + 7 O2(g) ---> 4 NO2(g) + 6 H2O(g) Use the following data for bond energies to determine the determine the enthalpy of the reaction. O−H : 459 kJ/mol N−H : 386 kJ/mol N−O : 201 kJ/mol N=O : 607 kJ/mol O=O : 494 kJ/mol The answer my professor gives is -4163 kJ/mol but I don't know how to get that answer especially with the bonds...
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