Answer to Question #318682 in General Chemistry for yow

Question

Given the following thermochemical equations:

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

2.) N₂O(g) + H₂(g) → N₂(g) + H₂O(0) ∆H°-367.4 k]

3.) H₂(0)+02 (0)→ H₂0 (1)

∆H°= -285.9 kJ

Find the value of AH" for the reaction:

2NH3(g) + 3N₂0(g) → 4N₂(g) + 3H₂O(1)

Answer 1

AH= -(1531+ 367.4+ 285.9)

= 2184.3kJ