Question

Given: N2H4(l) + O2(g) LaTeX: \longrightarrow⟶ N2(g) + 2H2O(g) ΔH°1 = –543 kJ·mol–1 2H2(g) + O2(g) LaTeX: \longrightarrow⟶ 2H2O(g) ΔH°2 = –484 kJ·mol–1 N2(g) + 3H2(g) LaTeX: \longrightarrow⟶ 2NH3(g) ΔH°3 = –92 kJ·mol–1 What is the standard enthalpy change for the following reaction? 2NH3(g) LaTeX: \longrightarrow⟶ N2H4(l) + H2(g) Group of answer choices

Answer 1

Step-by-step explanation:

Given reaction

N₂H₄ + O₂ ⇒ N₂ + 2H₂O ΔH₁ = -543 KJ ---------- ( 1 )

2H₂ + O₂ ⇒ 2H₂O ΔH₂ = -484 KJ ---------- ( 2 )

N₂ + 3 H₂ ⇒ 2NH₃ ΔH₃ = -92 KJ -----------( 3 )

( 1 ) - ( 2 ) +( 3 )

N₂H₄ + O₂ - 2H₂ - O₂ +N₂ + 3 H₂ ⇒ N₂ + 2H₂O - 2H₂O +2NH₃

ΔH = -543 + 484 -92 = -151 KJ

N₂H₄ + H₂ ⇒ 2NH₃ ΔH = -151 KJ .

2NH₃ ⇒ N₂H₄ + H₂ ΔH = + 151 KJ