What is the value of delta Hrxn for this equation: NH3(g)+2O2(g)⟶HNO3(g)+H2O(g)

Based on the molar bond enthalpies:
N−H 390
O−O 142
O=O 502
O−H. 464
N−O 201

The value of of the reaction can be 179 kJ/mol.

of the reaction can be given by the change in the energy for the bond formation and the energy for the dissociation of the bonds.

The energy for the dissociation of bonds are :

3 N-H bonds = 3 390

3 N-H bonds = 1170 kJ/mol

2 O-O bonds = 2 142

2 O-O bonds = 1004 kJ/ mol.

The total energy for the dissociation of bonds = 1170 + 1004 kJ/mol

The total energy for the dissociation of bonds = 2174 kJ/mol

The energy for the formation of bonds are :

3 N-O bonds = 3 201

3 N-O bonds = 603 kJ/mol

3 O-H bonds = 3 464

3 O-H bonds = 1392 kJ/mol

The total energy for the formation of bonds = 603 + 1392

The total energy for the formation of bonds = 1995 kJ/mol

of the reaction =  energy for the dissociation of bonds - energy for the formation of bonds

of the reaction = 2174 - 1995 kJ/mol

of the reaction = 179 kJ/mol.

The value of of the reaction can be 179 kJ/mol.

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The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

the dependant variable is spilled water

the independent is the heat