Question

Given: N_2(g) + 3H_2(g) → 2NH_3(g), ∆_rH⁰ = –92.4 KJ. mol^–1. What is the standard enthalpy of formation of NH_3(g).

Comments

Can u plz explain it

Answer 1

The given reaction is shown below.

Two moles of ammonia are formed during the reaction. Thus, the reaction for formation of one mole of ammonia is shown below.

The standard enthalpy of formation of ammonia is calculated as,

Therefore, the standard enthalpy of formation of ammonia gas is -46.2 kJ mol^-1.

Answer 2

_fH NH_3(g) = -92.4/2 = 46.2 KJ. mol^-1

Answer 3

For the given reaction,

N₂​(g) + 3H₂​(g) ⟶ 2NH₃​(g)

Enthalpy of reaction (Δr​H⁰) =  –92.4 kJ mol^–1

 N₂​(g) + H₂​(g) ⟶ NH₃​(g)

Hence, standard enthalpy of NH₃​ is equal to the 1/2 ​Δr​H⁰

As, Δr​H⁰ = –92.4 kJ mol^–1

∴ Standard enthalpy of NH₃ ​= \(\frac{-92.4}{2}\)

⇒ Standard enthalpy of NH₃​ = −46.2 kJ mol⁻¹