Question

Propane has structure ch3-ch2-ch3.Calculate the change in enthalpy for the reaction

Answer 1

Answer:

The standard reaction of enthalpies for [1] the hydrogenation of propene (CH2=CHCH3), [2] combustion of propane (CH3CH2CH3) and [3] formation of liquid water are given below:

[1] CH2=CHCH3 (g) + H2 (g) --> CH3CH2CH3 (g)                 Horxn = -124 kJ/mol

[2] CH3CH2CH3 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (l)  Horxn = -2220 kJ/mol

[3] H2 (g) + 1/2 O2 (g) --> H2O (l)                   Horxn = -286 kJ/mol

Calculate the standard enthalpy change accompanying the combustion of one mole of propene: CH2=CHCH3 (g) + 9/2 O2 (g) --> 3CO2 (g) + 3H2O (l)