Answer:
Overall enthalpy change for the formation of one mole nitric acid from nitrogen, hydrogen and oxygen, ΔH = -376 KJ
Note: the question is incomplete. The complete question is given below:
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92. kJ In the second step, ammonia and oxygen react to form nitric acid and water: NH3(g) + 2O2(g) → HNO3(g) + H2O(g) ΔH = -330. kJ Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest kJ.
Explanation:
From Hess's law of constant heat summation, the total enthalpy change for a reaction is the sum of all changes without regard to the number of multiple stages or steps involved in a reaction.
Enthalpy is a state function as it does not depend on the path taken to attain its value. Therefore, the summation of the enthalpy changes involved in the individual steps in the reaction of the formation of nitric acid will be equal to the enthalpy change of the overall reaction step.
For the first reaction step:
N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92. kJ
For the second reaction step:
NH₃(g) + 2O₂(g) → HNO3(g) + H2O(g) ΔH = -330. kJ
Overall reaction step:
[tex]\frac{1}{2}[/tex]N₂(g) + [tex]\frac{3}{2}[/tex]H₂(g) + 2O₂(g) → HNO₃ + H₂O ΔH = ?
The overall reaction for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen shows that the first reaction step should be divided by 2
[tex]\frac{1}{2}[/tex]N₂(g) + [tex]\frac{3}{2}[/tex]3H₂(g) → NH₃(g) ΔH = -46. kJ
Overall enthalpy change, ΔH = ΔH₁ + ΔH₂
Overall enthalpy change,ΔH = (-46 KJ) + (-330 KJ)
Overall enthalpy change,ΔH = -376 KJ
