calculate delta H for the formation of one mole of N2O5 from the elements at 25...

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

Solutions

Expert Solution

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

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

N2_(g) + 2.5O2_(g) ----------> N2O5_(g) H = 11.5 KJ

queen_honey_blossom answered 2 years ago

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