Question

at 298 K given that the enthalpy of formation at
Calculate the enthalpy change in the reaction
4NH,(g) + 30 g)
→ 2N.
(g) + 6H,00
298 K for NH, (g) and H,O (l) are -46.0 and
-286.0 kJ mol-l respectively.
Solution: A Hº for the reaction,​

Answer 1

Explanation:

According to Q.

4NH3 (g) + 3O2 (g) =2 N2 (g) + 6H2O (l)

delta H =

deltaH = H(products) -H (reactants)

= 6 (- 286) - 4 (-46)

= - 1532 KJ/mol