Question

8. The standard reaction enthalpy for the hydrogenation of propene, CH-CHCH2 + 5 02 → CH3CH2CH3p. is 124.0 kJ/mol. The standard reaction enthalpy for the formation of water, H ( + 7 O2(g) → H2O).is -286.0 kJ/mole. The standard reaction enthalpy for the combustion of propane, CH3CH2CH3 +5 O2(g) → 3 CO2 + H2O() is 2220 kJ/mol. Calculate the standard enthalpy of reaction for propene. (ANS: -2058 kJ/mol)

Answer 1

CH₂ = CH - CH3 (9) + H₂ (g) CH3 CH2 CH3 Dxn = -124 ks - 0 H2(g) + 102(3) Hoy DH=-286.0 Ks @ mol mol CH3 CH2 CH₂ + 50219) 3 02 (g) +4H₂0le) 34= 2220 K L - We have to calculate Standard ethalpy of combustion of propene CH₂=CH-CH3 (9) + 102 302 (g) + 34206) Dhe = ?

Subtract ep.2 from ep 3 cho cha +50262) 3002 at 4420 (e) on=-2220 kunoi - H₂ lg) + 2 O 2 (g) → H2011) Du= -286 Kolmoi

сиз си, сиз +4 о, (9) — 3 o, tə) + 34, оба) +, (9) Ди = –1939 Ка Adding ep. 8 ep. сн. = и-м, сә) + 10, (9) — 2е зco (9)+ зи, о (3) 04. —193- ач, - -20J Ko Duran = -2058 Kg Arones mol