Respuesta :
Answer:
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 value of [tex]\rm \Delta H_r_x_n[/tex] of the reaction can be 179 kJ/mol.
[tex]\rm \Delta H_r_x_n[/tex] 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 [tex]\times[/tex] 390
3 N-H bonds = 1170 kJ/mol
2 O-O bonds = 2 [tex]\times[/tex] 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 [tex]\times[/tex] 201
3 N-O bonds = 603 kJ/mol
3 O-H bonds = 3 [tex]\times[/tex] 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
[tex]\rm \Delta H_r_x_n[/tex] of the reaction = energy for the dissociation of bonds - energy for the formation of bonds
[tex]\rm \Delta H_r_x_n[/tex] of the reaction = 2174 - 1995 kJ/mol
[tex]\rm \Delta H_r_x_n[/tex] of the reaction = 179 kJ/mol.
The value of [tex]\rm \Delta H_r_x_n[/tex] of the reaction can be 179 kJ/mol.
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