Answer:
Explanation:
Given that;
The chemical equation for the reaction is:
[tex]\text {4NH}_3(g) +\text {5O}_2(g) }[/tex] ↔ [tex]\text {4NO}(g) +\text {6H}_2O}(g) }[/tex]
a). [tex]\Delta H^0 = \sum \Delta H^0_{reactant \ bonds \ broken} + \sum \Delta H^0_{product \ bonds \ formed}[/tex]
Using average bond energies;
For Bonds broken, we have:
4 × 3 N - H = 12 mol × 391 kJ/mol = 4692 kJ
5 × O = O = 5 mol × 498 kJ/mol = 2490 kJ
[tex]\mathbf{ \sum \Delta H^0_{reactant \ bonds \ broken} }[/tex] = 7182 kJ
For Bonds formed, we have:
4 × N = O = 4 mol × - 607 kJ/mol = -2428 kJ
6 × 2 O - H = 12 mol × - 467 kJ/mol = - 5604 kJ
[tex]\mathbf{ \sum \Delta H^0_{product \ bonds \ formed} }[/tex] = - 8032 kJ
Therefore;
[tex]\mathbf{ \Delta H^o = 7182 kJ + (- 8032 \ kJ)}[/tex]
[tex]\mathbf{ \Delta H^o =-850 \ kJ}[/tex]
b) Using Standard heats of formation;
[tex]\Delta H^0 = \sum m \Delta H^0_f (products) = \sum n \Delta H^0_f (reactants)[/tex]
[tex]\Delta H^0 = [ 4 \ mol \times \Delta H^0_f \ (NO(g)) + 6\ mol \times \Delta H^0_f(H_2O)] - [ 4 \ mol \times \Delta H^0_f \ (NH_3(g)) + 5 \ mol \times \Delta H^0_f \ (O_2)][/tex]
[tex]\Delta H^0 = [ 4 \ mol \times90.29 \ kJ/mol + 6\ mol \times -241.826 \ kJ/mol - [ 4 \ mol \times-45.9 \ kJ/mol + 5 \ mol \times 0][/tex]
[tex]\mathbf{\Delta H^0 = -906.196 \ kJ}[/tex]
c) The value of ΔH^o, calculated using average bond energies, is only considered to be an estimate of the standard enthalpy change for the reaction because there is a premise in the calculation that all the bonds broken in the reactant and all bonds are formed in products. Meanwhile, only little bonds are broken, and only little bonds are formed during the reaction.
