In: Chemistry
calculate delta H for the formation of one mole of N2O5 from the elements at 25 celsius using the following data.
2H2(g) + O2- 2H2O(l) delta H= -571.6 kj
N2O5(g) + H2O- 2HNO3(l) delta H= -73.7kj
1/2N2(g) + 3/2O2(g) + 1/2H2(g) - HNO3(l) delta H=-174.kj
first write the formation equation from its elements
N2(g) + 2.5O2(g) ---------> N2O5(g)
reverse the second equation
2HNO3(l) --------> N2O5(g) + H2O H = 73.7 kJ
double the third equation
N2(g) + 3O2(g) + H2(g) --------> 2HNO3(l) H = - 348 KJ
half and reverse the first equation
H2O(l) ----------> H2(g) + 1/2O2(g) H = + 285.8 KJ
add all the equations
H2O(l) ----------> H2(g) + 1/2O2(g) H = + 285.8 KJ
2HNO3(l) --------> N2O5(g) + H2O H = 73.7 kJ
N2(g) + 3O2(g) + H2(g) --------> 2HNO3(l) H = - 348 KJ
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N2(g) + 2.5O2(g) ----------> N2O5(g) H = 11.5 KJ