The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)(1)2NO(g) + O2(g) → 2NO2(g)(2)3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)(3)Use the values of ΔH°f to calculate the value of ΔH° for each of the preceding reactions.ΔH°f, kJ/molNH3(g)-46NO(g)90.H2O(g)-242NO 2(g)34H2O(l)-286HNO3(aq)-207ΔH°1 = kJΔH°2 = kJΔH°3 = kJWrite the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.)(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)

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!!!

i think it is d. e because the next one is igneous.

where's the diagram?