Final answer:
To calculate the ΔH for the given reaction, we can use Hess's Law and manipulate the given reactions to form the desired reaction. By adding reactions 4, 5, and 3, the ΔH for the reaction is found to be -1028.2 kJ.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) for the given reaction, we can use Hess's Law, which states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual reactions. We need to manipulate the given reactions so that when added together, they form the desired reaction. Let's start by manipulating the given reactions:
1) N₂ (g) + O₂ (g) → 2 NO (g) ΔH= -180.5 kJ
2) N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ΔH= -91.8 kJ
3) 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ΔH= -483.6 kJ
We can manipulate reaction 1 by multiplying it by 2 to get:
4) 2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ΔH= -361 kJ
We can manipulate reaction 2 by multiplying it by 2 to get:
5) 2 N₂ (g) + 6 H₂ (g) → 4 NH₃ (g) ΔH= -183.6 kJ
Finally, we can add reactions 4, 5, and 3 to get the desired reaction:
6) 2 N₂ (g) + 2 O₂ (g) + 6 H₂ (g) + O₂ (g) → 4 NO (g) + 4 NH₃ (g) + 2 H₂O (g) ΔH= -361 kJ -183.6 kJ -483.6 kJ = -1028.2 kJ
Therefore, the ΔH for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is -1028.2 kJ.