Question

Calculate the standard combustion enthalpy of propylene at 25 ° C using the standard formation enthalpy of liquid water (-286 kJ mol-1) and the following reactions:

1) CH2CHCH3 (g) + H2 (g) → CH3CH2CH3 (g); Δ? (1) = -125 kJ mol-1

2) CH3CH2CH3 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l); Δ? (2) = -2220 kJ mol-1

3) H 2 (g) + 0.5O2 (g) → H2O (l);Δ? ((3) = -286 kJ mol-1

Answer 1

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CH₂ CH CH 3 (g) + H₂(g) 11 21 - CH3 CH₂ CH3 (9) GHI = - 125 KT molt CH3 CH2 CH3 (9) + 502191 -> 300219) + 4H2010) SH - - 2220 KT mort 3) H2 (9) + 02 (9) ₂0121 SH₂ = -286 kJ molt On revensing (3) equation, we get 41 H2O(l) H₂ lgl & ļ02 19) Hy = - AH₂ = 286 KT On adding 1, 2) and 4 combustion of propylene CH₂ CH CH3 (9) + 90219) equations we get equation of 360₂19) + 3H2010) AH AHI + AH2 + 4H4 = (-125 +(-2220) + 286) KtImol SHC= - 2059 kJ Imol rience, standard combustion enthalpy of propylene ie Alte at 25.cis - 2059 KJ molt