Question

Consider the two reactions. 2 NH2(g) + 3 N2O(g) → 4N2(g) + 3 H2O(1) 4 NH,(g) + 30,(g) + 2 Ng(g) + 6H2O(1) AH° = -1010 kJ AH° = 1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N2(g) + 2O2(g) → N2O(g) AH° =

Answer 1

given 2NH2 + 3N20 (9) UNO(g) + 3H₂(1) < Hoa -100 kJ multiply love equator with @ (x2 NH3 G) + ux 3 NO2 cgl - cixy Ng CG) +423 Holl A Hor-yoyok T Aho = Ux_1010 kJ 8MHz 9) + 12 Ny 9 –1604.) +12H20 r)- yoyo kJ A Ho - Youo na - D o Then Equaton CINA 9) +34219) - 2N2 g) + OH goal AHO-1531 ET mutiply above Equation with & NH3 g) + 60209) - Unacg) + 12H0cd 4 H = 2x-1539 KT = - 3062 kJ Reverse The above equaton Naeg) + 12 H₂ W snith g) +602 (9) ② a Hot 3062 KT

Add Equalan + 8N9) + 12 No2g) — 16N2(9) + 121 AHO Clocok Nag) + Tatizace) - Sat 9) + 2 g AH. +3062 Nag) + 12 Noz 9) 12 N2 LON2g) +602 g A Ho= Cociua 3002 -978 KT - 12 N, 9)+602 glas ou a Reverse Equatan 12N2 g) + Go297 12 new g) 4H+ 978kT Divide with 12 N2 g) + Google 12 h New g), 4H = +97847 N2 g) + 02 Nag og [At= 815 . Anson AH = +81.5 kJ