Question

61. For each reaction, calculate AHX, ASixn, and AG x at 25 °C and state whether or not the reaction is spontaneous. If the reaction is not spontaneous, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25 °C? a. N204(8) —> 2 NO2(g) b. NH4Cl(s) → HCl(g) + NH3(g) c. 3 H2(g) + Fe2O3(s) — 2 Fe(s) + 3 H2O(g) d. N2(g) + 3 H2(g) — 2 NH3(g)

Answer 1

ves from my book @ I am taking standard values from m N204949) + 3 NOCI) AH = stipro duets - 14 reactants = 2042102(9) - 444. N20419) - 2x (33-2)-9-2 = 57.2 kJ /mnd Aso = isof products - sy reactante 225 N (9) - sof N2O4 (9) : 45º = 2 (240.1) – 30413 - 17509 J/kmd = 0.176 kJ/kmol AGO = AH-TASO T=2592=2986 = 57. 2xette 298 K x(0.176 kJ/kak = 4-752 kilmd As ago is the the reaction is not spontanery T= AH = 57.2 kJ /mnd 0.176 kJ/kurt 325k 1. It is spontaneous above 325k (52°c) So, the temperature is increased from 25°C Why clas) → Hel(9) + NH3(g) Akº = she ha tate with DHL WHte de = -92 komt 80 kJ/mol - 31404 - = -486-44 kiulush As = 18607 TK ml + 192.5 - quet - 284.6 Jik-mol = 0.2846kg/ lemon

A9° -486.4 - 298 * (0-2846) =-571-2 kolme As Ago is negative, the ren is spontaneous 3H2(g) + Fez Og(8) ► 2Fecs) + 3H019) SHO - 2x0+3x6-2418)-(360) +(-824-2) AH = 98.8 kJ/mol osº = 2x27015)+3 *(88-83) – (3*180.7+89-96) = 138.7 JIk mál = 0.1387 KJ/i.md o q° = 98.8 - 298 x 0-1387 = 57.46 kJ/mol As Ago is the the Ran is not spontaneous To set = 76.64 = 72.3k. (439.2°) above 2439.2°C , The reaction is spontaneous N269) +372 (9) 7.2 NHz (9) AHO = 2x1-46.1) -[o+3x0] 1.Z -92.2 Asº = 2x192.5 - [1916+131) = 62.4 l kmol = 0.0624 KJ/kmol A Gº = -92-2-298 x 0.0624 =-110.8 kJ/mol The ran is spontaneous @