Enthalpies of formation of CCl _ 4 ( g ) , H _ 2 O ( g ) , CO _ 2 (...

Using Hess's law of constant heat summation:
Enthalpy of reaction = Enthalpy of formation of products - Enthalpy of formation of reactants
$∴ {\Delta H}_{f} ° = {\Delta H}_{f} ° \left({CO}_{2}, g\right) + 4{\Delta H}_{f} ° ( HCl ) - {\Delta H}_{f} ° \left({CCl}_{4}\right) - 2{\Delta H}_{f} ° \left({H}_{2}O, r\right)$

$= [ - 394 ] + 4 [ - 92 ] + 105 - 2 \times [ 242 ]$

$= - 394 - 368 + 105 + 484$

$= - 173kJ / mole$