Question

For the combustion of cyclopropane at 25 ºC C3H6(g) + (9/2) O2(g) → 3 CO2(g) + 3 H2O(liq) ΔHºc(C3H6,g) = -2091 kJ / mol. What is the standard molar enthalpy of formation of cyclopropane(g) at 25 ºC if the standard molar enthalpies of formation of CO2(g) and H2O(liq) at that temperature are, respectively, -393.51 kJ / mol and -285.83 kJ / mol?

Answer 1

given combustion reaction of cyclopropane is given bellow

C3H6(g) + (9/2)O2  $\rightarrow$ 3CO2 +3H2O $\Delta$H⁰ = -2091KJ/mol

H⁰ C3H6= ? KJ/mol

$\Delta$_rxnH⁰ = -2091KJ/mol

_fH⁰ O_2(g) = 0 KJ/mol

_fH⁰ CO_2(g) = -393.51 KJ/mol

_fH⁰ H₂O_(g) = -285.83 KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰_rxn= [(3 _fH⁰ CO₂) + (3_fH⁰ H₂O)] - [(_fH⁰ C₃H₆) + (9/2_fH⁰ O₂)]

-2091 = [(3 (-393.15)) + (3 (-285.83))] - [$\Delta$_fH⁰C₃H₉ + (9/2 (0)]

-2091 = [-2036.94] - [$\Delta$_fHC₃H₉]

_fH⁰C₃H₉ = 2091 - 2036.94 = +54.06KJ/mol

ans standard molar enthalpy of formation of cyclopropane is + 54.06 KJ/mol