Answer
$\Delta$H° (initiation) = 242 kJ/mol (58 kcal/mol)
$\Delta$H° (first propagation step) = -21 kJ/mol (-5 kcal/mol)
$\Delta$H° (second propagation step) = -97 kJ/mol (-23 kcal/mol)
Work Step by Step
The initiation enthalpy is equal to the bond-dissociation energy of Cl-Cl which is 242 kJ/mol (58 kcal/mol).
The first propagation step enthalpy (-21 kJ/mol or -5 kcal/moL) can be calculated by subtracting the H-Cl bond-dissociation energy (431 kJ/mol or 103 kcal/mol) from the CH3CH2-H bond dissociation energy (410 kJ/mol or 98 kcal/mol).
The second propagation step enthalpy (-97 kJ/mol or -23 kcal/mol) can be calculated by subtracting the CH3CH2-Cl bond-dissociation energy (339 kJ/mol or 81 kcal/mol) from the Cl-Cl bond-dissociation energy (242 kJ/mol or 58 kcal/mol).
