Question

Data Tables

Heats of Formation

Compound delta Hf (kJ mol-1)

H20 (l) -285.83

CO2 (g) -391.51

Bond Enthalpies

Reaction delta H (298K)

C(s, graphite) to C(g) 716.7

H-H to 2H(g) 436

C-H to C(g) + H(g)

413 C-C to 2C(g) 348

*****For the word "to" it means arrow, or reacts to form this (not able to put an arrow on here)

a) If the standard enthalpy of combustion of gaseous cyclopropane, C3H6, is -2091.2kJ mol-1 at 25 C, calculate the standard enthalpy of formation of the compound. (The standard enthalphy of combustion is the heat of combustion of one mole of a compound to the most stable combustion products, which are CO2(g) and H2O(l) in the case of hydrocarbons.)

b) Calculate the standard enthalpy of formation for gaseous cyclopropane, as derived from the atomization pathway.

c) Comment on the difference between your answers for parts a and b based on your knowledge of the geometry of the molecule

Answer 1

a) Chemical reaction : 2C3H6 + 9O2 ---> 6CO2 + 6H2O

dHcomb = dHf(products) - dHf(reactants)

2 x -2091.2 = (6 x -391.51 + 6 x -285.83) - (2dHf[C3H6])

dHf[C3H6] = 59.18 kJ/mol  

b) chemical equation : 3C + 3H2 ---> C3H6

dHf[C3H6] = (3 x 348 + 6 x 413) - (3 x 716.7 + 3 x 436)

                  = 63.9 kJ/mol

c) The heat of formation calculate by bond energies is higher than the one calculated in part a). The bond energy of cyclopropane ring is higher and thus gives greater enthalpy value due to ring strain.