Question

Calculate the standard enthalpy change, ΔΗ0rn, în k] for the following chemical equation, using only the thermochemical equat

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Answer #1

multiply first equation with 12

6N2 + 6O2 ------------> 12NO  \DeltaH = 1084.8 kJ

revers and multiply with 12 for second equation

12NO2 ------------> 6N2 + 12O2 \DeltaH = - 405.6 kJ

multiply last equation with 3

12NH3 + 21O2 ------------> 12NO2 + 18H2O \DeltaH = - 3978.6 KJ

add all 3 above equations

6N2 + 6O2 ------------> 12NO  \DeltaH = 1084.8 kJ

12NO2 ------------> 6N2 + 12O2 \DeltaH = - 405.6 kJ

12NH3 + 21O2 ------------> 12NO2 + 18H2O \DeltaH = - 3978.6 KJ

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12NH3 + 15O2 -----------> 12NO + 18H2O  \DeltaH = - 3.30 x 103 KJ

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