Question

Using Thermodynamic Data (∆H_f), calculate the enthalpy change (∆H_rxn) for the following reactions:

Fe₂O₃ (s) + H₂ (g) → Fe (s) + H₂O (l)

N₂H₃CH₃ (l) + N₂O₄ (l) → H₂O (g) + N₂ (g) + CO₂ (g)

C₆H₁₂O₆ (s) + O₂ (g) → CO₂ (g) + H₂O (g)

Answer 1

vaas : () The given reaction is FeO3(s) +18 (9) fers) + 3010) Balancing the equation , FeO,Q) + 359 — 2 fe(s) + 33011) Standard enthalpies of formations are AHf of Feg 03 (3) = -824.2 kJ/mol -Alt of 12(g) = 0 AHj of Fels) = 0 altf of hod=-285.8 kJ/mol ..atran = Elif (products) - {OH ; (reactants) - 3x1-285:8) - (-824. 2) et mod = - 857.4 +829.2 kJ/mol =-332 kJ/mol Allan = -32.2 kJ/mol

The given reaction is - Nalbachy (1) + NaOg (1) 40 (9) + N₂ (g) + CO2(g) The balanced equation is 9 Na ty City (1) + 5 NaOg(1) ► 124069) +9N, (g) +4 CO2 (9) &H of Nah, Aby (1) = +54 kJ mo allf of Na Od (1) = + 9, 2 ks/mod Atj of 40(g) = -241.8 kt mo 44f of Nag) = 0 Ahf of co (g) = -393.5 kJ /mnd Alto xm = EAHj (products) - Ea Hj (reactants)

= [*(-0,41,8) + 9x0 + 4x (-3935)} - (549.2 +9x54] kilmul = [-2901.6 - 1574) - (46+216] kylmut = - 4475.6-262 kylmo = - 4737.6 kJ/mol Offron = - 4737.6 kg / and

m The given reaction is Call 06 (5) + 02 (4) (9) +14069) The balanced equation is GoP20 (3) + 60g (9) - 6 CO2(g) +6 40 (9) alf of C6H1₂O2 (s) = -1 260 kJ mol att of 10 (9) = -393.5 kJ mo 4H7 of $2009) = -241.8 kilmo) Aton = {a Hj (products) - EAHg (reactants) =[6*(-29.3.5) + 6*(- 241.5)) - -1260 +6x0] kJ/mnd =(-2361 - 1450.8) +1260 kolest = - 3811 8 + 1260 kJ / mo = -2550.8 kJ/mol Ahorxh = -255108 kJ/mol