Using the provided table, determine the enthalpy for the reaction 2 NH3 (g) + 3 N20 (g) 4 N2 (g) + 3 H20 (1)

ΔH°r = -1009.8 kJ

Explanation:

Let's consider the following balanced reaction.

2 NH₃(g) + 3 N₂O(g) ⇒ 4 N₂(g) + 3 H₂O(l)

We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression.

ΔH°r = [4 mol × ΔH°f(N₂(g)) + 3 mol × ΔH°f(H₂O(l))] - [2 mol × ΔH°f(NH₃(g)) + 3 mol × ΔH°f(N₂O(g))]

ΔH°r = [4 mol × 0 kJ/mol + 3 mol × (-285.8 kJ/mol)] - [2 mol × (-46.2 kJ/mol) + 3 mol × 81.6 kJ/mol]

ΔH°r = -1009.8 kJ

i think   the answer is c hope i'm right

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.