Respuesta :
Answer:
ΔH°1 = -908 kJ
ΔH°2 = -112 kJ
ΔH°3 = -140 kJ
Combination of the reactions for Ostwald Process' overall reaction
4NH₃ + 8O₂ → 4HNO₃ + 4H₂O
Total Heat of reaction = -1524 kJ
On a per mole basis
NH₃ + 2O₂ → HNO₃ + H₂O
Total Heat of reaction = -381 kJ/mol
Explanation:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
2NO(g) + O₂(g) → 2NO₂(g)
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)
ΔH°f, kJ/mol
NH3(g), -46
NO(g), 90.
H2O(g), -242
NO2(g), 34
H2O(l), -286
HNO3(aq), -207
We are to find the heat of reaction for the 3 leading reactions. ΔH°1, ΔH°2, ΔH°3
The heat of any reaction is given as the
ΔH(products) - ΔH(reactants)
For reaction 1, ΔH°1
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
ΔH°1 = ΔH(products) - ΔH(reactants)
ΔH(products) = (4×90) + (6×-242) = -1092 KJ
ΔH(reactants) = (4×-46) + (5×0) = -184 KJ
ΔH°1 = -1092 - (-184) = -908 KJ
For reaction 2, ΔH°2
2NO(g) + O₂(g) → 2NO₂(g)
ΔH°2 = ΔH(products) - ΔH(reactants)
ΔH(products) = (2×34) = 68 kJ
ΔH(reactants) = (2×90) + (1×0) = 180 kJ
ΔH°2 = 68 - 180 = -112 kJ
For reaction 3, ΔH°3
3NO2(g) + H₂O(l) → 2HNO₃(aq) + NO(g)
ΔH°3 = ΔH(products) - ΔH(reactants)
ΔH(products) = (2×-207) + (1×90) = -324 kJ
ΔH(reactants) = (3×34) + (1×-286) = -184 KJ
ΔH°3 = -324 - (-184) = -140 kJ
For the combination of the reactions,
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
2NO(g) + O₂(g) → 2NO₂(g) ×3
6NO(g) + 3O₂(g) → 6NO₂(g)
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g) ×2
6NO₂(g) + 2H₂O(l) → 4HNO₃(aq) + 2NO(g)
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
6NO(g) + 3O₂(g) → 6NO₂(g)
6NO₂(g) + 2H₂O(l) → 4HNO₃(aq) + 2NO(g)
Summing it all up, and cancelling the species that appear on both sides.
4NH₃ + 8O₂ → 4HNO₃ + 4H₂O
Total ΔH°f according to Born-Haber cycle and Hess' law.
Total ΔH°f = ΔH°1 + 3ΔH°2 + 2ΔH°3
= -908 + (3×-112) + (2×-140)
= -1524 kJ
On a per mole basis
ΔH°f = (-1524 ÷ 4) = -381 kJ/mol
Hope this Helps!!!
