Question

for the reaction,
3N2O (g) + 2NH3 (g) → 4N₂ (g) + 3 H₂O(g); ∆H° = -879.6 kJ
If ∆H°f[ NH3 (g)] = -45.9 KJ/mol
∆H°f[H2o (g)] = -241.8 KJ/mol
then ∆H°f[N2O (g)] will be :
(a), +246 kJ (b), +82 KJ (c),-82 KJ (d) - 246 KJ

Answer 1

Answer:

c

Explanation: