Question

3. Determine the change in enthalpy for the oxidation of ammonia as follows 4 NH (g)5 O2(g) 3 4 NO(g) + 6 H2O( using the foll

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Answer #1

reverse and double the first equation

4NH3(g)   -------------> 2N2(g) + 6H2(g) \DeltaH0 = 198.44 kJ

double the second equation

2N2(g) + 2O2(g) ---------> 4NO(g)   \DeltaH0 = 361 KJ

triple the third equation

3H2(g) + 3O2(g) ------------> 6H2O(l)   \DeltaH0 = - 1714.8 KJ

add all the 3 equatio ns

4NH3(g)   -------------> 2N2(g) + 6H2(g) \DeltaH0 = 198.44 kJ

2N2(g) + 2O2(g) ---------> 4NO(g)   \DeltaH0 = 361 KJ

3H2(g) + 3O2(g) ------------> 6H2O(l)   \DeltaH0 = - 1714.8 KJ

------------------------------------------------------------------------------------

4NH3(g) + 5O2(g) -------------> 4NO(g)   + 6H2O(l)  \DeltaH0 = - 1155.36 KJ

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