Question

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46. Using Hess's law, calculate the AH value for the following reaction: (8 marks) FeOce) + CO(g) → Fe(s) + CO2(g) Use these three reactions: Fe,036) + 3CO(g) + 2Fe6) + 3CO2(8) AH= -25.0 kJ 3Fe2O3(s) + CO(g) → 2Fe2O46) + CO2(8) AH = -47.0 kJ Fe30468) + CO(g) + 3FeO (0) + CO2(8) AH = +38.0 kJ Sodium bicarbonate (NaHCO3) is often used in the kitchen to extinguish fires. When heated, it decomposes into sodium carbonate, water vapour, and carbon dioxide. The AH for the reaction of 1 mole of sodium bicarbonate is +64.6 kJ. a) Write a balanced thermochemical equation for this reaction. (2 marks) b) Using the enthalpies of formation, calculate the enthalpy of formation of sodium bicarbonate. (5 marks) 47.

Answer 1

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46.

Feot co Fe co2 Rte t 3 C02 A 25-0kJ 3 Peao t o , 4M40 KT 304Co 3 Feot Uoz 0y t o (3) 44 58.0 us He las, Bui thi three euahin, we neid dlo wathench opens o o Same operahans is done wih their enesgies he Final eoba sep Hene Hue 3. dinde Reverse the gor we Sign of AM wi chan -38-9 Feo +글 AH O and slep Taes equahn nd divide tt b a, we ge e t 25 0 kT AH Now add 1 sep& and SteP euaha, Feot t Fet -38.o as o 3 2

the whou equahan b 6, we get Now 6x t Co2 kJ 38 An Gteot 3Fea 2 +7co 2tegOu +b Fe tWz step 3 Now Sub tout the eguahon fom th abore euaha get we Gfe0 tco +3re20 3D03 6 Feot GLo G fe t 6 CO2 - X6 ep 4 divide the abure eiahan b6 Now e t Co requnuahan Feo t co 38 AH- J-1.33 kJ 4H=-1.33kJ

Balcmeod thermo cheswiča equaten 47 a Natto3 ) Na Co3 Lt Co, g)+0 ) Frast all A for Na,co ) = -1130-7 k)/ma = 393-5 KJ/mol A Fr ocj)- -948 /mo we needl to calutali A,(NaH MAHpnduls Hy Griven, AHr 64.6 kJ EAectom 64 6 J= 1mol x A Lo, +1molx 4 (Nao) +1mol x AM ANaH) dividedyase to g 4moi Nans 64.6 JdmalX -130 1K t olX -393 TT mol t 1molx-a4)8 kT no1 A1 (Nan) - 883 kJ-64.6 T= 947 6 J 41 6 KI AH NaH As 1mol Natus is dumpoted