Answer to Question #319325 in General Chemistry for yow

Question

Given the following thermochemical equations:

1.) 4NH3(g) + 302(g) → 2N2(g) + 6H₂0(l) ∆H°-1531 kJ

2.) N₂O(g) + H₂(g) → N2(g) + H₂O (l) ∆H°-367.4 kJ

3.) H₂(g) + O2(g) → H₂0(l)

∆H°= -285.9 kJ

Find the value of ∆H° for the reaction:

2NH3(g) + 3N₂O(g) → 4N₂(g) + 3H₂0 (l)

Answer 1

The answer to your question is provided in the image: